Chapter 9: Differential Equations

The given differential equation is $\frac{d^4 y}{d x^4}+\sin y^{\prime \prime \prime}=0$

The highest-order derivative appearing in this equation is $\frac{d^4 y}{d x^4}$ and its order is 4 .

This D.E. is not a polynomial equation in its derivatives ( $\because$ The term $\sin y^{\prime \prime \prime}$ is a T-function of derivative $y^{\prime \prime \prime}$ ). Therefore, the degree of this D.E. is not defined.

Answer. Order 4 and degree not defined.

The given differential equation is $y^{\prime}+5 y=0$.

The highest-order derivative in this D.E. is $y^{\prime}\left(=\frac{d y}{d x}\right)$ and so its order is one. The given differential equation is a polynomial equation in derivatives ( $y^{\prime}$ here) and the highest power of the highest-order derivative $y^{\prime}$ is one, so its degree is one.

Answer. Order 1 and degree 1.

The given differential equation is $\left(\frac{d s}{d t}\right)^4+3 s \frac{d^2 s}{d t^2}=0$.

The highest-order derivative in this D.E. is $\frac{d^2 s}{d t^2}$ and its order is 2 . The given D.E is a polynomial equation in derivatives and the highest power of the highest-order derivative $\frac{d^2 s}{d t^2}$ is one. Therefore degree of D.E. is 1 .

Answer. Order 2 and degree 1.

The given differential equation is $\left(\frac{d^2 y}{d x^2}\right)^2+\cos \left(\frac{d y}{d x}\right)=0$.

The highest-order derivative appearing in this equation is $\frac{d^2 y}{d x^2}$ and its order is 2 .

This D.E. is not a polynomial equation in its derivatives

( ∵ The term $\cos \frac{d y}{d x}$ is a T-function of derivative $\frac{d y}{d x}$ ).

Therefore, the degree of this D.E. is not defined.

Answer. Order 2 and degree not defined.

The given differential equation is $\frac{d^2 y}{d x^2}=\cos 3 x+\sin 3 x$.

The highest-order derivative in this D.E. is $\frac{d^2 y}{d x^2}$ and its order is 2 .

This is a polynomial equation in its derivatives, and the highest power raised to highest order $\frac{d^2 y}{d x^2}=\left(\frac{d^2 y}{d x^2}\right)^1$ is one, so its degree is 1 .

Answer. Order 2 and degree 1.

Note: It may be remarked that the terms $\cos 3 x$ and $\sin 3 x$ present in the given D.E. are trigonometrical functions (but not T-functions of derivatives).

It may be noted that $\left(\cos 3 \frac{d y}{d x}\right)$ is not a polynomial function of derivatives.

The given differential equation is $\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right)^3+\left(y^{\prime}\right)^4+y^5=0$.

The highest-order derivative in this D.E. is $y^{\prime \prime \prime}$ and its order is 3.

The given differential equation is a polynomial equation in derivatives $y^{\prime \prime \prime}, y^{\prime \prime}$ and $y^{\prime}$ and the highest power of the highest-order derivative $y^{\prime \prime \prime}$ is two, so its degree is 2 .

Answer. Order 3 and degree 2.

The given differential equation is $y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0$.

The highest-order derivative in this D.E. is $y^{\prime \prime \prime}$ and its order is 3 .

The given differential equation is a polynomial equation in derivatives $y^{\prime \prime \prime}, y^{\prime \prime}$ and $y^{\prime}$ and the highest power of the highest-order derivative $y^{\prime \prime \prime}$ is one, so its degree is 1 .

Answer. Order 3 and degree 1.

The given differential equation is $y^{\prime}+y=e^x$.

The highest-order derivative in this D.E. is $y^{\prime}$ and its order is 1 .

The given differential equation is a polynomial equation in derivative $y^{\prime}$. (It may be noted that $e^x$ is an exponential function and not a polynomial function but is not an exponential function of derivatives) and the highest power of the highest-order derivative $y^{\prime}$ is one, so its degree is 1 .

Answer. Order 1 and degree 1.

The given differential equation is $y^{\prime \prime}+\left(y^{\prime}\right)^2+2 y=0$.

The highest-order derivative in this D.E. is $y^{\prime \prime}$ and its order is 2.

The given differential equation is a polynomial equation in derivatives $y^{\prime \prime}$ and $y^{\prime}$ and the highest power of the highest-order derivative $y^{\prime \prime}$ is one, so its degree is 1 .

Answer. Order 2 and degree 1.

The given differential equation is $y^{\prime \prime}+2 y^{\prime}+\sin y=0$.

The highest-order derivative in this D.E. is $y^{\prime \prime}$ and its order is 2 .

The given differential equation is a polynomial equation in derivatives $y^{\prime \prime}$ and $y^{\prime}$. (It may be noted that $\sin y$ is not a polynomial function of $y$, it is a T-function of $y$ but is not a T-function of derivatives) and the highest power of the highest-order derivative $y^{\prime \prime}$ is one, so its degree is one.

Answer. Order 2 and degree 1.

The given differential equation is

Notice that D.E. (i) is not a polynomial equation in its derivatives.

∴ Degree of D.E. (i) is not defined.

Answer. Option (D) is the correct answer.

The given differential equation is $2 x^2 \frac{d^2 y}{d x^2}-3 \frac{d y}{d x}+y=0$

The highest-order derivative appearing in this equation is $\frac{d^2 y}{d x^2}$ and its order is 2 .

Answer. Order of the given D.E. is 2.

Given: $y=e^x+1$

To prove: $y$ given, by (i) is a solution of the D.E. $y^{\prime \prime}-y^{\prime}=0$

From equation (i), $y^{\prime}=e^x+0=e^x$ and $y^{\prime \prime}=e^x$

∴ L.H.S. of D.E. (ii) $=y^{\prime \prime}-y^{\prime}=e^x-e^x=0=$ R.H.S. of D.E.

Therefore, $y$ given by (i) is a solution of D.E. (ii).

Given: $y=x^2+2 x+\mathrm{C}$

To prove: $y$ given by (i) is a solution of the D.E.

From equation (i), $y^{\prime}=2 x+2$

$\therefore \quad$ L.H.S. of D.E. $(i i)=y^{\prime}-2 x-2$

Therefore, $y$ given by (i) is a solution of D.E. (ii).

Given: $y=\cos x+\mathrm{C}$

To prove: $y$ given by (i) is a solution of D.E. $y^{\prime}+\sin x=0$

From equation (i), $y^{\prime}=-\sin x$

$\therefore \quad y$ given by ( $i$ ) is a solution of D.E. ( $i i$ ).

Given: $y=\sqrt{1+x^2}$

To prove: $y$ given by ( $i$ ) is a solution of D.E. $y^{\prime}=\frac{x y}{1+x^2}$

From $(i), y^{\prime}=\frac{d}{d x} \sqrt{1+x^2}=\frac{d}{d x}\left(1+x^2\right)^{1 / 2}$

$=\frac{1}{2}\left(1+x^2\right)^{-1 / 2} \frac{d}{d x}\left(1+x^2\right)=\frac{1}{2}\left(1+x^2\right)^{-1 / 2} \cdot 2 x=\frac{x}{\sqrt{1+x^2}}$

R.H.S. of D.E. (ii) $=\frac{x y}{1+x^2}=\frac{x}{1+x^2} \sqrt{1+x^2}$

(By (i))

Therefore, $y$ given by (i) is a solution of D.E. (ii).

Given: $y=\mathrm{A} x$

To prove: $y$ given by (i) is a solution of the D.E. $x y^{\prime}=y(x \neq 0)$

From $(i), y^{\prime}=\mathrm{A}(1)=\mathrm{A}$

L.H.S. of D.E. $(i i)=x y^{\prime}=x \mathrm{~A}$

Therefore, $y$ given by (i) is a solution of D.E. (ii).

Given: $y=x \sin x$

To prove: $y$ given by (i) is a solution of D.E.

From equation (i), $\frac{d y}{d x}\left(=y^{\prime}\right)=x \frac{d}{d x}(\sin x)+\sin x \frac{d}{d x} x=x \cos x+\sin x$

L.H.S. of D.E. $($ ii $)=x y^{\prime}=x(x \cos x+\sin x)$

R.H.S. of D.E. $(i i)=y+x \sqrt{x^2-y^2}$

Putting $y=x \sin x$ from (i),

From (iii) and (iv), L.H.S. of D.E. (ii) = R.H.S. of D.E. (ii)

$\therefore \quad y$ given by ( $i$ ) is a solution of D.E. ( $i i$ ).

In each of the Exercises 7 to 10, verify that the given functions (Explicit or Implicit) is a solution of the corresponding differential equation:

Given: $x y=\log y+\mathrm{C}$

We verify that the implicit function given by (i) satisfies the D.E. $y^{\prime}=\frac{y^2}{1-x y}$

Differentiating both sides of (i) w.r.t. $x$, we have

which matches the given differential equation (ii), i.e., Eqn. (ii) is proved.

Hence, the implicit function from (i) is a solution of D.E. (ii).

Given: $y-\cos y=x$

We check that the function given by (i) is a solution of the D.E.

Differentiating both sides of (i) w.r.t. $x$, we have

Putting the value of $x$ from (i) and value of $y^{\prime}$ from (iii) in L.H.S. of (ii), we have

L.H.S. $=(y \sin y+\cos y+x) y^{\prime}$

Thus, the function in (i) satisfies D.E. (ii).

Given: $x+y=\tan ^{-1} y$

We check that the function given by (i) is a solution of the D.E.

Differentiating both sides of (i), w.r.t. $x, 1+y^{\prime}=\frac{1}{1+y^2} y^{\prime}$

Cross-multiplying

$\Rightarrow y^2 y^{\prime}+y^2+1=0$ which is identical to D.E. (ii).

Hence, the function given by (i) is indeed a solution of D.E. (ii).

Given: $y=\sqrt{a^2-x^2}, x \in(-a, a)$

We check that the function given by (i) is a solution of the D.E.

From $(i), \quad \frac{d y}{d x}=\frac{1}{2}\left(a^2-x^2\right)^{-1 / 2} \frac{d}{d x}\left(a^2-x^2\right)$

Putting these values of $y$ and $\frac{d y}{d x}$ from (i) and (iii) in L.H.S. of (ii),

Hence, the function given by (i) is indeed a solution of D.E. (ii).

Option (D) 4 is the correct answer.

Result: The number of arbitrary constants ( $c_1, c_2, c_3$ etc.) in the general solution of a differential equation of $n$th order is $n$.

The number of arbitrary constants in a particular solution of a differential equation of any order is zero ( 0 ).

(By definition, a particular solution contains no arbitrary constant.)

Hence, Option (D) is the correct answer.

The given differential equation is

Integrating both sides, $\int d y=\int \frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x$

or $y=2 \tan \frac{x}{2}-x+c$

This is the required general solution.

The given differential equation is $\frac{d y}{d x}=\sqrt{4-y^2} \quad \Rightarrow d y=\sqrt{4-y^2} d x$

Separating the variables, $\frac{d y}{\sqrt{4-y^2}}=d x$

Integrating both sides, $\int \frac{d y}{\sqrt{2^2-y^2}} d y=\int 1 d x$

$\Rightarrow \quad y=2 \sin (x+c)$ This is the required general solution.

The given differential equation is $\frac{d y}{d x}+y=1$

$\Rightarrow \frac{d y}{d x}=1-y \quad \Rightarrow \quad d y=(1-y) d x \quad \Rightarrow d y=-(y-1) d x$

Separating the variables, $\frac{d y}{y-1}=-d x$

Integrating both sides, $\int \frac{d y}{y-1}=-\int 1 d x$

This is the required general solution.

The given differential equation is

Dividing by $\tan x \tan y$, we have

Integrating both sides, $\int \frac{\sec ^2 x}{\tan x} d x+\int \frac{\sec ^2 y}{\tan y} d y=\log c$

This is the required general solution.

For each of the differential equations in Exercises 5 to 7, find the general solution:

The given differential equation is $\left(e^x+e^{-x}\right) d y=\left(e^x-e^{-x}\right) d x$

Integrating both sides, $\int d y=\int\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) d x$

This is the required general solution.

The given differential equation is $\frac{d y}{d x}=\left(1+x^2\right)\left(1+y^2\right)$

Separating the variables,

Integrating both sides,

This is the required general solution.

The given differential equation is $y \log y d x-x d y=0$

Separating the variables, $\quad \frac{d y}{y \log y}=\frac{d x}{x}$

Integrating both sides $\quad \int \frac{d y}{y \log y}=\int \frac{d y}{x}$

To evaluate the left-hand integral, substitute $\log y=t$.

$\therefore y=e^{a x}$ This is the required general solution.

For each of the differential equations in Exercises 8 to 10, find the general solution:

The given differential equation is $x^5 \frac{d y}{d x}=-y^5$

Separating the variables, $\frac{d y}{\left(y^5\right)}=-\frac{d x}{\left(x^5\right)} \quad \Rightarrow y^{-5} d y=-x^{-5} d x$

Integrating both sides, $\int y^{-5} d y=-\int x^{-5} d x$

This is the required general solution.

The given differential equation is $\frac{d y}{d x}=\sin ^{-1} x$ or $\quad d y=\sin ^{-1} x d x$

Integrating both sides, $\quad \int 1 d y=\int \sin ^{-1} x d x$

or $\quad y=\int \sin ^{-1} x \cdot 1 d x$

Applying the product rule,

To evaluate $\int \frac{x}{\sqrt{1-x^2}} d x=-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} d x$

Substitute $1-x^2=t$. Differentiating, $-2 x d x=d t$

Remark: To explain * in eqn. (ii)

If all the terms in the solution of a D.E. involve logs, it is better to use $\log c$ or $\log |c|$ instead of $c$ in the solution.

Putting this value of $\int \frac{x}{\sqrt{1-x^2}} d x$ in ( $i$ ), the required general solution is

The given equation is $e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0$

Dividing each term by $\left(1-e^x\right) \tan y$, we have

Integrating both sides, $\int \frac{e^x}{1-e^x} d x+\int \frac{\sec ^2 y}{\tan y} d y=c$

or $\quad-\int \frac{-e^x}{1-e^x} d x+\log |\tan y|=c$

or $\quad-\log \left|1-e^x\right|+\log |\tan y|=c\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|\right]$

or $\log \frac{|\tan y|}{\left|1-e^x\right|}=\log c^{\prime} \quad$ (See Remark at the end of page 612)

or $\quad \frac{|\tan y|}{\left|1-e^x\right|}=c^{\prime}$

or $\quad \tan y=\mathrm{C}\left(1-e^x\right) . \quad\left[\because \quad|t|=c^{\prime} \quad \Rightarrow \quad t= \pm c^{\prime}=\mathrm{C}\right.$ (say)]

For each of the differential equations in Exercises 11 to 12, find a particular solution satisfying the given condition:

The given differential equation is $\left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x$

Separating the variables $d y=\frac{\left(2 x^2+x\right)}{x^3+x^2+x+1} d x$

or

Integrating both sides, we have

Let $\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}$ (via partial fraction decomposition)

Multiplying both sides by L.C.M. $=(x+1)\left(x^2+1\right)$, we have

Comparing coefficients of $x^2$ on both sides,

Comparing coefficients of $x$ on both sides,

Equating the constant terms $\mathrm{A}+\mathrm{C}=0$

Solving equations (iii), (iv), and (v) for A, B, C — subtracting (iv) from (iii) to eliminate B:

Adding (v) and (vi), $2 \mathrm{~A}=1$ or $\mathrm{A}=\frac{1}{2}$

From (v),

Putting $\mathrm{C}=-\frac{1}{2}$ in (iv), $\mathrm{B}-\frac{1}{2}=1$ or $\mathrm{B}=1+\frac{1}{2}=\frac{3}{2}$

Putting these values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ in (ii), we have

Substituting this result back into (i)

To find $\boldsymbol{c}$

When $x=0, y=1$ (given)

Putting $x=0$ and $y=1$ in (vii),

or $1=c \quad\left[\because \log 1=0\right.$ and $\left.\tan ^{-1} 0=0\right]$

Putting $c=1$ in eqn. (vii), the required solution is

This is the required particular solution.

The given differential equation is $x\left(x^2-1\right) \frac{d y}{d x}=1$

Integrating both sides, $\quad \int 1 d y=\int \frac{1}{x\left(x^2-1\right)} d x$

Let the integrand $\frac{1}{x(x+1)(x-1)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C}}{x-1}$

(using partial fractions)

Multiplying by L.C.M. $=x(x+1)(x-1)$,

or $1=\mathrm{A}\left(x^2-1\right)+\mathrm{B}\left(x^2-x\right)+\mathrm{C}\left(x^2+x\right)$

or $1=\mathrm{A} x^2-\mathrm{A}+\mathrm{B} x^2-\mathrm{B} x+\mathrm{C} x^2+\mathrm{C} x$

Equating coefficients of $x^2$, $x$ and constant terms on both sides,

Constants – $\mathrm{A}=1$ or $\mathrm{A}=-1$

Putting $\mathrm{A}=-1$ and $\mathrm{C}=\mathrm{B}$ from (iv) in (iii),

∴ From (iv), $\mathrm{C}=\mathrm{B}=\frac{1}{2}$

Substituting these values of A, B, C back into (ii),

Substituting this result back into (i),

To find $\boldsymbol{c}$ for the particular solution

Putting $y=0$, when $x=2$ (given) in ( $v$ ),

Putting this value of $c$ in ( $v$ ), the required particular solution is

OR

To evaluate $\int \frac{1}{x\left(x^2-1\right)} d x=\int \frac{x}{x^2\left(x^2-1\right)} d x=\frac{1}{2} \int \frac{2 x}{x^2\left(x^2-1\right)} d x$

Put $\quad x^2=t$.

For each of the differential equations in Exercises 13 to 14, find a particular solution satisfying the given condition:

The given differential equation is

Integrating both sides

To find $\boldsymbol{c}$ for particular solution

Putting $c=1$ in ( $i$ ), $y=x \cos ^{-1} a+1$

$\Rightarrow \quad \cos \left(\frac{y-1}{x}\right)=a$ This is the required particular solution.

The given differential equation is $\frac{d y}{d x}=y \tan x$

Separating the variables, $\frac{d y}{y}=\tan x d x$

Integrating both sides $\int \frac{1}{y} d y=\int \tan x d x$

where $\mathrm{C}= \pm c$

To find $\mathbf{C}$ for particular solution

Putting $y=1$ and $x=0$ in (i), $1=\mathrm{C} \sec 0=\mathrm{C}$

Putting $\mathbf{C}=1$ in ( $i$ ), the required particular solution is $y=\sec x$.

The given differential equation is $y^{\prime}=e^x \sin x$

Integrating both sides, $\int 1 d y=\int e^x \sin x d x$

or $\quad y=\mathrm{I}+\mathrm{C}$

where I $=\int_{\text {I }} e^x \sin x d x$

Again applying product rule,

[By (ii)]

Rearranging terms: $2 \mathrm{I}=e^x(\sin x-\cos x)$

Putting this value of I in ( $i$ ), the required solution is

To find $\boldsymbol{c}$. Given that required curve ( $i$ ) passes through the point $(0,0)$.

Putting $x=0$ and $y=0$ in (iii),

Putting $c=\frac{1}{2}$ in (iii), the required equation of the curve is

L.C.M. $=2 \therefore 2 y=e^x(\sin x-\cos x)+1$ or $2 y-1=e^x(\sin x-\cos x)$ which is the required equation of the curve.

The given differential equation is $x y \frac{d y}{d x}=(x+2)(y+2)$

To find $c$. Curve ( $i$ ) passes through the point ( $1,-1$ ).

Putting $x=1$ and $y=-1$ in $(i),-1-1=\log (1)+c$ or $-2=c$

$(\because \quad \log 1=0)$

Putting $c=-2$ in ( $i$ ), the particular solution curve is

or $y-x+2=\log \left((y+2)^2 x^2\right)$.

Let $\mathrm{P}(x, y)$ be any point on the required curve.

According to the question,

(Slope of the tangent to the curve at $\mathrm{P}(x, y)) \times y=x$

Now variables are separated.

Integrating both sides $\int y d y=\int x d x \quad \therefore \frac{y^2}{2}=\frac{x^2}{2}+c$

Multiplying by L.C.M. $=2, y^2=x^2+2 c$

or $y^2=x^2+\mathrm{A}$

where $\mathrm{A}=2 c$.

Given: Curve $(i)$ passes through the point $(0,-2)$.

Putting $x=0$ and $y=-2$ in (i), $4=\mathrm{A}$.

Putting $\mathrm{A}=4$ in (i), equation of required curve is

According to question, slope of the tangent at any point $\mathrm{P}(x, y)$ of the required curve.

$=2$. (Slope of the line joining the point of contact $\mathrm{P}(x, y)$ to the given point $\mathrm{A}(-4,-3)$ ).

Cross-multiplying, both sides, $(x+4) d y=2(y+3) d x$

Separating the variables, $\frac{d y}{y+3}=\frac{2}{x+4} d x$

Integrating both sides, $\int \frac{1}{y+3} d y=2 \int \frac{1}{x+4} d x$

(For $\log |c|$, see Foot Note page 612)

To find $\mathbf{C}$. Given that curve $(i)$ passes through the point $(-2,1)$.

Putting $x=-2$ and $y=1$ in ( $i$ ),

Putting $\mathrm{C}=1$ in (i), equation of required curve is

Let $x$ be the radius of the spherical balloon at time $t$.

Since the rate of change of volume of the spherical balloon is constant $=k$ (say)

Separating the variables, $\quad 4 \pi x^2 d x=k d t$

Integrating both sides, $4 \pi \int x^2 d x=k \int 1 d t$

To find $\boldsymbol{c}$ : Given: Initially radius is 3 units.

⇒ When $t=0, x=3$

Putting $t=0$ and $x=3$ in (i), we have

To find $\boldsymbol{k}$ : Given: When $t=3 \mathrm{sec}, x=6$ units

Putting $t=3$ and $x=6$ in $(i), \frac{4 \pi}{3}(6)^3=3 k+c$.

Putting $c=36 \pi$ from (ii), $\frac{4 \pi}{3}(216)=3 k+36 \pi$

or $4 \pi(72)-36 \pi=3 k \quad \Rightarrow \quad 288 \pi-36 \pi=3 k$

or $\quad 3 k=252 \pi \quad \Rightarrow k=84 \pi$

Putting values of $c$ and $k$ from (ii) and (iii) in (i), we have

Dividing both sides by $4 \pi, \frac{x^3}{3}=21 t+9$

$\Rightarrow \quad x^3=63 t+27 \quad \Rightarrow \quad x=(63 t+27)^{1 / 3}$.

Let P be the principal (amount) at the end of $t$ years. The given rate of increase of the principal per year $=r \%$ (of the principal)

Separating the variables, $\quad \frac{d \mathrm{P}}{\mathrm{P}}=\frac{r}{100} d t$

Integrating both sides, $\log \mathrm{P}=\frac{r}{100} t+c$

(Clearly P being principal is $>0$, and hence $\log |\mathrm{P}|=\log \mathrm{P}$ )

Finding c. Initial principal $=₹ 100$ (given)

i.e., When $t=0, \mathrm{P}=100$

Putting $t=0$ and $\mathrm{P}=100$ in (i), log $100=c$.

Putting $c=\log 100$ in $(i), \log \mathrm{P}=\frac{r}{100} t+\log 100$

Putting $\mathrm{P}=$ double of itself $=2 \times 100=₹ 200$

When $t=10$ years (given) in (ii),

$\Rightarrow r=10 \log 2=10(0.6931)=6.931 \%$ (given).

Let P be the principal (amount) at the end of $t$ years.

The rate of increase of the principal per year

$\Rightarrow 20 d \mathrm{P}=\mathrm{P} d t$

Separating the variables, $\frac{d \mathrm{P}}{\mathrm{P}}=\frac{d t}{20}$

Integrating both sides, we have

Finding c. Given: Initial principal deposited with the bank is $₹ 1000$.

⇒ When $t=0, \mathrm{P}=1000$

Putting $t=0$ and $\mathrm{P}=1000$ in (i), we have $\log 1000=c$

Putting $c=\log 1000$ in (i), $\log \mathrm{P}=\frac{t}{20}+\log 1000$

$\Rightarrow \log \mathrm{P}-\log 1000=\frac{t}{20} \quad \Rightarrow \log \frac{\mathrm{P}}{1000}=\frac{t}{20}$

Putting $t=10$ years (given), we have

Let $x$ be the bacteria present in the culture at time $t$ hours.

According to the given condition, the rate of bacterial growth is proportional to the current population.

i.e., $\frac{d x}{d t}$ is proportional to $x$.

$\therefore \quad \frac{d x}{d t}=k x$ where $k$ is the constant of proportionality $(k>0$ because rate of growth (i.e., increase) of bacteria is given.)

$\Rightarrow d x=k x d t \quad \Rightarrow \frac{d x}{x}=k d t$

Integrating both sides, $\int \frac{1}{x} d x=k \int 1 d t$

$\Rightarrow \quad \log x=k t+c$

To find $\boldsymbol{c}$. Given: Initially the bacteria count is $x_0($ say $)= 1,00,000$.

⇒ When $t=0, x=x_0$.

Putting these value in (i), $\log x_0=c$.

Putting $c=\log x_0$ in $(i), \log x=k t+\log x_0$

To find $k$ : According to given, the number of bacteria is increased by $10 \%$ in 2 hours.

∴ $\quad$ Increase in bacteria in 2 hours $=\frac{10}{100} \times 1,00,000=10,000$

$\therefore \quad x$, the amount of bacteria at $t=2$

Putting $x=x_1$ and $t=2$ in (ii),

Putting this value of $k$ in (ii), we have $\log \frac{x}{x_0}=\frac{1}{2}\left(\log \frac{11}{10}\right) t$

When $x=2,00,000$ (given);

then $\log \frac{2,00,000}{1,00,000}=\left(\frac{1}{2} \log \frac{11}{10}\right) t \quad \Rightarrow \log 2=\frac{1}{2} \log \left(\frac{11}{10}\right) t$

Cross-multiplying, $2 \log 2=\left(\log \frac{11}{10}\right) t \Rightarrow t=\frac{2 \log 2}{\left(\log \frac{11}{10}\right)}$ hours.

The given differential equation is $\frac{d y}{d x}=e^{x+y}$

Separating the variables, $\frac{d y}{\left(e^y\right)}=e^x d x \quad$ or $\quad e^{-y} d y=e^x d x$

Integrating both sides $\int e^{-y} d y=\int e^x d x$

Dividing by $-1, e^{-y}+e^x=-c$ or $e^x+e^{-y}=\mathrm{C}$ where $\mathrm{C}=-c$ This is the required solution.

Hence, Option (A) is the correct answer.

The given differential equation is

This D.E. looks to be homogeneous as degree of each coefficient of $d x$ and $d y$ is same throughout (here 2).

∴ The given differential equation is homogeneous.

Put $\frac{\boldsymbol{y}}{\boldsymbol{x}}=\boldsymbol{v}$. Therefore $y=v x$.

Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in (ii), we have

Rearranging terms: $v$ to R.H.S., $x \frac{d v}{d x}=\frac{1+v^2}{1+v}-v$

Cross-multiplying, $x(1+v) d v=(1-v) d x$

Separating the variables $\frac{1+v}{1-v} d v=\frac{d x}{x}$

Integrating both sides $\int \frac{1+v}{1-v} d v=\int \frac{1}{x} d x$

This is the required solution.

The given differential equation is $y^{\prime}=\frac{x+y}{x}$

∴ Differential equation (i) is homogeneous.

Put $\frac{y}{x}=v \quad \therefore \quad y=v x$

Putting these values of $\frac{d y}{d x}$ and $y$ in (i),

Separating the variables, $d v=\frac{d x}{x}$

Integrating both sides, $\int 1 d v=\int \frac{d x}{x} \quad v=\log |x|+c$

Putting $v=\frac{y}{x}, \frac{y}{x}=\log |x|+c \quad \therefore \quad y=x \log |x|+c x$

This is the required solution.

The given differential equation is

Differential equation (i) looks to be homogeneous because each coefficient of $d x$ and $d y$ is of degree 1 .

From $(i), \quad(x-y) d y=(x+y) d x$

∴ Differential equation (i) is homogeneous.

Put $\frac{y}{x}=v \quad \therefore y=v x$

Putting these values in (ii), $v+x \frac{d v}{d x}=\frac{1+v}{1-v}$

Shifting $v$ to R.H.S., $\quad x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v+v^2}{1-v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{1-v}$

Cross-multiplying, both sides, $\quad x(1-v) d v=\left(1+v^2\right) d x$

Separating the variables, $\quad \frac{(1-v)}{1+v^2} d v=\frac{d x}{x}$

Integrating both sides, $\int \frac{1-v}{1+v^2} d v=\int \frac{1}{x} d x+c$

The given differential equation is

This differential equation looks to be homogeneous because degree of each coefficient of $d x$ and $d y$ is same (here 2).

From equation (i), $2 x y d y=-\left(x^2-y^2\right) d x$

Dividing every term in the numerator and denominator of R.H.S. by $x^2$,

∴ The given differential equation is homogeneous.

Put $\frac{y}{x}=v$. Therefore $y=v x \therefore \quad \frac{d y}{d x}=v .1+x \frac{d v}{d x}=v+x \frac{d v}{d x}$ Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in differential equation (ii), we have

Integrating both sides, $\int \frac{2 v}{v^2+1} d v=-\int \frac{1}{x} d x$

Put $v=\frac{y}{x},\left(\frac{y^2}{x^2}+1\right) x=c \quad$ or $\quad\left(\frac{y^2+x^2}{x^2}\right) x=c$

or

This is the required solution.

The given differential equation is $x^2 \frac{d y}{d x}=x^2-2 y^2+x y$

The given differential equation looks to be Homogeneous as all terms in $x$ and $y$ are of same degree (here 2).

Dividing by $x^2, \frac{d y}{d x}=\frac{x^2}{x^2}-\frac{2 y^2}{x^2}+\frac{x y}{x^2}$

or

∴ Differential equation (i) is homogeneous.

So put $\frac{y}{x}=v \quad \therefore y=v x$

$\therefore \quad \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}=v+x \frac{d v}{d x}$

Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in (i),

$v+x \frac{d u}{d x}=1-2 v^2+v$ or $x \frac{d u}{d x}=1-2 v^2 \Rightarrow x d v=\left(1-2 v^2\right) d x$

Separating the variables, $\frac{d v}{1-2 v^2}=\frac{d x}{x}$

Integrating both sides, $\int \frac{1}{1^2-(\sqrt{2} v)^2} d v=\int \frac{1}{x} d x$

Putting $v=\frac{y}{x}, \quad \frac{1}{2 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \frac{y}{x}}{1-\sqrt{2} \frac{y}{x}}\right|=\log |x|+c$

Multiplying within logs by $x$ in L.H.S.,

In each of the Exercises 6 to 10, show that the given D.E. is homogeneous and solve each of them:

The given differential equation is

Dividing by $d x$

Dividing by $x, \frac{d y}{d x}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}=\mathrm{F}\left(\frac{y}{x}\right)$

∴ Given differential equation is homogeneous.

Put $\frac{y}{x}=v$ i.e., $y=v x$.

Differentiating w.r.t. $x, \frac{d y}{d x}=v+x \frac{d v}{d x}$

Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in (i), it becomes

Integrating both sides, $\int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x}$

Replacing $v$ by $\frac{y}{x}$, we have

This is the required solution.

The given differential equation is

Dividing every term in R.H.S. by $x^2$,

∴ The given differential equation is homogeneous.

So let us put $\frac{\boldsymbol{y}}{\boldsymbol{x}}=\boldsymbol{v}$. Therefore $y=v x$.

Putting these values in differential equation (i), we have

Cross-multiplying, both sides, $x(v \sin v-\cos v) d v=2 v \cos v d x$

Separating the variables, $\quad \frac{v \sin v-\cos v}{v \cos v} d v=2 \frac{d x}{x}$

Integrating both sides, $\quad \int \frac{v \sin v-\cos v}{v \cos v} d v=2 \int \frac{1}{x} d x$

Using $\frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}, \Rightarrow \int\left(\frac{v \sin v}{v \cos v}-\frac{\cos v}{v \cos v}\right) d v=2 \int \frac{1}{x} d x$

This is the required solution.

The given differential equation is $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$ or $\quad x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right)$

Dividing each term by $x, \frac{d y}{d x}=\frac{y}{x}-\sin \left(\frac{y}{x}\right)=\mathrm{F}\left(\frac{y}{x}\right)$

Since $\frac{d y}{d x}=\mathrm{F}\left(\frac{y}{x}\right)$, the given differential equation is homogeneous.

Putting $\frac{y}{x}=v$ i.e., $y=v x$ so that $\frac{d y}{d x}=v+x \frac{d v}{d x}$

Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in (i), we have

or $\quad x \frac{d v}{d x}=-\sin v \quad \therefore \quad x d v=-\sin v d x$

or $\quad \frac{d v}{\sin v}=\frac{-d x}{x} \quad$ or $\operatorname{cosec} v \mathrm{~d} v=-\frac{d x}{x}$

Integrating, $\quad \log |\operatorname{cosec} v-\cot v|=-\log |x|+\log |c|$

or $\log |\operatorname{cosec} v-\cot v|=\log \left|\frac{c}{x}\right|$

or $\quad \operatorname{cosec} v-\cot v= \pm \frac{c}{x}$

Replacing $v$ by $\frac{y}{x}, \operatorname{cosec} \frac{y}{x}-\cot \frac{y}{x}=\frac{\mathrm{C}}{x}$ where $\mathrm{C}= \pm c$

Cross-multiplying, both sides, $x\left(1-\cos \frac{y}{x}\right)=\mathrm{C} \sin \frac{y}{x}$ This is the required solution.

The given differential equation is $y d x+x\left(\log \frac{y}{x}\right) d y$

Since $\frac{d y}{d x}=\mathrm{F}\left(\frac{y}{x}\right)$, the given differential equation is homogeneous.

Putting $\frac{y}{x}=v$ i.e., $y=v x$ so that $\frac{d y}{d x}=v+x \frac{d v}{d x}$

Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in (i), we have

or $x \frac{d v}{d x}=\frac{v}{2-\log v}-v=\frac{v-2 v+v \log v}{2-\log v}=\frac{-v+v \log v}{2-\log v}$

or

$\therefore \quad x(2-\log v) d v=v(\log v-1) d x$

or $\quad \frac{2-\log v}{v(\log v-1)} d v=\frac{d x}{x} \quad$ or $\quad \frac{1-(\log v-1)}{v(\log v-1)} d v=\frac{d x}{x}$

or $\left[\frac{1}{v(\log v-1)}-\frac{1}{v}\right] d v=\frac{d x}{x}$

Second solution

The given differential equation is $y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0$

Dividing each term by $d y$,

Dividing each term by $y$,

∴ The given differential is homogeneous.

Put $\frac{x}{y}=v$ i.e. $x=v y$

so that $\frac{d x}{d y}=v+y \frac{d v}{d y}$

Putting these values in D. E. (i), we have

Cross-multiplying, $y d v=v(\log v+1) d y$

Separating the variables $\frac{d v}{v(\log v+1)}=\frac{d y}{y}$

Integrating both sides $\int \frac{\frac{1}{v}}{\log v+1} d v=\int \frac{1}{y} d y$

$\therefore \quad \log |\log v+1|=\log |y|+\log |c|=\log |c y|\left[\because \int \frac{f^{\prime}(v)}{f(v)} d v=\log |f(v)|\right]$

$\therefore \log v+1= \pm c y=\mathrm{C} y$ where $\mathrm{C}= \pm c$

Replacing $v$ by $\frac{x}{y}$, we have

$\log \frac{x}{y}+1=\mathrm{C} y$

or $-\log \frac{y}{x}+1=\mathrm{C} y \quad\left[\because \log \frac{x}{y}=-\log \frac{y}{x}\right.$ see page 632$]$

Dividing by $-1, \log \frac{y}{x}-1=-\mathrm{C} y$ or $=\mathrm{C}_1 y$ which is a primitive (solution) of the given D.E.

The given differential equation is $\left(1+e^{x / y}\right) d x+e^{x / y}\left(1-\frac{x}{y}\right) d y=0$

Dividing by $d y,\left(1+e^{x / y}\right) \frac{d x}{d y}+e^{x / y}\left(1-\frac{x}{y}\right)=0$

or $\left(1+e^{x / y}\right) \frac{d x}{d y}=-e^{x / y}\left(1-\frac{x}{y}\right)$ or $\frac{d x}{d y}=\frac{e^{x / y}\left(\frac{x}{y}-1\right)}{1+e^{x / y}}$

which is a differential equation of the form $\frac{d x}{d y}=f\left(\frac{x}{y}\right)$.

∴ The given differential equation is homogeneous.

Hence put $\frac{\boldsymbol{x}}{\boldsymbol{y}}=\boldsymbol{v}$ i.e., $x=v y$

Differentiating w.r.t. $y, \quad \frac{d x}{d y}=v+y \frac{d v}{d y}$

Putting these values of $\frac{x}{y}$ and $\frac{d x}{d y}$ in (i), we have

Now transposing $v$ to R.H.S.

Integrating, $\log \left|\left(v+e^v\right)\right|=-\log |y|+\log |c|$

Replacing $v$ by $\frac{x}{y}$, we have

Multiplying every term by $y$,

This is the required general solution.

For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition:

The given differential equation is

It looks to be a homogeneous differential equation because each coefficient of $d x$ and $d y$ is of same degree (here 1).

From $(i),(x+y) d y=-(x-y) d x$

∴ Given differential equation is homogeneous.

Put $\frac{y}{x}=v$. Therefore $y=v x$.

Putting these values in eqn. (ii), $\quad v+x \frac{d v}{d x}=\frac{v-1}{v+1}$

Separating the variables, $\frac{v+1}{v^2+1} d v=-\frac{d x}{x}$

To find $\boldsymbol{c}$ : Given: $y=1$ when $x=1$.

Putting $x=1$ and $y=1$ in (iii), $\quad \frac{1}{2} \log 2+\tan ^{-1} 1=c$ or $\quad c=\frac{1}{2} \log 2+\frac{\pi}{4} \quad\left(\because \tan \frac{\pi}{4}=1 \Rightarrow \tan ^{-1} 1=\frac{\pi}{4}\right)$ Putting this value of $c$ in (iii),

Multiplying by 2,

This is the required particular solution.

The given differential equation is

∴ The given differential equation is homogeneous.

Put $\frac{y}{x}=v$, i.e., $\quad y=v x$

Differentiating w.r.t. $x, \quad \frac{d y}{d x}=v+x \frac{d v}{d x}$

Putting these values of $\frac{y}{x}$ and $\frac{d y}{d x}$ in differential equation (i),

we have $v+x \frac{d v}{d x}=-v(1+v)=-v-v^2$

Rearranging terms: $v$ to R.H.S., $\quad x \frac{d v}{d x}=-v^2-2 v$

or $\quad x \frac{d v}{d x}=-v(v+2) \quad x d v=-v(v+2) d x$

or $\frac{d v}{v(v+2)}=-\frac{d x}{x}$

Integrating both sides, $\int \frac{1}{v(v+2)} d v=-\int \frac{1}{x} d x$

or $\frac{1}{2} \int \frac{2}{v(v+2)} d v=-\log |x|$ or $\frac{1}{2} \int \frac{(v+2)-v}{v(v+2)} d v=-\log |x|$

Separating terms

or $\quad \int\left(\frac{1}{v}-\frac{1}{v+2}\right) d v=-2 \log |x|$

or $\log |v|-\log |v+2|=\log x^{-2}+\log |c|$

or $\quad \log \left|\frac{v}{v+2}\right|=\log \left|c x^{-2}\right|$

$\therefore\left|\frac{v}{v+2}\right|=\left|\frac{c}{x^2}\right| \quad \therefore \frac{v}{v+2}= \pm \frac{c}{x^2}$

Replacing $v$ to $\frac{y}{x}$, we have

or $\quad x^2 y=\mathrm{C}(y+2 x)$

where $\mathrm{C}= \pm c$

To find $C$

Put $x=1$ and $y=1$ (given) in eqn. (ii), $1=3 \mathrm{C} \quad \therefore \mathrm{C}=\frac{1}{3}$

Putting $\mathrm{C}=\frac{1}{3}$ in eqn. (ii), required particular solution is

The given differential equation is

Dividing by $d x, \quad x \frac{d y}{d x}=-x \sin ^2 \frac{y}{x}+y$

Dividing by $x, \quad \frac{d y}{d x}=-\sin ^2 \frac{y}{x}+\frac{y}{x}$

∴ The given differential equation is homogeneous.

Put $\frac{\boldsymbol{y}}{\boldsymbol{x}}=\boldsymbol{v} \quad \therefore \quad y=v x \quad \therefore \quad \frac{d y}{d x}=v .1+x \frac{d v}{d x}=v+x \frac{d v}{d x}$

Putting these values in differential equation (i), we have

Separating the variables, $\frac{d v}{\sin ^2 v}=-\frac{d x}{x}$

Integrating, $\quad \int \operatorname{cosec}^2 v d v=-\int \frac{1}{x} d x$

$\Rightarrow \quad-\cot v=-\log |x|+c$

Dividing by $-1, \quad \cot v=\log |x|-c$

Putting $v=\frac{y}{x}, \quad \cot \frac{y}{x}=\log |x|-c$

To find c : $y=\frac{\pi}{4}$ when $x=1$ (given)

Putting $x=1$ and $y=\frac{\pi}{4}$ in (ii), $\cot \frac{\pi}{4}=\log 1-c$

or $1=0-c$ or $c=-1$

Putting $c=-1$ in (ii), required particular solution is

The given differential equation is

∴ Given differential equation (i) is homogeneous.

Put $\frac{y}{x}=v \quad \therefore \quad y=v x$

Putting these values in differential equation (i),

Separating the variables,

Integrating both sides,

Dividing by -1 ,

Putting $v=\frac{y}{x}$,

To find $\boldsymbol{c}$ : Given: $y=0$ when $x=1$

$\therefore \quad c=-1$

Putting $\quad c=-1$ in (ii), $\cos \frac{y}{x}=\log |x|+1=\log |x|+\log e$

$\Rightarrow \cos \frac{y}{x}=\log |e x|$ This is the required particular solution.

The given differential equation is

The given differential equation looks to be homogeneous because each coefficient of $d x$ and $d y$ is of same degree (2 here).

∴ The given differential equation is homogeneous.

Put $\frac{y}{x}=v \quad \therefore \quad y=v x \quad \therefore \quad \frac{d y}{d x}=v .1+x \frac{d v}{d x}=v+x \frac{d v}{d x}$

Putting these values in differential equation (ii), we have

Separating the variables,

$2 \frac{d v}{v^2}=\frac{d x}{x}$

Integrating both sides, $2 \int v^{-2} d v=\int \frac{1}{x} d x$

$\Rightarrow \quad 2 \frac{v^{-1}}{-1}=\log |x|+c \Rightarrow \frac{-2}{v}=\log |x|+c$

Putting $v=\frac{y}{x}, \quad \frac{-2}{\left(\frac{y}{x}\right)}=\log |x|+c$

To find $\boldsymbol{c}$ : Given: $y=2$, when $x=1$.

∴ From (iii), $\frac{-2}{2}=\log 1+c$ or $-1=c$

Putting $c=-1$ in (iii), the required particular solution is

We know that a homogeneous differential equation of the form $\frac{d x}{d y}=h\left(\frac{x}{y}\right)$ can be solved by the substitution $\frac{\boldsymbol{x}}{\boldsymbol{y}}=\boldsymbol{v}$ i.e., $x=v y$.

Hence, Option (C) is the correct answer.

Out of the four given options; option (D) is the only option in which all coefficients of $d x$ and $d y$ are of same degree (here 2). It may be noted that $x y$ is a term of second degree.

Hence differential equation in option (D) is Homogeneous differential equation.

The given differential equation is $\frac{d y}{d x}+2 y=\sin x$

(Standard form of a linear differential equation)

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have $\mathrm{P}=2$ and $\mathrm{Q}=\sin x$

Solution is $\quad y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

or $\quad y e^{2 x}=\int e^{2 x} \sin x d x+c$

or $\quad y e^{2 x}=\mathrm{I}+c$

where

Using integration by parts

Applying integration by parts again,

Substituting this value of I into (i), the required solution is

Dividing each term by $e^{2 x}, y=\frac{1}{5}(2 \sin x-\cos x)+\frac{c}{\left(e^{2 x}\right)}$

or $\quad y=\frac{1}{5}(2 \sin x-\cos x)+c e^{-2 x}$

This is the required general solution.

The given differential equation is $\frac{d y}{d x}+3 y=e^{-2 x}$

(Standard form of a linear differential equation)

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have $\mathrm{P}=3$ and $\mathrm{Q}=e^{-2 x}$

$\int \mathrm{P} d x=\int 3 d x=3 \int 1 d x=3 x \quad$ I.F. $=e^{\int \mathrm{P} d x}=e^{3 x}$

Solution is $y($ I.F. $)=\int \mathrm{Q}($ I.F. $) d x+c$

or $y e^{3 x}=\int e^{-2 x} e^{3 x} d x+c$ or $=\int e^{-2 x+3 x} d x+c=\int e^x d x+c$ or $\quad y e^{3 x}=e^x+c$

Dividing each term by $e^{3 x}$,

$y=\frac{e^x}{e^{3 x}}+\frac{c}{e^{3 x}} \quad$ or $\quad y=e^{-2 x}+c e^{-3 x}$

This is the required general solution.

The given differential equation is $\frac{d y}{d x}+\frac{y}{x}=x^2$

It is of the form $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$ Comparing $\mathrm{P}=\frac{1}{x}, \mathrm{Q}=x^2$

The general solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$ or $\quad y x=\int x^2 \cdot x d x+c=\int x^3 d x+c \quad$ or $\quad x y=\frac{x^4}{4}+c$.

The given differential equation is $\frac{d y}{d x}+(\sec x) y=\tan x$

It is of the form $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$.

Comparing $\quad \mathrm{P}=\sec x, \mathrm{Q}=\tan x$

The general solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

or $y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+c$

$=\int\left(\sec x \tan x+\tan ^2 x\right) d x+c=\int\left(\sec x \tan x+\sec ^2 x-1\right) d x+c$

$=\sec x+\tan x-x+c$

or $y(\sec x+\tan x)=\sec x+\tan x-x+c$.

For each of the following differential equations given in Exercises 5 to 8, find the general solution:

The given differential equation is $\cos ^2 x \frac{d y}{d x}+y=\tan x$

Dividing the entire equation by $\cos ^2 x$ to make the coefficient of $\frac{d y}{d x}$ unity,

It is of the form $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$.

Comparing $\mathrm{P}=\sec ^2 x, \mathrm{Q}=\sec ^2 x \tan x$

The general solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

Let $\tan x=t$. Then differentiating, $\sec ^2 x d x=d t$

I II

Applying integration by parts,

Putting this value in eqn. (i), $y e^{\tan x}=(\tan x-1) e^{\tan x}+c$

Dividing each term by $e^{\tan x}$,

$y=(\tan x-1)+c e^{-\tan x}$ This is the required general solution.

The given differential equation is $x \frac{d y}{d x}+2 y=x^2 \log x$

Dividing each term by $x$ (To make coeff. of $\frac{d y}{d x}$ unity)

It is of the form $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$.

The general solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

or

Dividing by $x^2, y=\frac{x^2}{4} \log x-\frac{x^2}{16}+\frac{c}{x^2}$

The given differential equation is $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

Dividing each term by $x \log x$ to make the coefficient of $\frac{d y}{d x}$ unity, $\frac{d y}{d x}+\frac{1}{x \log x} y=\frac{2}{x^2}$

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

The general solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

or $y \log x=\int \frac{2}{x^2} \log x d x=2 \int(\log x) x_{\text {II }}^{-2} d x+c$

Applying Product Rule of integration,

The given differential equation is $\left(1+x^2\right) d y+2 x y d x=\cot x d x$

Dividing each term by $d x,\left(1+x^2\right) \frac{d y}{d x}+2 x y=\cot x$

Dividing each term by $\left(1+x^2\right)$ to make coefficient of $\frac{d y}{d x}$ unity,

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

Solution is

Dividing by $1+x^2, \quad \quad y=\frac{\log |\sin x|}{1+x^2}+\frac{c}{1+x^2}$

or $y=\left(1+x^2\right)^{-1} \log |\sin x|+c\left(1+x^2\right)^{-1}$

This is the required general solution.

For each of the differential equations in Exercises 9 to 12, find the general solution:

The given differential equation is

Dividing each term by $x$ to make coefficient of $\frac{d y}{d x}$ unity,

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

Solution is $y($ I.F. $)=\int \mathrm{Q}($ I.F. $) d x+c$

or

This is the required general solution.

The given differential equation is

(Standard form of a linear differential equation) Comparing with $\frac{d x}{d y}+\mathrm{P} x=\mathrm{Q}$, we have, $\mathrm{P}=-1$ and $\mathrm{Q}=y$

Dividing each term by $e^{-y}, x=-y-1+\frac{c}{\left(e^{-y}\right)}$

or $\quad x+y+1=c e^y$

This is the required general solution.

The given differential equation is $y d x+\left(x-y^2\right) d y=0$

Dividing by $d y, y \frac{d x}{d y}+x-y^2=0$ or $y \frac{d x}{d y}+x=y^2$

Dividing each term by $y$ (to make coefficient of $\frac{d x}{d y}$ unity),

Comparing with $\frac{d x}{d y}+\mathrm{P} x=\mathrm{Q}$, we have

Solution is $x$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d y+c$

Dividing by $y, x=\frac{y^2}{3}+\frac{c}{y}$

This is the required general solution.

The given differential equation is $\left(x+3 y^2\right) \frac{d y}{d x}=y$

Dividing each term by $y$ (to make coefficient of $\frac{d x}{d y}$ unity), $\left.\frac{d x}{d y}-\frac{1}{y} x=3 y \quad \right\rvert\,$ (This is a linear D.E. in standard form) Comparing with $\frac{d x}{d y}+\mathrm{P} x=\mathrm{Q}$, we have $\mathrm{P}=\frac{-1}{y}$ and $\mathrm{Q}=3 y$

Solution is $x$ (I.F.) $=\int \mathrm{Q}($ I.F. $) d y+c$

Cross – Multiplying, $x=3 y^2+c y$

This is the required general solution.

For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfying the given condition:

The given differential equation is

(It is standard form of linear differential equation)

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

Finding c: $y=0$ when $x=\frac{\pi}{3}$ (given)

∴ From equation (i), $\quad 0=\cos \frac{\pi}{3}+c \cos ^2 \frac{\pi}{3}$

or $\quad 0=\frac{1}{2}+c\left(\frac{1}{2}\right)^2 \quad$ or $\quad 0=\frac{1}{2}+\frac{c}{4}$

Putting $c=-2$ in ( $i$ ), the required particular solution is

The given differential equation is

Dividing each term by $\left(1+x^2\right)$ to make coefficient of $\frac{d y}{d x}$ unity,

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

Solution is $y($ I.F. $)=\int \mathrm{Q}($ I.F. $) d x+c$

or

or

or

Finding c: $y=0$ when $x=1$

Putting $y=0$ and $x=1$ in $(i), \quad 0=\tan ^{-1} 1+c$

or $\quad 0=\frac{\pi}{4}+c \quad\left[\because \tan \frac{\pi}{4}=1\right] \quad \Rightarrow \quad c=-\frac{\pi}{4}$

Putting $c=-\frac{\pi}{4}$ in ( $i$ ), required particular solution is

The given differential equation is $\frac{d y}{d x}-3 y \cot x=\sin 2 x$

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

The general solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

or $y \frac{1}{\sin ^3 x}=\int \sin 2 x \cdot \frac{1}{\sin ^3 x} d x+c$

or $\quad \frac{y}{\sin ^3 x}=\int \frac{2 \sin x \cos x}{\sin ^3 x} d x+c \quad=2 \int \frac{\cos x}{\sin ^2 x} d x+c$

$=2 \int \frac{\cos x}{\sin x \cdot \sin x} d x+c=2 \int \operatorname{cosec} x \cot x d x=-2 \operatorname{cosec} x+c$

or $\quad \frac{y}{\sin ^3 x}=-\frac{2}{\sin x}+c$

Multiplying every term by L.C.M. $=\sin ^3 x$

To find $c$ : Putting $y=2$ and $x=\frac{\pi}{2}$ (given) in ( $i$ ),

$2=-2 \sin ^2 \frac{\pi}{2}+c \sin ^3 \frac{\pi}{2} \quad$ or $\quad 2=-2+c \quad$ or $\quad c=4$

Putting $c=4$ in (i), the required particular solution is

Given: Slope of the tangent to the curve at any point $(x, y)=$ Sum of coordinates of the point $(x, y)$.

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

$\int \mathrm{P} d x=\int-1 d x=-\int 1 d x=-x$

I.F. $=e^{\int \mathrm{P} d x}=e^{-x}$

Solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

i.e.,

To find $\boldsymbol{c}$ : Given: Curve $(i)$ passes through the origin $(0,0)$.

Putting $x=0$ and $y=0$ in (i), $0=0-1+c$

or $\quad-c=-1 \quad$ or $\quad c=1$

Putting $c=1$ in (i), equation of required curve is

According to question,

Sum of the coordinates of any point say $(x, y)$ on the curve. $=$ Magnitude of the slope of the tangent to the curve +5

(because of exceeds)

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

Solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

or $\quad y e^{-x}=\int(x-5) e^{-x} d x+c$

I I I

[Applying Product Rule: $\int$ I . II $d x=$ I $\int$ II $d x-\int \frac{d}{d x}($ I $)\left(\int\right.$ II $\left.\left.d x\right) d x\right]$

or

or

or

or

Multiplying both sides by L.C.M. $=e^x$

or $y=-x+5-1+c e^x$ or $x+y=4+c e^x$

To find $\boldsymbol{c}$ : Curve (i) passes through the point ( 0,2 ).

Putting $x=0$ and $y=2$ in (i),

Putting $c=-2$ in (i), required equation of the curve is

The given differential equation is

Dividing each term by $x$ to make coefficient of $\frac{d y}{d x}$ unity, $\left.\frac{d y}{d x}-\frac{1}{x} y=2 x \quad \right\rvert\,$ (This is a linear D.E. in standard form) Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have $\mathrm{P}=\frac{-1}{x}$ and $\mathrm{Q}=2 x$

Hence, Option (C) is the correct answer.

The given differential equation is

(i) The given differential equation is

The highest-order derivative in this equation is $\frac{d^2 y}{d x^2}$ and hence order of this differential equation is 2 .

This is a polynomial equation in its derivatives, and highest power of the highest order derivative $\frac{d^2 y}{d x^2}$ is 1 .

∴ Order 2, Degree 1.

(ii) The given differential equation is

The highest-order derivative in this equation is $\frac{d y}{d x}$ and hence order of this differential equation is 1 .

This is a polynomial equation in its derivatives, and highest power of the highest order derivative $\frac{d y}{d x}$ is $3 . \quad\left[\because\right.$ of $\left.\left(\frac{d y}{d x}\right)^3\right]$

∴ Order 1, Degree 3.

(iii) The given differential equation is $\frac{d^4 y}{d x^4}-\sin \left(\frac{d^3 y}{d x^3}\right)=0$. The highest-order derivative in this equation is $\frac{d^4 y}{d x^4}$ and hence order of this differential equation is 4 .

Degree of this differential equation is not defined because the given differential equation is not a polynomial equation in derivatives

∴ Order 4 and Degree not defined.

(i) The given function is

We verify that the function given by (i) satisfies the differential equation $\frac{d^2 y}{d x^2}+2 \frac{d y}{d x}-x y+x^2-2=0$

Differentiating both sides of (i), w.r.t. $x$,

Differentiating both sides once more, w.r.t. $x$

or $\quad x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}=a e^x+b e^{-x}+2$

∴ Putting $a e^x+b e^{-x}=x y-x^2$ from ( $i$ ), in R.H.S., we have

which matches the given differential equation (ii).

$\therefore \quad$ Function given by (i) is a solution of D.E. (ii).

(ii) The given function is

We verify that the function in (i) is a solution of the differential equation

From equation (i), $

\begin{aligned}

& \frac{d y}{d x}=\frac{d}{d x} e^x \cdot(a \cos x+b \sin x)+e^x \frac{d}{d x}(a \cos x+b \sin x) \\

& \text { or } \frac{d y}{d x}=e^x(a \cos x+b \sin x)+e^x(-a \sin x+b \cos x) \\

& \Rightarrow \quad \frac{d y}{d x}=y+e^x(-a \sin x+b \cos x)

\end{aligned}

y=x \sin 3 x

\frac{d^2 y}{d x^2}+9 y-6 \cos 3 x=0

\begin{aligned}

& =-9 x \sin 3 x+6 \cos 3 x \\

& =-9 y+6 \cos 3 x

\end{aligned}

x^2=2 y^2 \log y

\left(x^2+y^2\right) \frac{d y}{d x}-x y=0

2 x=2\left[y^2 \cdot \frac{1}{y} \frac{d y}{d x}+(\log y) 2 y \frac{d y}{d x}\right]

\therefore \quad \frac{d y}{d x}=\frac{x}{y+2 y \log y}=\frac{x}{y(1+2 \log y)}

\frac{d y}{d x}=\frac{x}{y\left(1+\frac{x^2}{y^2}\right)}=\frac{x}{y\left(\frac{y^2+x^2}{y^2}\right)}=\frac{x y^2}{y\left(x^2+y^2\right)}

\begin{aligned}

& \Rightarrow \quad \frac{d y}{d x}=\frac{x y}{x^2+y^2} \\

& \text { Cross-multiplying, both sides, }\left(x^2+y^2\right) \frac{d y}{d x}=x y \\

& \text { or } \quad\left(x^2+y^2\right) \frac{d y}{d x}-x y=0

\end{aligned}

The given differential equation is

Here each coefficient of $d x$ and $d y$ is of same degree (Here 3), therefore differential equation (i) looks to be homogeneous differential equation.

From equation (i), $\frac{d y}{d x}=\frac{\left(x^3-3 x y^2\right)}{y^3-3 x^2 y}$

Dividing every term in the numerator and denominator of R.H.S. by $x^3$,

Therefore the given differential equation is homogeneous.

Put $\frac{y}{x}=v$. Therefore $y=v x . \therefore \frac{d y}{d x}=v .1+x \frac{d v}{d x}=v+x \frac{d v}{d x}$ Putting these values in eqn. (ii),

Cross-multiplying, both sides, $\quad x\left(v^3-3 v\right) d v=\left(1-v^4\right) d x$

Separating the variables, $\frac{\left(v^3-3 v\right)}{1-v^4} d v=\frac{d x}{x}$

Integrating both sides,

Let us form partial fractions of

Multiplying both sides of (iv) by L.C.M. $=(1-v)(1+v)\left(1+v^2\right)$,

Comparing coefficients of like powers of $v$,

Constants $\mathrm{A}+\mathrm{B}+\mathrm{D}=0$

Let us solve eqns. (v), (vi), (vii), (viii) for $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$.

Eqn. (v) – eqn. (vii) gives, $-2 \mathrm{C}=4 \Rightarrow \mathrm{C}=\frac{-4}{2}=-2$

Eqn. (vi) – eqn. (viii) gives, $-2 \mathrm{D}=0$ or $\mathrm{D}=0$

Putting $\mathrm{C}=-2$ in ( $v$ ),

Putting $\mathrm{D}=0$ in (vi),

Adding (ix) and (x),

From (x), $\mathrm{B}=-\mathrm{A}=\frac{1}{2}$

Putting values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D in (iv), we have

Putting this value in eqn. (iii),

Squaring both sides and cross-multiplying, $1-v^2=c^2 x^2\left(1+v^2\right)^2$

Putting $v=\frac{y}{x}, 1-\frac{y^2}{x^2}=c^2 x^2\left(1+\frac{y^2}{x^2}\right)^2$

or $\frac{x^2-y^2}{x^2}=c^2 x^2 \frac{\left(x^2+y^2\right)^2}{x^4}$ or $\frac{x^2-y^2}{x^2}=\frac{c^2\left(x^2+y^2\right)^2}{x^2}$

or $x^2-y^2=\mathrm{C}\left(x^2+y^2\right)^2$ where $c^2=\mathrm{C}$

This is the required general solution.

The given differential equation is

Separating Variables, $\frac{d y}{\sqrt{1-y^2}}=\frac{-d x}{\sqrt{1-x^2}}$

Integrating both sides, $\int \frac{1}{\sqrt{1-y^2}} d y=-\int \frac{1}{\sqrt{1-x^2}} d x$

This is the required general solution.

The given differential equation is

Multiplying by $d x$ and dividing by $y^2+y+1$, we have

Integrating both sides,

Now, $y^2+y+1=y^2+y+\frac{1}{4}-\frac{1}{4}+1$

[To complete squares, add and subtract $\left(\frac{1}{2} \text { coeff. of } y\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}$ ]

Changing $y$ to $x, \int \frac{1}{x^2+x+1} d x=\frac{2}{\sqrt{3}} \tan ^{-1} \frac{2 x+1}{\sqrt{3}}$

Putting these values in eqn. (i),

Multiplying by $\frac{\sqrt{3}}{2}, \quad \tan ^{-1} \frac{2 y+1}{\sqrt{3}}+\tan ^{-1} \frac{2 x+1}{\sqrt{3}}=\frac{\sqrt{3}}{2} c$

or

$\left[\because \tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} \frac{a+b}{1-a b}\right.$ and replacing $\frac{\sqrt{3}}{2} c$ by $\left.\tan ^{-1} c^{\prime}\right]$

Multiplying every term in the numerator and denominator of L.H.S. by 3, we have

or $\quad \sqrt{3}(2 x+2 y+2)=c^{\prime}(2-2 x-2 y-4 x y)$

$\Rightarrow \quad 2 \sqrt{3}(x+y+1)=2 c^{\prime}(1-x-y-2 x y)$

Dividing each term by $2 \sqrt{3}, x+y+1=\frac{c^{\prime}}{\sqrt{3}}(1-x-y-2 x y)$ or $\quad(x+y+1)=\mathrm{A}(1-x-y-2 x y) \quad$ where $\mathrm{A}=\frac{c^{\prime}}{\sqrt{3}}$.

The given differential equation is

Integrating both sides,

To find $\boldsymbol{c}$ : Given: Curve (i) passes through $\left(0, \frac{\pi}{4}\right)$.

Putting $x=0$ and $y=\frac{\pi}{4}$ in (i), $\sec 0 \sec \frac{\pi}{4}=c \quad$ or $\quad \sqrt{2}=c$.

Putting $c=\sqrt{2}$ in (i), equation of required curve is

The given differential equation is

Dividing each term by $\left(1+y^2\right)\left(1+e^{2 x}\right)$, we have

Integrating both sides, we have

or

To evaluate $\int \frac{e^x}{1+e^{2 x}} d x$

Put $e^x=t \quad \therefore \quad e^x=\frac{d t}{d x} \quad$ or $e^x d x=d t$

$\therefore \quad \int \frac{e^x d x}{1+e^{2 x}}=\int \frac{d t}{1+t^2}=\tan ^{-1} t=\tan ^{-1} e^x$

Putting this value in $(i), \tan ^{-1} y+\tan ^{-1} e^x=c$

Finding c: $y=1$ when $x=0$ (given)

Putting $x=0$ and $y=1$ in (ii), $\quad \tan ^{-1} 1+\tan ^{-1} 1=c \quad(\because \stackrel{0}{e}=1)$

or $\quad \frac{\pi}{4}+\frac{\pi}{4}=c \quad\left[\because \tan \frac{\pi}{4}=1 \quad \therefore \tan ^{-1} 1=\frac{\pi}{4}\right]$

or $\quad c=\frac{2 \pi}{4}=\frac{\pi}{2}$

Putting $c=\frac{\pi}{2}$ in (ii), the particular solution is

The given differential equation is

$y \cdot e^{x / y} d x=\left(x \cdot e^{x / y}+y^2\right) d y, \quad y \neq 0$

or $\frac{d x}{d y}=\frac{x e^{x / y}+y^2}{y \cdot e^{x / y}}=\frac{x e^{x / y}}{y e^{x / y}}+\frac{y^2}{y e^{x / y}}$

or $\frac{d x}{d y}=\frac{x}{y}+y e^{-x / y}$

It is not a homogeneous differential equation (because of presence of only $y$ as a factor) yet it can be solved by putting $\frac{x}{y}=v$ i.e., $x=v y$.

so that $\frac{d x}{d y}=v+y \frac{d v}{d y}$

Putting these values of $x$ and $\frac{d x}{d y}$ in (i), we have

$v+y \frac{d v}{d y}=v+y e^{-v}$

or $\quad y \frac{d u}{d y}=y e^{-v} \quad$ or $\quad y \frac{d u}{d y}=\frac{y}{e^v}$

Cross-multiplying, and dividing both sides by $y$,

Integrating $e^v=y+c$ or $e^{x / y}=y+c$

This is the required general solution.

The given differential equation is

Put $x-y=t$

Differentiating w.r.t. $x, 1-\frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow-\frac{d y}{d x}=\frac{d t}{d x}-1 \Rightarrow \frac{d y}{d x}=\frac{-d t}{d x}+1$

Putting these values in (i), $\frac{-d t}{d x}+1=-\left(\frac{t-1}{t+1}\right)$

$\Rightarrow \quad-\frac{d t}{d x}=-1-\left(\frac{t-1}{t+1}\right)$

Multiplying by $-1, \frac{d t}{d x}=1+\frac{t-1}{t+1}=\frac{t+1+t-1}{t+1}$

Integrating both sides, $\int\left(\frac{t+1}{t}\right) d t=2 \int 1 d x$

or $\int\left(\frac{t}{t}+\frac{1}{t}\right) d t=2 x+c \quad$ or $\int\left(1+\frac{1}{t}\right) d t=2 x+c$

Putting $t=x-y, x-y+\log |x-y|=2 x+c$

Finding c: $y=-1$ when $x=0$

Putting $x=0, y=-1$ in (ii),

Putting $c=1$ in (ii), required particular solution is

The given differential equation is

Multiplying both sides by $\frac{d y}{d x}$,

It is of the form $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$.

Comparing, $\mathrm{P}=\frac{1}{\sqrt{x}}$ and $\mathrm{Q}=\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$

The general solution is

Multiplying both sides by $e^{-2 \sqrt{x}}$, we have

$y=e^{-2 \sqrt{x}}(2 \sqrt{x}+c)$ is the required general solution.

The given differential equation is

(It is standard form of linear differential equation.)

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

Solution is $\quad y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+c$

or

To find $c$ : Given that $y=0$, when $x=\frac{\pi}{2}$.

Putting $\quad x=\frac{\pi}{2}$ and $y=0$ in $(i), \quad 0=2 \cdot \frac{\pi^2}{4}+c$

or $0=\frac{\pi^2}{2}+c \quad \Rightarrow c=\frac{-\pi^2}{2}$

Putting $c=-\frac{\pi^2}{2}$ in (i), the required particular solution is

The given differential equation is

Cross-multiplying, both sides, $(x+1) e^y d y=\left(2-e^y\right) d x$

Separating the variables, $\quad \frac{e^y d y}{2-e^y}=\frac{d x}{x+1}$

Integrating both sides, $\int \frac{e^y}{2-e^y} d y=\int \frac{1}{x+1} d x$

When $x=0, y=0$ (given)

Putting $\mathrm{C}=1$ in (i) the required particular solution is

Note: The particular solution may be written as

which expresses $y$ as an explicit function of $x$.

The given differential equation is $\frac{y d x-x d y}{y}=0$

Cross-multiplying, both sides, $\quad y d x-x d y=0$

Separating the variables, $\quad \frac{d x}{x}=\frac{d y}{y}$

Integrating both sides, $\log |x|=\log |y|+\log |c|$

or $y=\mathrm{C} x$ where $\mathrm{C}= \pm \frac{1}{c}$

This is the required solution.

Hence, Option (C) is the correct answer.

We know that general solution of differential equation of the type

Hence, Option (C) is the correct answer.

The given differential equation is

Dividing each term by $d x$,

or $\quad e^x \frac{d y}{d x}+y e^x=-2 x$

Dividing each term by $e^x$ to make coefficient of $\frac{d y}{d x}$ unity,

$\frac{d y}{d x}+y=\frac{-2 x}{e^x}$ ((This is a linear D.E. in standard form))

Comparing with $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, we have

Solution is $y$ (I.F.) $=\int \mathrm{Q}$ (I.F.) $d x+\mathrm{C}$

or $\quad y e^x=\int \frac{-2 x}{e^x} e^x d x+\mathrm{C}$

or $\quad y e^x=-2 \int x d x+\mathrm{C} \quad$ or $\quad y e^x=-2 \frac{x^2}{2}+\mathrm{C}$

or $\quad y e^x=-x^2+\mathrm{C} \quad$ or $y e^x+x^2=\mathrm{C}$

Hence, Option (C) is the correct answer.