# Miscellaneous Exercise Differential Equations Solution - NCERT Class 12

## Chapter 9 Ex.9.ME Question 1

For each of the differential equations given below, indicate its order and degree (if defined).

(i) $$\frac{{{d^2}y}}{{d{x^2}}} + 5x{\left( {\frac{{dy}}{{dx}}} \right)^2} - 6y = \log x$$

(ii) $${\left( {\frac{{dy}}{{dx}}} \right)^3} - 4{\rm{ }}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 7y = \sin x$$

(iii) $$\frac{{{d^4}y}}{{d{x^4}}} - \sin \left( {\frac{{{d^3}y}}{{d{x^3}}}} \right) = 0$$

### Solution

(i) $$\frac{{{d^2}y}}{{d{x^2}}} + 5x{\left( {\frac{{dy}}{{dx}}} \right)^2} - 6y = \log x$$

$$\Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} + 5x{\left( {\frac{{dy}}{{dx}}} \right)^2} - 6y - \log x = 0$$

Highest order derivative present in differential equation is $$\frac{{{d^2}y}}{{d{x^2}}}$$. Its order is two.

Highest power raised to $$\frac{{{d^2}y}}{{d{x^2}}}$$ is one. Its degree is one.

(ii) $${\left( {\frac{{dy}}{{dx}}} \right)^3} - 4{\rm{ }}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 7y = \sin x$$

$$\Rightarrow \; {\left( {\frac{{dy}}{{dx}}} \right)^3} - 4{\rm{ }}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 7y - \sin x = 0$$

Highest order derivative in differential equation is $$\frac{{dy}}{{dx}}$$. its order is one.

Highest power raised to $$\frac{{dy}}{{dx}}$$ is three. Its degree is three.

(iii) $$\frac{{{d^4}y}}{{d{x^4}}} - \sin \left( {\frac{{{d^3}y}}{{d{x^3}}}} \right) = 0$$

Highest order derivative in differential equation is $$\frac{{{d^4}y}}{{d{x^4}}}$$. Order is four.

The given differential equation is not a polynomial equation. Degree is not defined.

## Chapter 9 Ex.9.ME Question 2

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i) $$xy = a{e^x} + b{e^{ - x}} + {x^2} \quad : \quad x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$$

(ii) $$y = {e^x}\left( {a\cos x + b\sin x} \right) \quad : \quad \frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y = 0$$

(iii) $$y = x\sin 3x \quad : \quad \frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0$$

(iv) $${x^2} = 2{y^2}\log y \quad : \quad \left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy = 0$$

### Solution

(i) $$xy = a{e^x} + b{e^{ - x}} + {x^2} \quad : \quad x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$$

$$xy = a{e^x} + b{e^{ - x}} + {x^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Differentiating both sides with respect to $$x$$, we get:

\begin{align}&x\frac{{dy}}{{dx}} + y.1 = a\frac{d}{{dx}}\left( {{e^x}} \right) + b\frac{d}{{dx}}\left( {{e^{ - x}}} \right) + \frac{d}{{dx}}\left( {{x^2}} \right)\\ &\Rightarrow \; x\frac{{dy}}{{dx}} + y = a{e^x} + b{e^{ - x}} + 2x\end{align}

Again, differentiating both sides with respect to $$x$$, we get:

\begin{align} &\Rightarrow \; x\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} = a{e^x} + b{e^{ - x}} + 2\\& \Rightarrow \; x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} = a{e^x} + b{e^{ - x}} + 2\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Now, we have $$x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$$

\begin{align}LHS &= x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2\\ &= a{e^x} + b{e^{ - x}} + 2 - \left( {a{e^x} + b{e^{ - x}} + {x^2}} \right) + {x^2} - 2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{using }}\left( 1 \right){\text{ and }}\left( 2 \right)} \right]\\& = a{e^x} + b{e^{ - x}} + 2 - a{e^x} - b{e^{ - x}} - {x^2} + {x^2} - 2\\ &= 0\\ &= RHS\end{align}

Thus, the given function is a solution of the corresponding differential equation.

(ii) $$y = {e^x}\left( {a\cos x + b\sin x} \right) \quad : \quad \frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y = 0$$

$$y = {e^x}\left( {a\cos x + b\sin x} \right)\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Differentiating both sides with respect to $$x$$, we get:

\begin{align}y &= {e^x}\left( {a\cos x + b\sin x} \right) = a{e^x}\cos x + b{e^x}\sin x\\ &\Rightarrow \; \frac{{dy}}{{dx}} = a.\frac{d}{{dx}}\left( {{e^x}\cos x} \right) + b.\frac{d}{{dx}}\left( {{e^x}\sin x} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = a\left( {{e^x}\cos x - {e^x}\sin x} \right) + b\left( {{e^x}\sin x + {e^x}\cos x} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = \left( {a + b} \right){e^x}\cos x + \left( {b - a} \right){e^x}\sin x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Again, differentiating both sides with respect to $$x$$, we get:

\begin{align} &\Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = \left( {a + b} \right)\frac{d}{{dx}}\left( {{e^x}\cos x} \right) + \left( {b - a} \right)\frac{d}{{dx}}\left( {{e^x}\sin x} \right)\\ &\Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = (a + b).\left( {{e^x}\cos x - {e^x}\sin x} \right) + \left( {b - a} \right)\left( {{e^x}\sin x + {e^x}\cos x} \right)\\ &\Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = {e^x}\left[ {\left( {a + b} \right)\left( {\cos x - \sin x} \right) + \left( {b - a} \right)\left( {\sin x + \cos x} \right)} \right]\\ & \Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = {e^x}\left[ {a\cos x - a\sin x + b\cos x - b\sin x + b\sin x + b\cos x - a\sin x - a\cos x} \right]\\& \Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = 2{e^x}\left( {b\cos x - a\sin x} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Now, we have $$\frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y = 0$$

\begin{align}LHS & = \frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y\\ &= 2{e^x}\left( {b\cos x - a\sin x} \right) - 2\left[ {\left( {a + b} \right){e^x}\cos x + \left( {b - a} \right){e^x}\sin x} \right] + 2{e^x}\left( {a\cos x + b\sin x} \right)\\& \qquad \qquad \left[ {{\text{using }}\left( 1 \right){\text{, }}\left( 2 \right){\text{ and }}\left( 3 \right)} \right]\\ &= {e^x}\left[ {\left( {2b\cos x - 2a\sin x} \right) - \left( {2a\cos x + 2b\cos x} \right) - \left( {2b\sin x - 2a\sin x} \right) + \left( {2a\cos x + 2b\sin x} \right)} \right]\\ &= {e^x}\left[ {2b\cos x - 2a\sin x - 2a\cos x - 2b\cos x - 2b\sin x + 2a\sin x + 2a\cos x + 2b\sin x} \right]\\ &= {e^x}\left[ 0 \right]\\& = 0\\ &= RHS\end{align}

Thus, the given function is a solution of the corresponding differential equation.

(iii) $$y = x\sin 3x \quad : \quad \frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0$$

$$y = x\sin 3x\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Differentiating both sides with respect to $$x$$, we get:

\begin{align}& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {x\sin 3x} \right) = \sin 3x + x.\cos 3x.3\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \sin 3x + 3x\cos 3x\end{align}

Again, differentiating both sides with respect to $$x$$, we get:

\begin{align}&\Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\sin 3x} \right) + 3\frac{d}{{dx}}\left( {x\cos 3x} \right)\\& \Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = 3\cos 3x + 3\left[ {\cos 3x + x\left( { - \sin 3x} \right).3} \right]\\ &\Rightarrow \; \frac{{{d^2}y}}{{d{x^2}}} = 6\cos 3x - 9x\sin 3x\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Now, we have $$\frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0$$

\begin{align}LHS &= \frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x\\ &= \left( {6\cos 3x - 9x\sin 3x} \right) + 9x\sin 3x - 6\cos 3x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{using }}\left( 1 \right){\text{ and }}\left( 2 \right)} \right]\\& = 0\\ &= RHS\end{align}

Thus, the given function is a solution of the corresponding differential equation.

(iv) $${x^2} = 2{y^2}\log y \quad : \quad \left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy = 0$$

$${x^2} = 2{y^2}\log y\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Differentiating both sides with respect to $$x$$, we get:

\begin{align} &\Rightarrow \; 2x = 2\frac{d}{{dx}}\left[ {{y^2}\log y} \right]\\ &\Rightarrow \; x = \left[ {2y.\log y.\frac{{dy}}{{dx}} + {y^2}.\frac{1}{y}.\frac{{dy}}{{dx}}} \right]\\ &\Rightarrow \; x = \frac{{dy}}{{dx}}\left( {2y\log y + y} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{x}{{y\left( {1 + 2\log y} \right)}}\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Now, we have $$\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy = 0$$

\begin{align}LHS &= \left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy\\ &= \left( {2{y^2}\log y + {y^2}} \right).\frac{x}{{y\left( {1 + 2\log y} \right)}} - xy\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{using }}\left( 1 \right){\text{ and }}\left( 2 \right)} \right]\\& = {y^2}\left( {1 + 2\log y} \right).\frac{x}{{y\left( {1 + 2\log y} \right)}} - xy\\& = xy - xy\\ &= 0\\& = RHS\end{align}

Thus, the given function is a solution of the corresponding differential equation.

## Chapter 9 Ex.9.ME Question 3

Form the differential equation representing the family of curves given by $${\left( {x - a} \right)^2} + 2{y^2} = {a^2}$$ where $$a$$ is an arbitrary constant.

### Solution

\begin{align}&{\left( {x - a} \right)^2} + 2{y^2} = {a^2}\\& \Rightarrow \; {x^2} + {a^2} - 2ax + 2{y^2} = {a^2}\\ &\Rightarrow \; 2{y^2} = 2ax - {x^2}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Differentiating both sides with respect to $$x$$, we get:

\begin{align}& \Rightarrow \; 4y\frac{{dy}}{{dx}} = 2a - 2x\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{2a - 2x}}{{4y}}\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{2ax - 2{x^2}}}{{4xy}}\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{2{y^2} + {x^2} - 2{x^2}}}{{4xy}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{from}}\left( 1 \right),{\text{ }}2ax = 2{y^2} + {x^2}} \right]\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{2{y^2} - {x^2}}}{{4xy}}\end{align}

Thus, the differential equation of the family of curves is given as $$\frac{{dy}}{{dx}} = \frac{{2{y^2} - {x^2}}}{{4xy}}$$.

## Chapter 9 Ex.9.ME Question 4

Prove that $${x^2} - {y^2} = c{\left( {{x^2} + {y^2}} \right)^2}$$ is the general solution of differential equation $$\left( {{x^3} - 3x{y^2}} \right)dx = \left( {{y^3} - 3{x^2}y} \right)dy$$, where $$c$$ is a parameter.

### Solution

\begin{align}&\left( {{x^3} - 3x{y^2}} \right)dx = \left( {{y^3} - 3{x^2}y} \right)dy\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{{x^3} - 3x{y^2}}}{{{y^3} - 3{x^2}y}}\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

This is a homogeneous equation, to simplify it, let $$y = vx$$

\begin{align} &\Rightarrow \; \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {vx} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Using $$\left( 1 \right)$$ and $$\left( 2 \right)$$

\begin{align}& \Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{{x^3} - 3x{{\left( {vx} \right)}^2}}}{{{{\left( {vx} \right)}^3} - 3{x^2}\left( {vx} \right)}}\\& \Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{1 - 3{v^2}}}{{{v^3} - 3v}}\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{1 - 3{v^2}}}{{{v^3} - 3v}} - v\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{1 - 3{v^2} - v\left( {{v^3} - 3v} \right)}}{{{v^3} - 3v}}\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{1 - {v^4}}}{{{v^3} - 3v}}\\& \Rightarrow \; \frac{{{v^3} - 3v}}{{1 - {v^4}}}dv = \frac{{dx}}{x}\end{align}

Integrating both sides, we get:

\begin{align}& \Rightarrow \; \int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = \log x + \log C'\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\\& \Rightarrow \; \int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = \int {\frac{{{v^3}dv}}{{1 - {v^4}}}} - 3\int {\frac{{vdv}}{{1 - {v^4}}}} \\& \Rightarrow \; \int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = {I_1} - 3{I_2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\\&{\text{ }}\left( {{\text{where}},{\text{ }}{I_1} = \int {\frac{{{v^3}dv}}{{1 - {v^4}}}} {\text{ and }}{I_2} = \int {\frac{{vdv}}{{1 - {v^4}}}} {\text{ }}} \right)\end{align}

Let $$1 - {v^4} = t$$

Therefore,

\begin{align}& \Rightarrow \; \frac{d}{{dv}}\left( {1 - {v^4}} \right) = \frac{{dt}}{{dv}}\\& \Rightarrow \; - 4{v^3} = \frac{{dt}}{{dv}}\\& \Rightarrow \; {v^3}dv = - \frac{{dt}}{4}\end{align}

Now,

\begin{align}{I_1} &= \int { - \frac{{dt}}{{4t}}} \\& = - \frac{1}{4}\log t\\& = - \frac{1}{4}\log \left( {1 - {v^4}} \right)\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 5 \right)\end{align}

And

\begin{align}{I_2}& = \int {\frac{{vdv}}{{1 - {v^4}}}} \\& = \int {\frac{{vdv}}{{1 - {{\left( {{v^2}} \right)}^2}}}}\end{align}

Let $${v^2} = p$$

Therefore,

\begin{align}& \Rightarrow \; \frac{d}{{dv}}\left( {{v^2}} \right) = \frac{{dp}}{{dv}}\\ &\Rightarrow \; 2v = \frac{{dp}}{{dv}}\\& \Rightarrow \; vdv = \frac{{dp}}{2}\end{align}

Now,

\begin{align}{I_2} &= \frac{1}{2}\int {\frac{{dp}}{{1 - {p^2}}}} \\ &= \frac{1}{{2 \times 2}}\log \left| {\frac{{1 + p}}{{1 - p}}} \right|\\ &= \frac{1}{4}\log \left| {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right|\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 6 \right)\end{align}

Using $$\left( 4 \right)$$, $$\left( 5 \right)$$ and $$\left( 6 \right)$$

$$\int {\left( {\frac{{{v^3} - 3v}}{{1 - {v^4}}}} \right)} dv = - \frac{1}{4}\log \left( {1 - {v^4}} \right) - \frac{3}{4}\log \left| {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right|\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 7 \right)$$

Using $$\left( 2 \right)$$ and $$\left( 7 \right)$$

\begin{align}& - \frac{1}{4}\log \left( {1 - {v^4}} \right) - \frac{3}{4}\log \left| {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right| = \log x + \log C'\\ &\Rightarrow \; - \frac{1}{4}\log \left[ {\left( {1 - {v^4}} \right){{\left( {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right)}^3}} \right] = \log C'x\end{align}

\begin{align}& \Rightarrow \; - \frac{1}{4}\log \left[ {\left( {1 - {v^2}} \right)\left( {1 + {v^2}} \right){{\left( {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right)}^3}} \right] = \log C'x\\ &\Rightarrow \; \frac{{{{\left( {1 + {v^2}} \right)}^4}}}{{{{\left( {1 - {v^2}} \right)}^2}}} = {\left( {C'x} \right)^{ - 4}}\\ &\Rightarrow \; \frac{{{{\left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right)}^4}}}{{{{\left( {1 - \frac{{{y^2}}}{{{x^2}}}} \right)}^2}}} = \frac{1}{{{{C'}^4}{x^4}}}\\& \Rightarrow \; \frac{{{{\left( {{x^2} + {y^2}} \right)}^4}}}{{{x^4}{{\left( {{x^2} - {y^2}} \right)}^2}}} = \frac{1}{{{{C'}^4}{x^4}}}\\ &\Rightarrow \; {\left( {{x^2} - {y^2}} \right)^2} = {{C'}^4}{\left( {{x^2} + {y^2}} \right)^4}\end{align}

Taking square root on both sides

\begin{align} &\Rightarrow \; \left( {{x^2} - {y^2}} \right) = {{C'}^2}{\left( {{x^2} + {y^2}} \right)^2}\\ &\Rightarrow \; \left( {{x^2} - {y^2}} \right) = C{\left( {{x^2} + {y^2}} \right)^2}\;\;\;\;\;\;\;\;\;\;\left( {where,{\text{ }}C = {{C'}^2}} \right)\end{align}

## Chapter 9 Ex.9.ME Question 5

Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

### Solution

Equation of a circle in first quadrant with centre $$(a,a)$$ and radius $$(a)$$ which touches coordinate axes is:

$${\left( {x - a} \right)^2} + {\left( {y - a} \right)^2} = {a^2}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Differentiating both sides with respect to $$x$$, we get:

\begin{align} &\Rightarrow \; 2\left( {x - a} \right) + 2\left( {y - a} \right)\frac{{dy}}{{dx}} = 0\\& \Rightarrow \; \left( {x - a} \right) + \left( {y - a} \right)y' = 0\\& \Rightarrow \; \left( {x - a} \right) + yy' - ay' = 0\\ &\Rightarrow \; x + yy' - a\left( {1 + y'} \right) = 0\\& \Rightarrow \; a = \frac{{x + yy'}}{{1 + y'}}\end{align}

Substituting this value in equation $$\left( 1 \right)$$, we get:

\begin{align}&{\left[ {x - \left( {\frac{{x + yy'}}{{1 + y'}}} \right)} \right]^2} + {\left[ {y - \left( {\frac{{x + yy'}}{{1 + y'}}} \right)} \right]^2} = {\left( {\frac{{x + yy'}}{{1 + y'}}} \right)^2}\\ &\Rightarrow \; {\left[ {\frac{{\left( {x - y} \right)y'}}{{\left( {1 + y'} \right)}}} \right]^2} + {\left[ {\frac{{y - x}}{{1 + y'}}} \right]^2} = {\left[ {\frac{{x + yy'}}{{1 + y'}}} \right]^2}\\& \Rightarrow \; {\left( {x - y} \right)^2}.{{y'}^2} + {\left( {x - y} \right)^2} = {\left( {x + yy'} \right)^2}\\ &\Rightarrow \; {\left( {x - y} \right)^2}\left[ {1 + {{\left( {y'} \right)}^2}} \right] = {\left( {x + yy'} \right)^2}\end{align}

Hence, the differential equation of the family of circles is $${\left( {x - y} \right)^2}\left[ {1 + {{\left( {y'} \right)}^2}} \right] = {\left( {x + yy'} \right)^2}$$

## Chapter 9 Ex.9.ME Question 6

Find the general solution of the differential equation $$\frac{{dy}}{{dx}} + \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} = 0$$

### Solution

\begin{align}&\frac{{dy}}{{dx}} + \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} = 0\\ &\Rightarrow \; \frac{{dy}}{{dx}} = - \frac{{\sqrt {1 - {y^2}} }}{{\sqrt {1 - {x^2}} }}\\& \Rightarrow \; \frac{{dy}}{{\sqrt {1 - {y^2}} }} = - \frac{{dx}}{{\sqrt {1 - {x^2}} }}\end{align}

Integrating both sides, we get:

\begin{align}& \Rightarrow \; {\sin ^{ - 1}}y = - {\sin ^{ - 1}}x + C\\ &\Rightarrow \; {\sin ^{ - 1}}x + {\sin ^{ - 1}}y = C\end{align}

## Chapter 9 Ex.9.ME Question 7

Show that the general solution of the differential equation $$\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$$ is given by $$\left( {x + y + 1} \right) = A\left( {1 - x - y - 2xy} \right)$$, where $$A$$ is parameter.

### Solution

\begin{align}&\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0\\ &\Rightarrow \; \frac{{dy}}{{dx}} = - \left( {\frac{{{y^2} + y + 1}}{{{x^2} + x + 1}}} \right)\\ &\Rightarrow \; \frac{{dy}}{{{y^2} + y + 1}} = - \frac{{dx}}{{{x^2} + x + 1}}\end{align}

Integrating both sides, we get:

\begin{align}&\int {\frac{{dy}}{{{y^2} + y + 1}}} = - \int {\frac{{dx}}{{{x^2} + x + 1}}} \\& \Rightarrow \; \int {\frac{{dy}}{{{{\left( {y + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} = - } \int {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} \\ &\Rightarrow \; \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left[ {\frac{{y + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right] = - \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left[ {\frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right] + C\\ &\Rightarrow \; {\tan ^{ - 1}}\left[ {\frac{{2y + 1}}{{\sqrt 3 }}} \right] + {\tan ^{ - 1}}\left[ {\frac{{2x + 1}}{{\sqrt 3 }}} \right] = \frac{{\sqrt 3 C}}{2}\\ &\Rightarrow \; {\tan ^{ - 1}}\left[ {\frac{{\frac{{2y + 1}}{{\sqrt 3 }} + \frac{{2x + 1}}{{\sqrt 3 }}}}{{1 - \frac{{\left( {2y + 1} \right)}}{{\sqrt 3 }}\frac{{\left( {2x + 1} \right)}}{{\sqrt 3 }}}}} \right] = \frac{{\sqrt 3 C}}{2}\\& \Rightarrow \; {\tan ^{ - 1}}\left[ {\frac{{\frac{{2x + 2y + 2}}{{\sqrt 3 }}}}{{1 - \frac{{\left( {4xy + 2x + 2y + 1} \right)}}{3}}}} \right] = \frac{{\sqrt 3 C}}{2}\\& \Rightarrow \; \left[ {\frac{{\frac{{2x + 2y + 2}}{{\sqrt 3 }}}}{{1 - \frac{{\left( {4xy + 2x + 2y + 1} \right)}}{3}}}} \right] = \tan \left( {\frac{{\sqrt 3 C}}{2}} \right)\\ &\Rightarrow \; \left[ {\frac{{\frac{{2x + 2y + 2}}{{\sqrt 3 }}}}{{\frac{{3 - \left( {4xy + 2x + 2y + 1} \right)}}{3}}}} \right] = {C_1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{where}},\;{C_1} = \tan \left( {\frac{{\sqrt 3 C}}{2}} \right)} \right]\\& \Rightarrow \; \frac{{\sqrt 3 \left( {2x + 2y + 2} \right)}}{{3 - \left( {4xy + 2x + 2y + 1} \right)}} = {C_1}\\& \Rightarrow \; 2\sqrt 3 \left( {x + y + 1} \right) = {C_1}\left( {3 - 4xy - 2x - 2y - 1} \right)\\ &\Rightarrow \; 2\sqrt 3 \left( {x + y + 1} \right) = {C_1}\left( {2 - 4xy - 2x - 2y} \right)\\& \Rightarrow \; 2\sqrt 3 \left( {x + y + 1} \right) = {C_1} \times 2\left( {1 - 2xy - x - y} \right)\\ &\Rightarrow \; \sqrt 3 \left( {x + y + 1} \right) = {C_1}\left( {1 - x - y - 2xy} \right)\\ &\Rightarrow \; \left( {x + y + 1} \right) = \frac{{{C_1}}}{{\sqrt 3 }}\left( {1 - x - y - 2xy} \right)\\ &\Rightarrow \; \left( {x + y + 1} \right) = A\left( {1 - x - y - 2xy} \right)\;\;\;\;\;\;\;\;\;\;\;\left[ {{\text{where }}A = \frac{{{C_1}}}{{\sqrt 3 }}} \right]\end{align}

## Chapter 9 Ex.9.ME Question 8

Find the equation of the curve passing through the point $$\left( {0,\frac{\pi }{4}} \right)$$ whose differential equation is $$\sin x\cos ydx + \cos x\sin ydy = 0$$.

### Solution

\begin{align}&\sin x\cos ydx + \cos x\sin ydy = 0\\ &\Rightarrow \; \frac{{\sin x\cos ydx + \cos x\sin ydy}}{{\cos x\cos y}} = 0\\ &\Rightarrow \; \tan xdx + \tan ydy = 0\\& \Rightarrow \; \log \left( {\sec x} \right) + \log \left( {\sec y} \right) = \log C\\& \Rightarrow \; \log \left( {\sec x.\sec y} \right) = \log C\\ &\Rightarrow \; \sec x.\sec y = C\end{align}

The curve passes through the point $$\left( {0,\frac{\pi }{4}} \right)$$

Therefore,

\begin{align}& \Rightarrow \; 1 \times \sqrt 2 = C\\ &\Rightarrow \; C = \sqrt 2 \\&\sec x.\sec y = \sqrt 2 \\& \Rightarrow \; \sec x.\frac{1}{{\cos y}} = \sqrt 2 \\& \Rightarrow \; \cos y = \frac{{\sec x}}{{\sqrt 2 }}\end{align}

## Chapter 9 Ex.9.ME Question 9

Find the particular solution of the differential equation $$\left( {1 + {e^{2x}}} \right)dy + \left( {1 + {y^2}} \right){e^x}dx = 0$$, given that $$y = 1$$ when $$x = 0$$.

### Solution

\begin{align}&\left( {1 + {e^{2x}}} \right)dy + \left( {1 + {y^2}} \right){e^x}dx = 0\\& \Rightarrow \; \frac{{dy}}{{1 + {y^2}}} + \frac{{{e^x}}}{{1 + {e^{2x}}}} = 0\end{align}

Integrating both sides, we get:

$\Rightarrow \; {\tan ^{ - 1}}y + \int {\frac{{{e^x}dx}}{{1 + {e^{2x}}}} = C} \qquad \ldots \left( 1 \right)$

Let $${e^x} = t \Rightarrow \; {e^{2x}} = {t^2}$$

\begin{align}&\frac{d}{{dx}}\left( {{e^x}} \right) = \frac{{dt}}{{dx}}\\ &\Rightarrow \; {e^x} = \frac{{dt}}{{dx}}\\ &\Rightarrow \; {e^x}dx = dt\end{align}

Substituting this value in equation $$\left( 1 \right)$$, we get:

\begin{align}&{\tan ^{ - 1}}y + \int {\frac{{dt}}{{1 + {t^2}}} = C} \\ &\Rightarrow \; {\tan ^{ - 1}}y + {\tan ^{ - 1}}t = C\\& \Rightarrow \; {\tan ^{ - 1}}y + {\tan ^{ - 1}}\left( {{e^x}} \right) = C\end{align}

When $$x = 0;\;\;y = 1$$

Hence,

\begin{align}&{\tan ^{ - 1}}1 + {\tan ^{ - 1}}1 = C\\& \Rightarrow \; \frac{\pi }{4} + \frac{\pi }{4} = C\\ &\Rightarrow \; C = \frac{\pi }{2}\end{align}

Thus, $${\tan ^{ - 1}}y + {\tan ^{ - 1}}\left( {{e^x}} \right) = \frac{\pi }{2}$$

## Chapter 9 Ex.9.ME Question 10

Solve the differential equation $$y{e^{\frac{x}{y}}}dx = \left( {x{e^{\frac{x}{y}}} + {y^2}} \right)dy\;\left( {y \ne 0} \right)$$.

### Solution

\begin{align}&y{e^{\frac{x}{y}}}dx = \left( {x{e^{\frac{x}{y}}} + {y^2}} \right)dy\\& \Rightarrow \; y{e^{\frac{x}{y}}}\frac{{dx}}{{dy}} = x{e^{\frac{x}{y}}} + {y^2}\\& \Rightarrow \; {e^{\frac{x}{y}}}\left[ {y.\frac{{dx}}{{dy}} - x} \right] = {y^2}\\& \Rightarrow \; {e^{\frac{x}{y}}}\frac{{\left[ {y.\frac{{dx}}{{dy}} - x} \right]}}{{{y^2}}} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Let $${e^{\frac{x}{y}}} = z$$

Differentiating it with respect to $$y$$, we get:

\begin{align}&\frac{d}{{dy}}\left( {{e^{\frac{x}{y}}}} \right) = \frac{{dz}}{{dy}}\\& \Rightarrow \; {e^{\frac{x}{y}}}.\frac{d}{{dy}}\left( {\frac{x}{y}} \right) = \frac{{dz}}{{dy}}\\& \Rightarrow \; {e^{\frac{x}{y}}}\left[ {\frac{{y.\frac{{dx}}{{dy}} - x}}{{{y^2}}}} \right] = \frac{{dz}}{{dy}}\\& \Rightarrow \; \frac{{dz}}{{dy}} = 1 \qquad \left[ {from{\text{ }}\left( 1 \right)} \right]\\& \Rightarrow \; dz = dy\end{align}

Integrating both sides, we get

\begin{align}& \Rightarrow \; z = y + C\\ &\Rightarrow \; {e^{\frac{x}{y}}} = y + C\end{align}

## Chapter 9 Ex.9.ME Question 11

Find a particular solution of the differential equation $$\left( {x - y} \right)\left( {dx + dy} \right) = \left( {dx - dy} \right)$$, given that $$y = - 1$$, when $$x = 0$$. (Hint: put $$x - y = t$$)

### Solution

\begin{align}&\left( {x - y} \right)\left( {dx + dy} \right) = \left( {dx - dy} \right)\\& \Rightarrow \; \left( {x - y + 1} \right)dy = \left( {1 - x + y} \right)dx\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{1 - x + y}}{{x - y + 1}}\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{1 - \left( {x - y} \right)}}{{1 + \left( {x - y} \right)}} \qquad \ldots \left( 1 \right)\end{align}

Let $$x - y = t \quad \ldots \left( 2 \right)$$

\begin{align}& \Rightarrow \; \frac{d}{{dx}}\left( {x - y} \right) = \frac{{dt}}{{dx}}\\& \Rightarrow \; 1 - \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}\\& \Rightarrow \; 1 - \frac{{dt}}{{dx}} = \frac{{dy}}{{dx}} \qquad \ldots \left( 3 \right)\end{align}

Using $$\left( 1 \right)$$, $$\left( 2 \right)$$ and $$\left( 3 \right)$$

\begin{align}\\ &\Rightarrow \; 1 - \frac{{dt}}{{dx}} = \frac{{1 - t}}{{1 + t}}\\&\Rightarrow \; \frac{{dt}}{{dx}} = 1 - \left( {\frac{{1 - t}}{{1 + t}}} \right)\\& \Rightarrow \; \frac{{dt}}{{dx}} = \frac{{\left( {1 + t} \right) - \left( {1 - t} \right)}}{{1 + t}}\\ &\Rightarrow \; \frac{{dt}}{{dx}} = \frac{{2t}}{{1 + t}}\\& \Rightarrow \; \left( {\frac{{1 + t}}{t}} \right)dt = 2dx\\& \Rightarrow \; \left( {1 + \frac{1}{t}} \right)dt = 2dx\end{align}

Integrating both sides, we get:

\begin{align}& \Rightarrow \; t + \log \left| t \right| = 2x + C\\ &\Rightarrow \; \left( {x - y} \right) + \log \left| {x - y} \right| = 2x + C\\ &\Rightarrow \; \log \left| {x - y} \right| = x + y + C\end{align}

When $$x = 0;\;\;y = - 1$$

\begin{align}&\log 1 = 0 - 1 + C\\& \Rightarrow \; C = 1\\&\log \left| {x - y} \right| = x + y + 1\end{align}

## Chapter 9 Ex.9.ME Question 12

Solve the differential equation $$\left[ {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}} \right]\frac{{dx}}{{dy}} = 1\;\left( {x \ne 0} \right)$$

### Solution

\begin{align}&\left[ {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}} \right]\frac{{dx}}{{dy}} = 1\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}\\ &\Rightarrow \; \frac{{dy}}{{dx}} + \frac{y}{{\sqrt x }} = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }}\end{align}

This equation is a linear differential equation of the form

$$\frac{{dy}}{{dx}} + Py = Q$$ where $$P = \frac{1}{{\sqrt x }}$$ and $$Q = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }}$$

Now, $$I.F. = {e^{\int {Pdx} }} = {e^{\int {\frac{1}{{\sqrt x }}dx} }} = {e^{2\sqrt x }}$$

The general solution of the given differential equation is given by,

\begin{align}&y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx + C} \\ &\Rightarrow \; y{e^{2\sqrt x }} = \int {\left( {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} \times {e^{2\sqrt x }}} \right)} dx + C\\ &\Rightarrow \; y{e^{2\sqrt x }} = \int {\frac{1}{{\sqrt x }}dx + C} \\ &\Rightarrow \; y{e^{2\sqrt x }} = 2\sqrt x + C\end{align}

## Chapter 9 Ex.9.ME Question 13

Find a particular solution of the differential equation $$\frac{{dy}}{{dx}} + y\cot x = 4x\cos {\text{ec}}x\left( {x \ne 0} \right)$$, given that $$y = 0$$ when $$x = \frac{\pi }{2}$$

### Solution

$$\frac{{dy}}{{dx}} + y\cot x = 4x\cos ecx$$

This equation is a linear differential equation of the form

$$\frac{{dy}}{{dx}} + Py = Q$$ where $$P = \cot x$$ and $$Q = 4x{\text{cosec}}x$$

Now, $$I.F. = {e^{\int {Pdx} }} = {e^{\int {\cot xdx} }} = {e^{\log \left| {\sin x} \right|}} = \sin x$$

The general solution of the given differential equation is given by,

\begin{align}&y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx + C} \\& \Rightarrow \; y\sin x = \int {\left( {4x{\text{cosec}}x.\sin x} \right)dx + C} \\& \Rightarrow \; y\sin x = 4\int {xdx} + C\\ &\Rightarrow \; y\sin x = 4.\frac{{{x^2}}}{2} + C\\ &\Rightarrow \; y\sin x = 2{x^2} + C\end{align}

When $$x = \frac{\pi }{2};\;\;y = 0$$

Therefore,

\begin{align} &\Rightarrow \; 0 = 2 \times \frac{{{\pi ^2}}}{4} + C\\& \Rightarrow \; C = - \frac{{{\pi ^2}}}{2}\end{align}

Thus, $$y\sin x = 2{x^2} - \frac{{{\pi ^2}}}{2}$$

## Chapter 9 Ex.9.ME Question 14

Find a particular solution of the differential equation $$\left( {x + 1} \right)\frac{{dy}}{{dx}} = 2{e^{ - y}} - 1$$, given that $$y = 0$$ when $$x = 0$$.

### Solution

\begin{align}&\left( {x + 1} \right)\frac{{dy}}{{dx}} = 2{e^{ - y}} - 1\\& \Rightarrow \; \frac{{dy}}{{2{e^{ - y}} - 1}} = \frac{{dx}}{{x + 1}}\\& \Rightarrow \; \frac{{{e^y}dy}}{{2 - {e^y}}} = \frac{{dx}}{{x + 1}}\end{align}

Integrating both sides, we get:

$$\int {\frac{{{e^y}dy}}{{2 - {e^y}}}} = \log \left| {x + 1} \right| + \log C \qquad \ldots \left( 1 \right)$$

Let $$2 - {e^y} = t$$

\begin{align}& \Rightarrow \; \frac{d}{{dy}}\left( {2 - {e^y}} \right) = \frac{{dt}}{{dy}}\\ &\Rightarrow \; - {e^y} = \frac{{dt}}{{dy}}\\ &\Rightarrow \; - {e^y}dy = dt\end{align}

Substituting this value in equation (1), we get:

\begin{align}& \Rightarrow \; \int {\frac{{ - dt}}{t}} = \log \left| {x + 1} \right| + \log C\\& \Rightarrow \; - \log \left| t \right| = \log \left| {C\left( {x + 1} \right)} \right|\\ &\Rightarrow \; - \log \left| {2 - {e^y}} \right| = \log \left| {C\left( {x + 1} \right)} \right|\\ &\Rightarrow \; \frac{1}{{2 - {e^y}}} = C\left( {x + 1} \right)\\& \Rightarrow \; 2 - {e^y} = \frac{1}{{C\left( {x + 1} \right)}}\end{align}

When $$x = 0;\;\;y = 0$$

Therefore,

\begin{align}& \Rightarrow \; 2 - 1 = \frac{1}{C}\\ &\Rightarrow \; C = 1\end{align}

Hence,

\begin{align}&2 - {e^y} = \frac{1}{{x + 1}}\\& \Rightarrow \; {e^y} = 2 - \frac{1}{{x + 1}}\\ &\Rightarrow \; {e^y} = \frac{{2x + 2 - 1}}{{x + 1}}\\& \Rightarrow \; {e^y} = \frac{{2x + 1}}{{x + 1}}\\ &\Rightarrow \; y = \log \left| {\frac{{2x + 1}}{{x + 1}}} \right|,\;\;\;\left( {x \ne - 1} \right)\end{align}

## Chapter 9 Ex.9.ME Question 15

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was $$20,000$$ in 1999 and $$25,000$$ in the year 2004, what will be the population of the village in 2009?

### Solution

Let the population at any instant $$(t)$$ be $$y$$.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

\begin{align}& \Rightarrow \; \frac{{dy}}{{dx}} \propto y\\& \Rightarrow \; \frac{{dy}}{{dt}} = ky\\ &\Rightarrow \; \frac{{dy}}{y} = kdt\\&\log y = kt + C\end{align}

In 1999, $$t = 0$$ and $$y = 20000$$

$$\log 20000 = C$$

In $$2004$$, $$t = 5$$ and $$y = 25000$$

$$\log 25000 = k.5 + C$$

\begin{align}& \Rightarrow \; \log 25000 = 5k + \log 20000\\ &\Rightarrow \; 5k = \log \left( {\frac{{25000}}{{20000}}} \right) = \log \left( {\frac{5}{4}} \right)\\& \Rightarrow \; k = \frac{1}{5}\log \left( {\frac{5}{4}} \right)\end{align}

In $$2009$$, $$t = 10$$ years

\begin{align}&\log y = 10 \times \frac{1}{5}\log \left( {\frac{5}{4}} \right) + \log \left( {20000} \right)\\& \Rightarrow \; \log y = \log \left[ {20000 \times {{\left( {\frac{5}{4}} \right)}^2}} \right]\\& \Rightarrow \; y = 20000 \times \frac{5}{4} \times \frac{5}{4}\\& \Rightarrow \; y = 31250\end{align}

Therefore, population of village in 2009 is $$31250.$$

## Chapter 9 Ex.9.ME Question 16

The general solution of the differential equation $$\frac{{ydx - xdy}}{y} = 0$$ is

(A) $$xy = C$$

(B) $$x = C{y^2}$$

(C) $$y = Cx$$

(D) $$y = C{x^2}$$

### Solution

\begin{align}&\frac{{ydx - xdy}}{y} = 0\\ &\Rightarrow \; \frac{{ydx - xdy}}{{xy}} = 0\\ &\Rightarrow \; \frac{1}{x}dx - \frac{1}{y}dy = 0\\&\log \left| x \right| - \log \left| y \right| = \log k\\& \Rightarrow \; \log \left| {\frac{x}{y}} \right| = \log k\\ &\Rightarrow \; \frac{x}{y} = k\\& \Rightarrow \; y = \frac{1}{k}x\\& \Rightarrow \; y = Cx\;\;\;\;\;\;\;\;\;\left( {{\text{where, }}C = \frac{1}{k}} \right)\end{align}

Thus, the correct option is C.

## Chapter 9 Ex.9.ME Question 17

The general solution of a differential equation of the type $$\frac{{dx}}{{dy}} + {P_1}x = {Q_1}$$ is

(A) $$y.{e^{\int {{P_1}dy} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dy} }}} \right)} dy + C$$

(B) $$y.{e^{\int {{P_1}dx} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dx} }}} \right)} dx + C$$

(C) $$x.{e^{\int {{P_1}dy} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dy} }}} \right)} dy + C$$

(D) $$x.{e^{\int {{P_1}dx} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dx} }}} \right)} dx + C$$

### Solution

I.F. for $$\frac{{dx}}{{dy}} + {P_1}x = {Q_1}$$ is $${e^{\int {{P_1}dy} }}$$

\begin{align}&x\left( {I.F.} \right) = \left( {\int {{Q_1} \times I.F.} } \right)dy + C\\&x.{e^{\int {{P_1}dy} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dy} }}} \right)} dy + C\end{align}

Thus, the correct option is C.

## Chapter 9 Ex.9.ME Question 18

The general solution of the differential equation $${e^x}dy + (y{e^x} + 2x)dx = 0$$ is

(A) $$x{e^y} + {x^2} = C$$

(B) $$x{e^y} + {y^2} = C$$

(C) $$y{e^x} + {x^2} = C$$

(D) $$y{e^y} + {x^2} = C$$

### Solution

\begin{align}{e^x}&dy + \left( {y{e^x} + 2x} \right)dx = 0\\ &\Rightarrow \; {e^x}\frac{{dy}}{{dx}} + y{e^x} + 2x = 0\\ &\Rightarrow \; \frac{{dy}}{{dx}} + y = - 2x{e^{ - x}}\end{align}

This is a linear differential equation of the form

$$\frac{{dy}}{{dx}} + Py = Q$$ where, $$P = 1$$ and $$Q = - 2x{e^{ - x}}$$

Now,

\begin{align}&I.F. = {e^{\int {Pdx} }} = {e^{\int {dx} }} = {e^x}\\&y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)} dx + C\\& \Rightarrow \; y{e^x} = \int {\left( { - 2x{e^{ - x}}.{e^x}} \right)} dx + C\\& \Rightarrow \; y{e^x} = - \int {2x} dx + C\\ &\Rightarrow \; y{e^x} = - {x^2} + C\\ &\Rightarrow \; y{e^x} + {x^2} = C\end{align}

Thus, the correct option is C.

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