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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