NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5

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Chapter 9 Ex.9.5 Question 1

Show that the given differential equation is homogeneous and solve them.

$$\left( {{x^2} + xy} \right)dy = \left( {{x^2} + {y^2}} \right)dx$$

Solution

$$\left( {{x^2} + xy} \right)dy = \left( {{x^2} + {y^2}} \right)dx$$ can be written as:

\begin{align}\frac{{dy}}{{dx}} &= \frac{{{x^2} + {y^2}}}{{{x^2} + xy}}\\F\left( {x,y} \right) &= \frac{{{x^2} + {y^2}}}{{{x^2} + xy}}\\F\left( {\lambda x,\lambda y} \right) &= \frac{{{{\left( {\lambda x} \right)}^2} + {{\left( {\lambda y} \right)}^2}}}{{{{\left( {\lambda x} \right)}^2} + \left( {\lambda x} \right)\left( {\lambda y} \right)}} = \frac{{{x^2} + {y^2}}}{{{x^2} + xy}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Equation is a homogeneous equation.

Let $$y = vx$$

\begin{align}&\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{{x^2} + {{\left( {vx} \right)}^2}}}{{{x^2} + x\left( {vx} \right)}}\\& \Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{1 + v}}\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{1 + v}} - v = \frac{{\left( {1 + {v^2}} \right) - v\left( {1 + v} \right)}}{{1 + v}}\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{1 - v}}{{1 + v}}\\& \Rightarrow \; \left( {\frac{{1 + v}}{{1 - v}}} \right)dv = \frac{{dx}}{x}\\& \Rightarrow \; \left( {\frac{{2 - 1 + v}}{{1 - v}}} \right)dv = \frac{{dx}}{x}\\& \Rightarrow \; \left( {\frac{2}{{1 - v}} - 1} \right)dv = \frac{{dx}}{x}\\ &\Rightarrow \; - 2\log \left( {1 - v} \right) - v = \log x - \log C\\& \Rightarrow \; v = - 2\log \left( {1 - v} \right) - \log x + \log C\\ &\Rightarrow \; v = \log \left[ {\frac{C}{{x{{\left( {1 - v} \right)}^2}}}} \right]\\ &\Rightarrow \; \frac{y}{x} = \log \left[ {\frac{C}{{x{{\left( {1 - \frac{y}{x}} \right)}^2}}}} \right]\\& \Rightarrow \; \frac{y}{x} = \log \left[ {\frac{{Cx}}{{{{\left( {x - y} \right)}^2}}}} \right]\\& \Rightarrow \; \frac{{Cx}}{{{{\left( {x - y} \right)}^2}}} = {e^{\frac{y}{x}}}\\& \Rightarrow \; {\left( {x - y} \right)^2} = Cx{e^{\frac{{ - y}}{x}}}\end{align}

Chapter 9 Ex.9.5 Question 2

Show that the given differential equation is homogeneous and solve them.

$$y' = \frac{{x + y}}{x}$$

Solution

\begin{align}&y' = \frac{{x + y}}{x}\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{{x + y}}{x}\\F\left( {x,y} \right) &= \frac{{x + y}}{x}\\F\left( {\lambda x,\lambda y} \right) &= \frac{{\lambda x + \lambda y}}{{\lambda x}} = \frac{{x + y}}{x} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Equation is a homogeneous equation.

\begin{align}{\text{Let }}&y = vx\\&\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{x + vx}}{x}\\& \Rightarrow \; v + x\frac{{dv}}{{dx}} = 1 + v\\x\frac{{dv}}{{dx}} &= 1\\ &\Rightarrow \; dv = \frac{{dx}}{x}\\& \Rightarrow \; \int {dv} = \int {\frac{{dx}}{x}} \\&v = \log \left| x \right| + C\\ &\Rightarrow \; \frac{y}{x} = \log \left| x \right| + C\\ &\Rightarrow \; y = x\log \left| x \right| + Cx\end{align}

Chapter 9 Ex.9.5 Question 3

Show that the given differential equation is homogeneous and solve them.

$$\left( {x - y} \right)dy - \left( {x + y} \right)dx = 0$$

Solution

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

\begin{align}{\text{Let }}F\left( {x,y} \right) &= \frac{{x + y}}{{x - y}}\\\therefore F\left( {\lambda x,\lambda y} \right) &= \frac{{\lambda x + \lambda y}}{{\lambda x - \lambda y}} = \frac{{x + y}}{{x - y}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {vx} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{x + vx}}{{x - vx}} = \frac{{1 + v}}{{1 - v}}\\&x\frac{{dv}}{{dx}} = \frac{{1 + v}}{{1 - v}} - v = \frac{{1 + v - v\left( {1 - v} \right)}}{{1 - v}}\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{1 - v}}\\ &\Rightarrow \; \frac{{1 - v}}{{\left( {1 + {v^2}} \right)}}dv = \frac{{dx}}{x}\\& \Rightarrow \; \left( {\frac{1}{{1 + {v^2}}} - \frac{v}{{1 + {v^2}}}} \right)dv = \frac{{dx}}{x}\\&{\tan ^{ - 1}}v - \frac{1}{2}\log \left( {1 + {v^2}} \right) = \log x + C\\& \Rightarrow \; {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) - \frac{1}{2}\log \left[ {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right] = \log x + C\\ &\Rightarrow \; {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) - \frac{1}{2}\log \left[ {\frac{{{x^2} + {y^2}}}{{{x^2}}}} \right] = \log x + C\\ &\Rightarrow \; {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) - \frac{1}{2}\left[ {\log \left( {{x^2} + {y^2}} \right) - \log {x^2}} \right] = \log x + C\\& \Rightarrow \; {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = \frac{1}{2}\log \left( {{x^2} + {y^2}} \right) + C\end{align}

Chapter 9 Ex.9.5 Question 4

Show that the given differential equation is homogeneous and solve them.

$$\left( {{x^2} - {y^2}} \right)dx + 2xydy = 0$$

Solution

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

\begin{align}{\text{Let}}\;F\left( {x,y} \right) &= - \frac{{\left( {{x^2} - {y^2}} \right)}}{{2xy}}\\\therefore F\left( {\lambda x,\lambda y} \right) &= \left[ {\frac{{{{\left( {\lambda x} \right)}^2} - {{\left( {\lambda y} \right)}^2}}}{{2\left( {\lambda x} \right)\left( {\lambda y} \right)}}} \right] = \frac{{ - \left( {{x^2} - {y^2}} \right)}}{{2xy}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {vx} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = - \left[ {\frac{{{x^2} - {{\left( {vx} \right)}^2}}}{{2x.\left( {vx} \right)}}} \right]\\&v + x\frac{{dv}}{{dx}} = \frac{{{v^2} - 1}}{{2v}}\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{{v^2} - 1}}{{2v}} - v = \frac{{{v^2} - 1 - 2{v^2}}}{{2v}}\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{ - \left( {1 + {v^2}} \right)}}{{2v}}\\ &\Rightarrow \; \frac{{2v}}{{1 + {v^2}}}dv = - \frac{{dx}}{x}\\ &\Rightarrow \; \log \left( {1 + {v^2}} \right) = - \log x + \log C = \log \frac{C}{x}\\ &\Rightarrow \; 1 + {v^2} = \frac{C}{x}\\& \Rightarrow \; \left[ {1 + \frac{{{y^2}}}{{{x^2}}}} \right] = \frac{C}{x}\\ &\Rightarrow \; {x^2} + {y^2} = Cx\end{align}

Chapter 9 Ex.9.5 Question 5

Show that the given differential equation is homogeneous and solve them.

$${x^2}\frac{{dy}}{{dx}} = {x^2} - 2{y^2} + xy$$

Solution

\begin{align}{x^2}\frac{{dy}}{{dx}} &= {x^2} - 2{y^2} + xy\\\frac{{dy}}{{dx}} &= \frac{{{x^2} - 2{y^2} + xy}}{{{x^2}}}\end{align}

\begin{align}{\text{Let }}F\left( {x,y} \right) &= \frac{{{x^2} - 2{y^2} + xy}}{{{x^2}}}\\\therefore F\left( {\lambda x,\lambda y} \right)& = \frac{{{{\left( {\lambda x} \right)}^2} - 2{{\left( {\lambda y} \right)}^2} + \left( {\lambda x} \right)\left( {\lambda y} \right)}}{{{{\left( {\lambda x} \right)}^2}}} = \frac{{{x^2} - 2{y^2} + xy}}{{{x^2}}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\& \Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{{x^2} - 2{{\left( {vx} \right)}^2} + x.\left( {vx} \right)}}{{{x^2}}}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = 1 - 2{v^2} + v\\& \Rightarrow \; x\frac{{dv}}{{dx}} = 1 - 2{v^2}\\ &\Rightarrow \; \frac{{dv}}{{1 - 2{v^2}}} = \frac{{dx}}{x}\\& \Rightarrow \; \frac{{dv}}{{2\left( {\frac{1}{2} - {v^2}} \right)}} = \frac{{dx}}{x}\\& \Rightarrow \; \frac{1}{2}\left[ {\frac{{dv}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} - {v^2}}}} \right] = \frac{{dx}}{x}\\&\frac{1}{2}\frac{1}{{2 \times \frac{1}{{\sqrt 2 }}}}\log \left| {\frac{{\frac{1}{{\sqrt 2 }} + v}}{{\frac{1}{{\sqrt 2 }} - v}}} \right| = \log \left| x \right| + C\\ &\Rightarrow \; \frac{1}{{2\sqrt 2 }}\log \left| {\frac{{\frac{1}{{\sqrt 2 }} + \frac{y}{x}}}{{\frac{1}{{\sqrt 2 }} - \frac{y}{x}}}} \right| = \log \left| x \right| + C\\ &\Rightarrow \; \frac{1}{{2\sqrt 2 }}\log \left| {\frac{{x + \sqrt 2 y}}{{x - \sqrt 2 y}}} \right| = \log \left| x \right| + C\end{align}

Chapter 9 Ex.9.5 Question 6

Show that the given differential equation is homogeneous and solve them.

$$xdy - ydx = \sqrt {{x^2} + {y^2}} dx$$

Solution

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

\begin{align}{\text{Let, }}F\left( {x,y} \right) &= \frac{{y + \sqrt {{x^2} + {y^2}} }}{x}\\\therefore F\left( {\lambda x,\lambda y} \right) &= \frac{{\left( {\lambda x} \right) + \sqrt {{{\left( {\lambda x} \right)}^2}{{\left( {\lambda y} \right)}^2}} }}{{\lambda x}} = \frac{{y + \sqrt {{x^2} + {y^2}} }}{x} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {vx} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{vx + \sqrt {{x^2} + {{\left( {vx} \right)}^2}} }}{x}\\& \Rightarrow \; v + x\frac{{dv}}{{dx}} = v + \sqrt {1 + {v^2}} \\& \Rightarrow \; \frac{{dv}}{{\sqrt {1 + {v^2}} }} = \frac{{dx}}{x}\\&\log \left| {v + \sqrt {1 + {v^2}} } \right| = \log \left| x \right| + \log C\\& \Rightarrow \; \log \left| {\frac{y}{x} + \sqrt {1 + \frac{{{y^2}}}{{{x^2}}}} } \right| = \log \left| {Cx} \right|\\& \Rightarrow \; \log \left| {\frac{{y + \sqrt {{x^2} + {y^2}} }}{x}} \right| = \log \left| {Cx} \right|\\ &\Rightarrow \; y + \sqrt {{x^2} + {y^2}} = C{x^2}\end{align}

Chapter 9 Ex.9.5 Question 7

Show that the given differential equation is homogeneous and solve them.

$$\left\{ {x\cos \left( {\frac{y}{x}} \right) + y\sin \left( {\frac{y}{x}} \right)} \right\}ydx = \left\{ {y\sin \left( {\frac{y}{x}} \right) - x\cos \left( {\frac{y}{x}} \right)} \right\}xdy$$

Solution

\begin{align}\left\{ {x\cos \left( {\frac{y}{x}} \right) + y\sin \left( {\frac{y}{x}} \right)} \right\}ydx &= \left\{ {y\sin \left( {\frac{y}{x}} \right) - x\cos \left( {\frac{y}{x}} \right)} \right\}xdy\\\frac{{dy}}{{dx}}& = \frac{{\left\{ {x\cos \left( {\frac{y}{x}} \right) + y\sin \left( {\frac{y}{x}} \right)} \right\}y}}{{\left\{ {y\sin \left( {\frac{y}{x}} \right) - x\cos \left( {\frac{y}{x}} \right)} \right\}x}}\end{align}

\begin{align}{\text{Let }}F\left( {x,y} \right) &= \frac{{\left\{ {x\cos \left( {\frac{y}{x}} \right) + y\sin \left( {\frac{y}{x}} \right)} \right\}y}}{{\left\{ {y\sin \left( {\frac{y}{x}} \right) - x\cos \left( {\frac{y}{x}} \right)} \right\}x}}\\\therefore F\left( {\lambda x,\lambda y} \right) &= \frac{{\left\{ {\lambda x\cos \left( {\frac{{\lambda y}}{{\lambda x}}} \right) + \lambda y\sin \left( {\frac{{\lambda y}}{{\lambda x}}} \right)} \right\}\lambda y}}{{\left\{ {\lambda y\sin \left( {\frac{{\lambda y}}{{\lambda x}}} \right) - \lambda x\cos \left( {\frac{{\lambda y}}{{\lambda x}}} \right)} \right\}\lambda x}} = \frac{{\left\{ {x\cos \left( {\frac{y}{x}} \right) + y\sin \left( {\frac{y}{x}} \right)} \right\}y}}{{\left\{ {y\sin \left( {\frac{y}{x}} \right) - x\cos \left( {\frac{y}{x}} \right)} \right\}x}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\& \Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{\left( {x\cos v + vx\sin v} \right).vx}}{{\left( {vx\sin v - x\cos v} \right)x}}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{v\cos v + {v^2}\sin v}}{{v\sin v - \cos v}}\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{v\cos v + {v^2}\sin v}}{{v\sin v - \cos v}} - v\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{v\cos v + {v^2}\sin v - {v^2}\sin v + v\cos v}}{{v\sin v - \cos v}}\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{2v\cos v}}{{v\sin v - \cos v}}\\ &\Rightarrow \; \left[ {\frac{{v\sin v - \cos v}}{{v\cos v}}} \right]dv = \frac{{2dx}}{x}\\ &\Rightarrow \; \left( {\tan v - \frac{1}{v}} \right)dv = \frac{{2dx}}{x}\\&\log \left( {\sec v} \right) - \log v = 2\log x + \log C\\ &\Rightarrow \; \log \left( {\frac{{\sec v}}{v}} \right) = \log \left( {C{x^2}} \right)\\& \Rightarrow \; \left( {\frac{{\sec v}}{v}} \right) = C{x^2}\\& \Rightarrow \; \sec v = C{x^2}v\\& \Rightarrow \; \sec \left( {\frac{y}{x}} \right) = C.{x^2}.\frac{y}{x}\\ &\Rightarrow \; \sec \left( {\frac{y}{x}} \right) = Cxy\\ &\Rightarrow \; \cos \left( {\frac{y}{x}} \right) = \frac{1}{{Cxy}} = \frac{1}{C}.\frac{1}{{xy}}\\ &\Rightarrow \; xy\cos \left( {\frac{y}{x}} \right) = k{\text{ }}\left( {k = \frac{1}{C}} \right)\end{align}

Chapter 9 Ex.9.5 Question 8

Show that the given differential equation is homogeneous and solve them.

$$x\frac{{dy}}{{dx}} - y + x\sin \left( {\frac{y}{x}} \right) = 0$$

Solution

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

\begin{align}&{\text{Let }}F\left( {x,y} \right) = \frac{{y - x\sin \left( {\frac{y}{x}} \right)}}{x}\\&\therefore F\left( {\lambda x,\lambda y} \right) = \frac{{\lambda y - \lambda x\sin \left( {\frac{{\lambda y}}{{\lambda x}}} \right)}}{{\lambda x}} = \frac{{y - x\sin \left( {\frac{y}{x}} \right)}}{x} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {vx} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{vx - x\sin v}}{x}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = v - \sin v\\ &\Rightarrow \; \frac{{dv}}{{\sin v}} = - \frac{{dx}}{x}\\& \Rightarrow \; \cos ecvdv = - \frac{{dx}}{x}\\&\log \left| {\cos ecv - \cot v} \right| = - \log x + \log C = \log \frac{C}{x}\\& \Rightarrow \; \cos ec\left( {\frac{y}{x}} \right) - \cot \left( {\frac{y}{x}} \right) = \frac{C}{x}\\ &\Rightarrow \; \frac{1}{{\sin \left( {\frac{y}{x}} \right)}} - \frac{{\cos \left( {\frac{y}{x}} \right)}}{{\sin \left( {\frac{y}{x}} \right)}} = \frac{C}{x}\\& \Rightarrow \; x\left[ {1 - \cos \left( {\frac{y}{x}} \right)} \right] = C\sin \left( {\frac{y}{x}} \right)\end{align}

Chapter 9 Ex.9.5 Question 9

Show that the given differential equation is homogeneous and solve them.

$$ydx + x\log \left( {\frac{y}{x}} \right)dy - 2xdy = 0$$

Solution

\begin{align}&ydx + x\log \left( {\frac{y}{x}} \right)dy - 2xdy = 0\\ &\Rightarrow \; ydx = \left[ {2x - x\log \left( {\frac{y}{x}} \right)} \right]dy\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{y}{{2x - x\log \left( {\frac{y}{x}} \right)}}\end{align}

\begin{align}{\text{Let }}F\left( {x,y} \right)& = \frac{y}{{2x - x\log \left( {\frac{y}{x}} \right)}}\\\therefore F\left( {\lambda x,\lambda y} \right) &= \frac{{\lambda y}}{{2\left( {\lambda x} \right) - \left( {\lambda x} \right)\log \left( {\frac{{\lambda y}}{{\lambda x}}} \right)}} = \frac{y}{{2x - x\log \left( {\frac{y}{x}} \right)}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {vx} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{vx}}{{2x - x\log v}}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{v}{{2 - \log v}}\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{v}{{2 - \log v}} - v\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{v - 2v + v\log v}}{{2 - \log v}}\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{v\log v - v}}{{2 - \log v}}\\ &\Rightarrow \; \frac{{2 - \log v}}{{v\left( {\log v - 1} \right)}}dv = \frac{{dx}}{x}\\& \Rightarrow \; \left[ {\frac{{1 + \left( {1 - \log v} \right)}}{{v\left( {\log v - 1} \right)}}} \right]dv = \frac{{dx}}{x}\\ &\Rightarrow \; \left[ {\frac{1}{{v\left( {\log v - 1} \right)}} - \frac{1}{v}} \right]dv = \frac{{dx}}{x}\\&\int {\frac{1}{{v\left( {\log v - 1} \right)}}} dv - \int {\frac{1}{v}} dv = \int {\frac{1}{x}} dx\\& \Rightarrow \; \int {\frac{{dv}}{{v\left( {\log v - 1} \right)}}} - \log v = \log x + \log C\\&{\text{Let, }}\log v - 1 = t\\& \Rightarrow \; \frac{d}{{dv}}\left( {\log v - 1} \right) = \frac{{dt}}{{dv}}\\ &\Rightarrow \; \frac{1}{v} = \frac{{dt}}{{dv}}\\ &\Rightarrow \; \frac{{dv}}{v} = dt\\ &\Rightarrow \; \int {\frac{{dt}}{t}} - \log v = \log x + \log C\\& \Rightarrow \; \log t - \log \left( {\frac{y}{x}} \right) = \log \left( {Cx} \right)\\& \Rightarrow \; \log \left[ {\log \left( {\frac{y}{x}} \right) - 1} \right] - \log \left( {\frac{y}{x}} \right) = \log \left( {Cx} \right)\\& \Rightarrow \; \log \left[ {\frac{{\log \left( {\frac{y}{x}} \right) - 1}}{{\frac{y}{x}}}} \right] = \log \left( {Cx} \right)\\& \Rightarrow \; \frac{x}{y}\left[ {\log \left( {\frac{y}{x}} \right) - 1} \right] = Cx\\& \Rightarrow \; \log \left( {\frac{y}{x}} \right) - 1 = Cy\end{align}

Required solution of the given differential equation.

Chapter 9 Ex.9.5 Question 10

Show that the given differential equation is homogeneous and solve them.

$$\left( {1 + {e^{\frac{x}{y}}}} \right)dx + {e^{\frac{x}{y}}}\left( {1 - \frac{x}{y}} \right)dy = 0$$

Solution

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

\begin{align}{\text{Let }}F\left( {x,y} \right)& = \frac{{ - {e^{\frac{x}{y}}}\left( {1 - \frac{x}{y}} \right)}}{{1 + {e^{\frac{x}{y}}}}}\\\therefore F\left( {\lambda x,\lambda y} \right)& = \frac{{ - {e^{\frac{{\lambda x}}{{\lambda y}}}}\left( {1 - \frac{{\lambda x}}{{\lambda y}}} \right)}}{{1 + {e^{\frac{{\lambda x}}{{\lambda y}}}}}} = \frac{{ - {e^{\frac{{\lambda x}}{{\lambda y}}}}\left( {1 - \frac{x}{y}} \right)}}{{1 + {e^{\frac{x}{\lambda }}}}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let}}\;x = vy\\ &\Rightarrow \; \frac{d}{{dy}}\left( x \right) = \frac{d}{{dy}}\left( {vy} \right)\\& \Rightarrow \; \frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}\\&v + y\frac{{dv}}{{dy}} = \frac{{ - {e^v}(1 - v)}}{{1 + {e^v}}}\\& \Rightarrow \; y\frac{{dv}}{{dy}} = \frac{{ - {e^v} + v{e^v}}}{{1 + {e^v}}} - v\\&\Rightarrow \; y\frac{{dv}}{{dy}} = \frac{{ - {e^v} + v{e^v} - v - v{e^v}}}{{1 + {e^v}}}\\ &\Rightarrow \; y\frac{{dv}}{{dy}} = - \left[ {\frac{{v + {e^v}}}{{1 + {e^v}}}} \right]\\ &\Rightarrow \; \left[ {\frac{{1 + {e^v}}}{{v + {e^v}}}} \right]dv = - \frac{{dy}}{y}\\& \Rightarrow \; \log \left( {v + {e^v}} \right) = - \log y + \log C\\ &\Rightarrow \; \log \left( {v + {e^v}} \right) = \log \left( {\frac{C}{y}} \right)\\ &\Rightarrow \; \left[ {\frac{x}{y} + {e^{\frac{x}{y}}}} \right] = \frac{C}{y}\\& \Rightarrow \; x + y{e^{\frac{x}{y}}} = C\end{align}

Chapter 9 Ex.9.5 Question 11

For the below differential equation find the particular solution satisfying the given condition:

$$\left( {x + y} \right)dy + \left( {x - y} \right)dx = 0;{\text{ }}y = 1{\text{ when }}x = 1$$

Solution

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

\begin{align}&{\text{Let }}F\left( {x,y} \right) = \frac{{ - \left( {x - y} \right)}}{{x + y}}\\&\therefore F\left( {\lambda x,\lambda y} \right) = \frac{{ - \left( {\lambda x - \lambda y} \right)}}{{\lambda x + \lambda y}} = \frac{{ - \left( {x - y} \right)}}{{x + y}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\& \Rightarrow \; \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {vx} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{ - \left( {x - vx} \right)}}{{x + vx}}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{v - 1}}{{v + 1}}\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{v - 1}}{{v + 1}} - v = \frac{{v - 1 - v\left( {v + 1} \right)}}{{v + 1}}\\& \Rightarrow \; x\frac{{dv}}{{dx}} = \frac{{v - 1 - {v^2} - v}}{{v + 1}} = \frac{{ - \left( {1 + {v^2}} \right)}}{{v + 1}}\\& \Rightarrow \; \frac{{v + 1}}{{1 + {v^2}}}dv = - \frac{{dx}}{x}\\ &\Rightarrow \; \left[ {\frac{v}{{1 + {v^2}}} + \frac{1}{{1 + {v^2}}}} \right]dv = - \frac{{dx}}{x}\\&\frac{1}{2}\log \left( {1 + {v^2}} \right) + {\tan ^{ - 1}}v = - \log x + k\\&\Rightarrow \; \log \left( {1 + {v^2}} \right) + 2{\tan ^{ - 1}}v = - 2\log x + 2k\\ &\Rightarrow \; \log \left[ {\left( {1 + {v^2}} \right){x^2}} \right] + 2{\tan ^{ - 1}}v = 2k\\& \Rightarrow \; \log \left[ {\left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right){x^2}} \right] + 2{\tan ^{ - 1}}\frac{y}{x} = 2k\\ &\Rightarrow \; \log \left( {{x^2} + {y^2}} \right) + 2{\tan ^{ - 1}}\frac{y}{x} = 2k\\&{\text{Now, }}y = 1{\text{ at }}x = 1\\& \Rightarrow \; \log 2 + 2{\tan ^{ - 1}}1 = 2k\\ &\Rightarrow \; \log 2 + 2 \times \frac{\pi }{4} = 2k\\& \Rightarrow \; \frac{\pi }{2} + \log 2 = 2k\\&\log \left( {{x^2} + {y^2}} \right) + 2{\tan ^{ - 1}}\frac{y}{x} = \frac{\pi }{2} + \log 2\end{align}

Chapter 9 Ex.9.5 Question 12

For the below differential equation find the particular solution satisfying the given condition:

$${x^2}dy + \left( {xy + {y^2}} \right)dx = 0;{\text{ }}y = 1{\text{ when }}x = 1$$

Solution

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

\begin{align}&F\left( {x,y} \right) = \frac{{ - \left( {xy + {y^2}} \right)}}{{{x^2}}}\\&\therefore F\left( {\lambda x,\lambda y} \right) = \frac{{\left[ {\lambda x.\lambda y + {{\left( {\lambda y} \right)}^2}} \right]}}{{{{\left( {\lambda x} \right)}^2}}} = \frac{{ - \left( {xy + {y^2}} \right)}}{{{{\left( {\lambda x} \right)}^2}}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&y = vx\\& \Rightarrow \; \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {vx} \right)\\ &\Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{ - \left[ {x.vx + {{\left( {vx} \right)}^2}} \right]}}{{{x^2}}} = - v - {v^2}\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = - {v^2} - 2v = - v\left( {v + 2} \right)\\& \Rightarrow \; \frac{{dv}}{{v\left( {v + 2} \right)}} = - \frac{{dx}}{x}\\ &\Rightarrow \; \frac{1}{2}\left[ {\frac{2}{{v\left( {v + 2} \right)}}} \right]dv = - \frac{{dx}}{x}\\& \Rightarrow \; \frac{1}{2}\left[ {\frac{{\left( {v + 2} \right) - v}}{{v\left( {v + 2} \right)}}} \right]dv = - \frac{{dx}}{x}\\& \Rightarrow \; \frac{1}{2}\left[ {\frac{1}{v} - \frac{1}{{v + 2}}} \right]dv = - \frac{{dx}}{x}\\& \Rightarrow \; \frac{1}{2}\left[ {\log v - \log \left( {v + 2} \right)} \right] = - \log x + \log C\\& \Rightarrow \; \frac{1}{2}\log \left( {\frac{v}{{v + 2}}} \right) = \log \frac{C}{x}\\ &\Rightarrow \; \frac{v}{{v + 2}} = {\left( {\frac{C}{x}} \right)^2}\\ &\Rightarrow \; \frac{{\frac{y}{x}}}{{\frac{y}{x} + 2}} = {\left( {\frac{C}{x}} \right)^2}\\ &\Rightarrow \; \frac{y}{{y + 2x}} = \frac{{{C^2}}}{{{x^2}}}\\ &\Rightarrow \; \frac{{{x^2}y}}{{y + 2x}} = {C^2}\\&y = 1{\text{ at }}x = 1\\ &\Rightarrow \; \frac{1}{{1 + 2}} = {C^2}\\& \Rightarrow \; {C^2} = \frac{1}{3}\\&\frac{{{x^2}y}}{{y + 2x}} = \frac{1}{3}\\ &\Rightarrow \; y + 2x = 3{x^2}y\end{align}

Chapter 9 Ex.9.5 Question 13

For the below differential equation find the particular solution satisfying the given condition:

\begin{align}\left[ {x\,{{\sin }^2}\left( {\frac{y}{x} - y} \right)} \right]dx + xdy = 0;{\text{ }}y = \frac{\pi }{4}{\text{ when }}x = 1\end{align}

Solution

\begin{align}&\left[ {x{{\sin }^2}\left( {\frac{y}{x} - y} \right)} \right]dx + xdy = 0\\& \Rightarrow \; \frac{{dy}}{{dx}} = \frac{{ - \left[ {x{{\sin }^2}\left( {\frac{y}{x}} \right) - y} \right]}}{x}\end{align}
\begin{align}&{\text{Let }}F\left( {x,y} \right) = \frac{{ - \left[ {x{{\sin }^2}\left( {\frac{y}{x}} \right) - y} \right]}}{x}\\&\therefore F\left( {\lambda x,\lambda y} \right) = \frac{{ - \left[ {\lambda x{{\sin }^2}\left( {\frac{{\lambda y}}{{\lambda x}}} \right) - \lambda y} \right]}}{x} = \frac{{ - \left[ {x{{\sin }^2}\left( {\frac{y}{x}} \right) - y} \right]}}{x} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\& \Rightarrow \; \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {vx} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = \frac{{ - \left[ {x{{\sin }^2}v - vx} \right]}}{x}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = - \left[ {{{\sin }^2}v - v} \right] = v - {\sin ^2}v\\ &\Rightarrow \; x\frac{{dv}}{{dx}} = - {\sin ^2}v\\ &\Rightarrow \; \frac{{dv}}{{{{\sin }^2}v}} = - \frac{{dx}}{x}\\ &\Rightarrow \; \cos e{c^2}dv = - \frac{{dx}}{x}\\ &\Rightarrow \; - \cot v = - \log \left| x \right| - \log C\\ &\Rightarrow \; \cot v = \log \left| x \right| + \log C\\ &\Rightarrow \; \cot \left( {\frac{y}{x}} \right) = \log \left| x \right| + \log C\\ &\Rightarrow \; \cot \left( {\frac{y}{x}} \right) = \log \left| {Cx} \right|\\&y = \frac{\pi }{4}{\text{ at }}x = 1\\ &\Rightarrow \; \cot \left( {\frac{\pi }{4}} \right) = \log \left| C \right|\\& \Rightarrow \; 1 = \log C\end{align}

\begin{align} &\Rightarrow \; C = {e^1} = e\\&\cot \left( {\frac{y}{x}} \right) = \log \left| {ex} \right|\end{align}

Chapter 9 Ex.9.5 Question 14

For the below differential equation find the particular solution satisfying the given condition:

$$\frac{{dy}}{{dx}} - \frac{y}{x} + {\text{cosec}}\left( {\frac{y}{x}} \right) = {\text{0; }}y = 0{\text{ when }}x = 1$$

Solution

\begin{align}&\frac{{dy}}{{dx}} - \frac{y}{x} + {\text{cosec}}\left( {\frac{y}{x}} \right) = {\text{0}}\\ &\Rightarrow \; \frac{{dy}}{{dx}} = \frac{y}{x} - \cos ec\left( {\frac{y}{x}} \right)\end{align}

\begin{align}&{\text{Let }}F\left( {x,y} \right) = \frac{y}{x} - \cos ec\left( {\frac{y}{x}} \right)\\&\therefore F\left( {\lambda x,\lambda y} \right) = \frac{{\lambda y}}{{\lambda x}} - \cos ec\left( {\frac{{\lambda y}}{{\lambda x}}} \right) = \frac{y}{x} - \cos ec\left( {\frac{y}{x}} \right) = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\& \Rightarrow \; \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {vx} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\&v + x\frac{{dv}}{{dx}} = v - \cos ecv\\ &\Rightarrow \; - \frac{{dv}}{{\cos ecv}} = \frac{{dx}}{x}\\& \Rightarrow \; - \sin vdv = \frac{{dx}}{x}\\& \Rightarrow \; \cos v = \log x + \log C = \log \left| {Cx} \right|\\ &\Rightarrow \; \cos \left( {\frac{y}{x}} \right) = \log \left| {Cx} \right|\end{align}

\begin{align}&y = 0{\text{ at }}x = 1\\& \Rightarrow \; \cos \left( 0 \right) = \log C\\& \Rightarrow \; C = {e^1} = e\\&\cos \left( {\frac{y}{x}} \right) = \log \left| {\left( {ex} \right)} \right|\end{align}

Required solution of the given differential equation.

Chapter 9 Ex.9.5 Question 15

For the below differential equation find the particular solution satisfying the given condition:

$$2xy + {y^2} - 2{x^2}\frac{{dy}}{{dx}} = 0;y = 2{\text{ when }}x = 1$$

Solution

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

\begin{align}&{\text{Let }}F\left( {x,y} \right) = \frac{{2xy + {y^2}}}{{2{x^2}}}\\&\therefore F\left( {\lambda x,\lambda y} \right) = \frac{{2\left( {\lambda x} \right)\left( {\lambda y} \right) + {{\left( {\lambda y} \right)}^2}}}{{2{{\left( {\lambda x} \right)}^2}}} = \frac{{2xy + {y^2}}}{{2{x^2}}} = {\lambda ^0}F\left( {x,y} \right)\end{align}

Given differential equation is a homogeneous equation.

\begin{align}&{\text{Let }}y = vx\\ &\Rightarrow \; \frac{d}{{dx}}\left( y \right) = \frac{d}{{dx}}\left( {vx} \right)\\& \Rightarrow \; \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\\& \Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{2x\left( {vx} \right) + {{\left( {vx} \right)}^2}}}{{2{x^2}}}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = \frac{{2v + {v^2}}}{2}\\ &\Rightarrow \; v + x\frac{{dv}}{{dx}} = v + \frac{{{v^2}}}{2}\\& \Rightarrow \; \frac{2}{{{v^2}}}dv = \frac{{dx}}{x}\\ &\Rightarrow \; 2.\frac{{{v^{ - 2 + 1}}}}{{ - 2 + 1}} = \log \left| x \right| + C\\ &\Rightarrow \; - \frac{2}{v} = \log \left| x \right| + C\\ &\Rightarrow \; - \frac{2}{{\left( {\frac{y}{x}} \right)}} = \log \left| x \right| + C\\ &\Rightarrow \; - \frac{{2x}}{y} = \log \left| x \right| + C\\&y = 2{\text{ at }}x = 1\\& \Rightarrow \; - 1 = \log \left( 1 \right) + C\\& \Rightarrow \; C = - 1\\ &\Rightarrow \; - \frac{{2x}}{y} = \log \left| x \right| - 1\\& \Rightarrow \; \frac{{2x}}{y} = 1 - \log \left| x \right|\\ &\Rightarrow \; y = \frac{{2x}}{{1 - \log \left| x \right|}},\left( {x \ne 0,x \ne e} \right)\end{align}

Chapter 9 Ex.9.5 Question 16

A homogeneous differential equation of the form $$\frac{{dx}}{{dy}} = h\left( {\frac{x}{y}} \right)$$ can be solved by making the substitution

(A) $$y = vx$$

(B) $$v = yx$$

(C) $$x = vy$$

(D) $$x = v$$

Solution

For solving homogeneous equation of form $$\frac{{dx}}{{dy}} = h\left( {\frac{x}{y}} \right)$$, we need to make substitution as $$x = vy$$.

Thus, the correct option is C.

Chapter 9 Ex.9.5 Question 17

Which of the following is a homogeneous differential equation?

(A) $$\left( {4x + 6y + 5} \right)dy - \left( {3y + 2x + 4} \right)dx = 0$$

(B) $$\left( {xy} \right)dx - \left( {{x^3} + {y^3}} \right)dy = 0$$

(C) $$\left( {{x^3} + 2{y^2}} \right)dx + 2xydy = 0$$

(D) $${y^2}dx + \left( {{x^2} - x{y^2} - {y^2}} \right)dy = 0$$

Solution

$$F\left( {x,y} \right)$$ is homogeneous function of degree $$n$$, if $$F\left( {\lambda x,\lambda v} \right) = \lambda 'F\left( {x,y} \right)$$for non-zero constant $$\left( \lambda \right)$$.

Consider equation given in D:

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

\begin{align}F\left( {\lambda x,\lambda y} \right)& = \frac{{{{\left( {\lambda y} \right)}^2}}}{{{{\left( {\lambda y} \right)}^2} + \left( {\lambda x} \right){{\left( {\lambda y} \right)}^2} - {{\left( {\lambda x} \right)}^2}}}\\& = \frac{{{\lambda ^2}{y^2}}}{{{\lambda ^2}\left( {{y^2} + x{y^2} - {x^2}} \right)}}\\ &= {\lambda ^0}\left( {\frac{{{y^2}}}{{{y^2} + x{y^2} - {x^2}}}} \right)\\ &= {\lambda ^0}F\left( {x,y} \right)\end{align}

Differential equation given in D is a homogeneous equation.

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