Class 12 NCERT Solutions- Mathematics Part ii – Chapter 9– Differential Equations Exercise 9.5

In the article, we will solve Exercise 9.5 from Chapter 9, “Differential Equations” in the NCERT. Exercise 9.5 covers Homogeneous differential equations.

STEP 1: First of all, prove that the given differential equation is a Homogeneous differential equation.

STEP 2: Then put

y = vx

in the differential equation.

STEP 3: Then solve by using the variable separation method.

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

Q.1:

Solution:

We have

(x^2+xy)dy = (x^2+y^2)dx

This equation can be written as –

\frac{dy}{dx} = \frac{(x^2+y^2)}{(x^2+xy)}

Let

F(x,y) = \frac{(x^2+y^2)}{(x^2+xy)}

F(\lambda x,\lambda y) = \frac{(\lambda x)^2+(\lambda y)^2}{(\lambda x)^2+(\lambda x)(\lambda y)} = \frac{ x^2+ y^2}{ x^2+ xy} = \lambda ^oF( x,  y)

Thus, Equation is a homogeneous equation.

Let

y = vx

Differentiating both sides with respect to x we get:

\frac{dy}{dx} = v +x\frac{dv}{dx}

v +x\frac{dv}{dx} = \frac{x^2+(vx)^2}{x^2+x(vx)}

v +x\frac{dv}{dx} = \frac{1+v^2}{1+v}

x\frac{dv}{dx} = \frac{1+v^2}{1+v}-v= \frac{(1+v^2)-v(1+v)}{1+v}

(\frac{1+v}{1-v})dv= \frac{dx}{x}

(\frac{2-1+v}{1-v})dv= \frac{dx}{x}

(\frac{2}{1-v}-1)dv= \frac{dx}{x}

Integrating both sides:

-2log(1-v)- v  = logx - logc

v  = -2log(1-v)-logx +logc

v  = log[\frac{c}{x(1-v)^2}]

\frac{y}{x}  = log[\frac{c}{x(1-\frac{y}{x})^2}]

\frac{y}{x}  = log[\frac{cx}{(x-y)^2}]

[\frac{cx}{(x-y)^2}]=  e^\frac{y}{x}

Hence the required solution is

(x-y)^2 = cxe^\frac{-y}{x}

Q.2:

Solution:

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

can be written as

\frac{dy}{dx} = \frac{x+y}{x}

Let

F(x,y) = \frac{x+y}{x}

F( \lambda x,\lambda y) = \frac{ \lambda x+\lambda y}{ \lambda x} =  \frac{x+y}{x} = \lambda ^oF(x,y)

Thus, Equation is a homogeneous equation.

Let

y = vx

Differentiating both sides we get:

\frac{dy}{dx} = v +x\frac{dv}{dx}

v +x\frac{dv}{dx} = \frac{x+vx}{x}

v +x\frac{dv}{dx} = 1+v

x\frac{dv}{dx} = 1

dv = \frac{dx}{x}

On Integrating:

\int dv = \int \frac{dx}{x}

v= log|x|+c

\frac{y}{x}= log|x|+c

Hence the required solution is

y= xlog|x|+cx

Q.3:

Solution:

(x-y)dy - (x+y)dx = 0

can be written as

\frac{dy}{dx} = \frac{x+y}{x-y}

F(x,y) =\frac{x+y}{x-y}

F( \lambda x, \lambda y ) = \frac{\lambda x+\lambda y}{\lambda x-\lambda  y} = \frac{x+y}{x-y} = \lambda ^oF(x,y)

Thus, Equation is a homogeneous equation.

Let

y = vx

On Differentiating:

\frac{dy}{dx} = v +x\frac{dv}{dx}

From above, we have

\frac{dy}{dx} = \frac{x+y}{x-y} \ and \ y=vx

, Substituting these values:

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(1-v)}{1-v}

x\frac{dv}{dx} = \frac{1+v^2}{1-v}

( \frac{1-v}{1+v^2})dv = \frac{dx}{x}

( \frac{1}{1+v^2}- \frac{v}{1+v^2})dv = \frac{dx}{x}

\tan^{-1}v -\frac{1}{2}\log(1+v^2) = \log x +c

\tan^{-1}(\frac{y}{x}) -\frac{1}{2}\log[1+(\frac{y}{x})^2] = \log x +c

\tan^{-1}(\frac{y}{x}) -\frac{1}{2}\log[\frac{x^2+y^2}{x^2}] = \log x +c

\tan^{-1}(\frac{y}{x}) -\frac{1}{2}[\log({x^2+y^2})-\log{x^2}] = \log x +c

So,

\tan^{-1}(\frac{y}{x}) =\frac{1}{2}\log({x^2+y^2}) +c

Q.4:

Solution:

(x^2-y^2)dx+2xydy=0

can be written as

\frac{dy}{dx} = \frac{-(x^2-y^2)}{2xy}

Let

F(x,y) =  \frac{-(x^2-y^2)}{2xy}

F(\lambda x,\lambda y) = - [\frac{ (\lambda x)^2- (\lambda  y)^2}{2 (\lambda x )(\lambda  y)}] = -\frac{(x^2-y^2)}{2xy} = \lambda ^oF(x,y)

Thus, differential equation is a homogeneous equation.

Let

y = vx

\frac{dy}{dx} = v +x\frac{dv}{dx}

v +x\frac{dv}{dx} = -[\frac{x^2-(vx)^2}{2x(vx)}]

v +x\frac{dv}{dx} = \frac{v^2-1}{2v}

x\frac{dv}{dx} = \frac{v^2-1}{2v}-v =  \frac{v^2-1- 2v^2}{2v}

x\frac{dv}{dx} = \frac{-(1+v^2)}{2v}

\frac{2v}{1+v^2} =-\frac{dx}{x}

\log(1+v^2)=-\log x+\log c = \log(\frac{c}{x})

1+v^2 =\frac{c}{x}

1+(\frac{y}{x})^2 =\frac{c}{x}

x^2+y^2 = cx

Q.5:

Solution:

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

can be written as

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

F(x,y) =  \frac{x^2-2y^2+xy}{x^2}

F( \lambda x, \lambda y) =  \frac{(\lambda  x)^2-2 (\lambda  y)^2+ (\lambda x)( \lambda  y)}{ (\lambda  x)^2} = \frac{x^2-2y^2+xy}{x^2}= \lambda ^oF(x,y)

Thus, given differential equation is a homogeneous equation.

y = vx

\frac{dy}{dx} = v +x\frac{dv}{dx}

v +x\frac{dv}{dx} =  \frac{x^2-2(vx)^2+x(vx)}{x^2}

v +x\frac{dv}{dx} = 1-2v^2+v

x\frac{dv}{dx} = 1-2v^2

\frac{dv}{ 1-2v^2} =\frac{dx}{x}

\frac{dv}{ 2(\frac{1}{2}-v^2)} =\frac{dx}{x}

\frac{1}{2}[\frac{dv}{ (\frac{1}{\sqrt2})^2-v^2}] =\frac{dx}{x}

\frac{1}{2}\frac{1}{2\times\frac{1}{\sqrt2}}\log|\frac{\frac{1}{\sqrt2}+v}{\frac{1}{\sqrt2}-v}| = \log|x|+c

\frac{1}{2\sqrt2}\log|\frac{\frac{1}{\sqrt2}+\frac{y}{x}}{\frac{1}{\sqrt2}-\frac{y}{x}}| = \log|x|+c

\frac{1}{2\sqrt2}\log|\frac{x+\sqrt2y}{{x-\sqrt2y}}| = \log|x|+c

Q.6:

Solution:

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

cab be written as

\frac{dy}{dx} = \frac{y+\sqrt{x^2+y^2}}{x}

Let

F(x,y) =  \frac{y+\sqrt{x^2+y^2}}{x}

F(\lambda x,\lambda  y) =  \frac{(\lambda y)+\sqrt{(\lambda x)^2+(\lambda y)^2}}{\lambda  x} = \frac{y+\sqrt{x^2+y^2}}{x} = \lambda ^oF(x,y)

Thus, Given differential equation is a homogeneous equation.

Let

y = vx

\frac{dy}{dx} = v +x\frac{dv}{dx}

v+x\frac{dy}{dx} = \frac{vx+\sqrt{x^2+(vx)^2}}{x}

v+x\frac{dy}{dx} = v+\sqrt{1+v^2}

x\frac{dy}{dx} =\sqrt{1+v^2}

\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}

\log|v+\sqrt{1+v^2}| = \log|x|+\log c

\log|\frac{y}{x}+\sqrt{1+(\frac{y}{x})^2}| = \log|cx|

\log|\frac{y+\sqrt{x^2+y^2}}{x}|= \log|cx|

{y+\sqrt{x^2+y^2}}= cx^2

Q.7:

Solution:

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

can be written as

\frac{dy}{dx} = \frac{\{{x\cos (\frac{y}{x})+ y \sin(\frac{y}{x})}\}y}{ \{{y\sin (\frac{y}{x})-x \cos(\frac{y}{x})}\}x}

F(x,y)= \frac{\{{x\cos (\frac{y}{x})+ y \sin(\frac{y}{x})}\}y}{ \{{y\sin (\frac{y}{x})-x \cos(\frac{y}{x})}\}x}

F(\lambda  x,\lambda  y)= \frac{\{{\lambda  x\cos (\frac{\lambda  y}{\lambda x})+ \lambda  y \sin(\frac{\lambda y}{\lambda x})}\}\lambda  y}{ \{{\lambda  y\sin (\frac{\lambda y}{\lambda x})-\lambda  x \cos(\frac{\lambda y}{\lambda x})}\}\lambda  x} = \frac{\{{x\cos (\frac{y}{x})+ y \sin(\frac{y}{x})}\}y}{ \{{y\sin (\frac{y}{x})-x \cos(\frac{y}{x})}\}x} = \lambda ^oF(x,y)

Thus, Given differential equation is a homogeneous equation.

Let

y = vx

\frac{dy}{dx} = v +x\frac{dv}{dx}

v+x\frac{dv}{dx}=\frac{\{{x\cos v+ vx \sin v}\}vx}{ \{{vx\sin v-x \cos v}\}x}

v+x\frac{dv}{dx}=\frac{{v\cos v+ v^2 \sin v}}{ {v\sin v- \cos v}}

x\frac{dv}{dx}=\frac{{v\cos v+ v^2 \sin v}}{ {v\sin v- \cos v}}-v

x\frac{dv}{dx}=\frac{{v\cos v+ v^2 \sin v -{v^2\sin v+ v\cos v}}}{ {v\sin v- \cos v}}

x\frac{dv}{dx}=\frac{2v\cos v}{ {v\sin v- \cos v}}

[\frac{ {v\sin v- \cos v}}{v\cos v}]dv = \frac{2dx}{x}

[ \tan v-\frac{1}{v} ]dv = \frac{2dx}{x}

\log (\sec v) - \log v = 2\log x+\log c

\log (\frac{\sec v}{v}) =\log(cx^2)

(\frac{\sec v}{v}) =cx^2

{\sec v} =cx^2v

{\sec (\frac{y}{x})} =cx^2(\frac{y}{x})

{\sec (\frac{y}{x})} =cxy

\cos(\frac{y}{x}) = \frac{1}{cxy}

xy\cos(\frac{y}{x}) = k  , where \space k = \frac{1}{c}

Q.8:

Solution:

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

can be written as

\frac{dy}{dx} = \frac{y -x\sin(\frac{y}{x})}{x}

F(x,y)= \frac{y -x\sin(\frac{y}{x})}{x}

F(\lambda x,\lambda y)= \frac{\lambda y -\lambda x\sin(\frac{\lambda y}{\lambda x})}{\lambda  x}= \frac{y -x\sin(\frac{y}{x})}{x} =\lambda ^oF(x,y)

Thus, Given differential equation is a homogeneous equation.

Let

y = vx

\frac{dy}{dx} = v +x\frac{dv}{dx}

v +x\frac{dv}{dx} = \frac{vx - x\sin v}{x}

v +x\frac{dv}{dx} = v - \sin v

x\frac{dv}{dx} = - \sin v

\cosec vdv =-\frac{dx}{x}

\log|\cosec v-\cot v| = -\log x+ \log c = \log\frac{c}{x}

\cosec \frac{y}{x}-\cot  \frac{y}{x} =  \frac{c}{x}

\frac{1}{\sin (\frac{y}{x})}-\frac{\cos  (\frac{y}{x})}{\sin ( \frac{y}{x})} =  \frac{c}{x}

x[1-\cos  (\frac{y}{x})] = c\sin (\frac{y}{x})

Q.9:

Solution:

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

can be written as

\frac{dy}{dx} =  \frac{y}{2x-x\log(\frac{y}{x})}

Let

F(x,y) =  \frac{y}{2x-x\log(\frac{y}{x})}

F(\lambda x,\lambda y) =  \frac{\lambda y}{2(\lambda x)-(\lambda x)\log(\frac{\lambda y}{\lambda  x})} = \frac{y}{2x-x\log(\frac{y}{x})} = \lambda ^oF(x,y)

Thus, Given differential equation is a homogeneous equation

Let

y = vx

\frac{dy}{dx} = v +x\frac{dv}{dx}

v+x\frac{dv}{dx} =  \frac{vx}{2x-x\log(\frac{vx}{x})}

v+x\frac{dv}{dx} =  \frac{v}{2-\log v}

x\frac{dv}{dx} =  \frac{v}{2-\log v}-v

x\frac{dv}{dx} =  \frac{v-2v+v\log v}{2-\log v}

x\frac{dv}{dx} =  \frac{v\log v- v}{2-\log v}

\frac{2-\log v}{v(\log v- 1)}dv = \frac{dx}{x}

[ \frac{1+(1-\log v)}{v(\log v- 1)}]dv = \frac{dx}{x}

[ \frac{1}{v(\log v- 1)}- \frac{1}{v}]dv = \frac{dx}{x}

\int \frac{1}{v(\log v- 1)}-\int \frac{1}{v}dv = \int\frac{dx}{x}

\int \frac{1}{v(\log v- 1)}-\log v= \log x+\log c

Let

(\log v -1) = t

\frac{d}{dv}(\log v - 1) = \frac{dt}{dv}

\frac{1}{v} = \frac{dt}{dv}

\frac{dv}{v} = dt

So,

\int \frac{dt}{t}-\log v= \log x+\log c

\log t-\log v= \log x+\log c

\log [\log (\frac{y}{x})-1]-\log (\frac{y}{x}) = \log (cx)

\log [\frac{\log (\frac{y}{x})-1}{\frac{y}{x}}]-\log (\frac{y}{x}) = \log (cx)

\frac{x}{y} [{\log (\frac{y}{x})-1}] =  cx

{\log (\frac{y}{x})-1} =  cy

Q.10:

Solution:

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

can be written as

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

F(x,y) =  \frac{-e^\frac{x}{y}(1-\frac{x}{y})}{(1+e^\frac{x}{y})}

F(\lambda x,\lambda y) =  \frac{-e^\frac{\lambda x}{\lambda y}(1-\frac{\lambda x}{\lambda y})}{(1+e^\frac{\lambda x}{\lambda  y})} = \frac{-e^\frac{x}{y}(1-\frac{x}{y})}{(1+e^\frac{x}{y})} = \lambda ^oF(x,y)

Thus, Given differential equation is a homogeneous equation.

Let

x = vy

\frac{dx}{dy} = v +y\frac{dv}{dy}

v+y\frac{dv}{dy} = \frac{-e^v(1-v)}{(1+e^v)}

y\frac{dv}{dy} = \frac{-e^v+ve^v}{(1+e^v)}-v

y\frac{dv}{dy} = \frac{-e^v+ve^v-v-ve^v}{1+e^v}

y\frac{dv}{dy} = -[\frac{v+e^v}{1+e^v}]

[\frac{v+e^v}{1+e^v}]dv = -\frac{dy}{y}

\log(v+e^v) = \log( \frac{c}{y})

(\frac{x}{y}+e^\frac{x}{y}) = \frac{c}{y}

x+ye^\frac{x}{y} = c

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

Q.11:

Solution:

(x+y)dy+(x-y)dx = 0 ;

can be written as

\frac{dy}{dx}=\frac{-(x-y)}{x+y}

F(x,y) =\frac{-(x-y)}{x+y}

F(\lambda x,\lambda y) =\frac{-(\lambda x-\lambda y)}{\lambda x+\lambda  y} = \frac{-(x-y)}{x+y} = \lambda ^o F(x,y)

Thus, Given differential equation is a homogeneous equation.

Let

y = vx

\frac{dy}{dx} = v +x\frac{dv}{dx}

v +x\frac{dv}{dx} = \frac{-(x-vx)}{x+vx}

v +x\frac{dv}{dx} =  \frac{v-1}{v+1}

x\frac{dv}{dx} = \frac{v-1}{v+1}-v = \frac{v-1-v(v+1)}{v+1}

x\frac{dv}{dx}= \frac{v-1-v^2-v}{v+1} = \frac{-(1+v^2)}{v+1}

\frac{v+1}{1+v^2}dv = -\frac{dx}{x}

[  \frac{v}{1+v^2}+\frac{1}{1+v^2}]dv = -\frac{dx}{x}

\frac{1}{2}\log(1+v^2)+\tan^{-1}v = -\log x +k

\log(1+v^2)+2\tan^{-1}v = -2\log x +2k

\log[(1+v^2)x^2]+2\tan^{-1}v = 2k

\log[(1+(\frac{y}{x})^2)x^2]+2\tan^{-1}v = 2k

\log(x^2+y^2)+2\tan^{-1}\frac{y}{x} = 2k

Now, put

y = 1\space  and \space  x = 1

\log(2)+2\tan^{-1}1 = 2k

\log 2 + 2\times\frac{\pi}{4} = 2k

we get,

\log(x^2+y^2)+2\tan^{-1}\frac{y}{x} = \frac{\pi}{2}+\log2