In the article, we will solve Exercise 9.5 from Chapter 9, “Differential Equations” in the NCERT. Exercise 9.5 covers Homogeneous differential equations.
Basic Concept to Solve Exercise 9.5
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.
Exercise 9.5 Solution
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
Q.12:
Solution:
x^2dy+(xy+y^2)dx = 0 ; can be written as
\frac{dy}{dx} = \frac{-(xy+y^2)}{x^2}
F(x,y) = \frac{-(xy+y^2)}{x^2}
F(\lambda x,\lambda y) = \frac{-(\lambda x.\lambda y+(\lambda y)^2)}{(\lambda x)^2} = \frac{-(xy+y^2)}{x^2} = \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{-[x.vx+(vx)^2]}{x^2} = -v-v^2
x \frac{dy}{dx} = -v^2-2v = -v(v+2)
\frac{dv} {v(v+2)} = - \frac{dx}{x}
\frac{1}{2}[\frac{2} {v(v+2)}]dv = - \frac{dx}{x}
\frac{1}{2}[\frac{(v+2)-v} {v(v+2)}]dv = - \frac{dx}{x}
\frac{1}{2}[\frac{1}{v} -\frac{1}{v+2}]dv = - \frac{dx}{x}
\frac{1}{2}[\log{v} -\log{(v+2)}] = - \log{x} +\log c
\frac{1}{2}\log(\frac{v}{v+2}) = \log \frac{c}{x}
\frac{v}{v+2} = (\frac{c}{x})^2
\frac{\frac{y}{x}}{\frac{y}{x}+2} = (\frac{c}{x})^2
\frac{y}{y+2x} = \frac{c^2}{x^2}
\frac{x^2y}{y+2x} = {c^2} Now, put y = 1 and x = 1
\frac{1}{1+2} = {c^2}
{c^2} = \frac{1}{3}
\frac{x^2y}{y+2x} = \frac{1}{3}
{y+2x} = 3x^2y
Q.13:
Solution:
[x\sin^2(\frac{y}{x}-y)]dx+xdy = 0 ; can be written as
\frac{dy}{dx} = \frac{-[x\sin^2(\frac{y}{x})-y]}{x}
F(x,y)= \frac{-[x\sin^2(\frac{y}{x})-y]}{x}
F(\lambda x,\lambda y)= \frac{-[\lambda x\sin^2(\frac{\lambda y}{\lambda x})-\lambda y]}{\lambda x} = \frac{-[x\sin^2(\frac{y}{x})-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{-[x\sin^2v-vx]}{x}
v+x\frac{dv}{dx} = {-[\sin^2v-v]} = v-\sin^2v
x\frac{dv}{dx} = -\sin^2v
\frac{dv}{\sin^2v } = -\frac{dx}{x}
\cosec^2v {dv} = -\frac{dx}{x}
-\cot v = -\log|x| -\log c
\cot v = \log|x| +\log c
\cot (\frac{y}{x}) = \log|x| +\log c
\cot (\frac{y}{x}) = \log|cx| Now, put
y = \frac{\pi}{4} \space at \space x = 1
\cot (\frac{\pi}{4}) = \log|c|
\log c = 1 \space; \space i.e \space c = e
\cot (\frac{y}{x}) = \log|ex|
Q.14:
Solution:
\frac{dy}{dx} - \frac{y}{x}+\cosec{(\frac{y}{x})} = 0 ; can be written as
\frac{dy}{dx} = \frac{y}{x}-\cosec{(\frac{y}{x})}
F(x,y) = \frac{y}{x}-\cosec{(\frac{y}{x})}
F(\lambda x,\lambda y) = \frac{\lambda y}{\lambda x}-\cosec{(\frac{\lambda y}{\lambda x})} = \frac{ y}{ x}-\cosec{(\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} = v-\cosec v
-\frac{dv}{\cosec v} = \frac{dx}{x}
-\sin vdv = \frac{dx}{x}
\cos v = \log x +\log c = \log|cx|
\cos (\frac{y}{x}) = \log|cx|
y = 0 \space at \space x= 1
\cos (0) = \log c
c = e^1 = e
\cos (\frac{y}{x}) = \log|ex|
Q.15:
Solution:
2xy+y^2-2x^2\frac{dy}{dx} = 0 can be written as
\frac{dy}{dx} = \frac{2xy+y^2}{2x^2} Let
F(x,y) = \frac{2xy+y^2}{2x^2}
F(\lambda x,\lambda y) = \frac{2(\lambda x)(\lambda y)+(\lambda y)^2}{2(\lambda x)^2} = \frac{2xy+y^2}{2x^2} = \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{2x(vx)+(vx)^2}{2x^2}
v+x\frac{dy}{dx} = \frac{2v+v^2}{2}
v+x\frac{dy}{dx} = v+\frac{v^2}{2}
\frac{2}{v^2}dv = \frac{dx}{x}
2. \frac{v^{-2+1}}{-2+1} = \log|x|+c
-\frac{2}{v} = \log|x|+c
-\frac{2}{(\frac{y}{x})} = \log|x|+c
-\frac{2x}{y} = \log|x|+c
y = 1 \space at \space x= 1
-1 = \log(1)+c
c = -1
-\frac{2x}{y} = \log|x|-1
\frac{2x}{y} = 1-\log|x|
y = \frac{2x}{1-\log|x| } , \space (x \neq 0 , x\neq e)
Q.16: A homogeneous differential equation of the form
(A)
(B)
(C)
(D)
Solution:
For solving homogeneous equation of form
\frac{dx}{dy} = h(\frac{x}{y}) , we need to make substitution asx = vy .Thus, the correct option is C.
Q.17: Which of the following is a homogeneous differential equation?
(A)
(B)
(C)
(D)
Solution:
F(x,y) is homogeneous function of degree n , ifF(\lambda x,\lambda y ) = \lambda^{'} F(x,y) for non-zero constant\lambda .Consider equation given in D
y^2dx+(x^2-xy^2-y^2)dy = 0 which can be written as
\frac{dy}{dx} = \frac{-y^2}{x^2-xy^2-y^2} = \frac{y^2}{xy^2+y^2-x^2}
F(x,y)= \frac{y^2}{xy^2+y^2-x^2}
F(\lambda x,\lambda y)= \frac{(\lambda y)^2}{(\lambda x)(\lambda y)^2+(\lambda y)^2-(\lambda x)^2} = \frac{y^2}{xy^2+y^2-x^2} = F(x, y ) Thus, Differential equation given in D is a homogeneous equation.
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