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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of $y^2dx+(x^2-xy+y^2)dy=0$

$\begin{array}{1 1}(A)\;\tan ^{-1}\bigg(\large\frac{y}{x}\bigg)+\log y=c \\ (B)\;\tan ^{-1}\bigg(\large\frac{x}{y}\bigg)-\log y=c \\(C)\;\tan ^{-1}\bigg(\large\frac{x}{y}\bigg)+\log y=c \\(D)\;\tan ^{-1}\bigg(\large\frac{y}{x}\bigg)-\log y=c \end{array} $

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Toolbox:
  • A linear differential equation of the form.$ \large\frac{dx}{dy}$$=f(y,x)$ can be solved by substituting $x=vy$ and $\large\frac{dx}{dy}$$=v+y \large\frac{dv}{dy}.$
  • $ \int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}\bigg(\large\frac{x}{a}\bigg)+c$
Step 1:
Given $y^2dx+(x^2-xy+y^2)dy=0$
This can be written as
$\large\frac{dx}{dy}=\frac{-(x^2-xy+y^2)}{y^2}$
Dividing by $y^2$ we get,
$\large\frac{dx}{dy}=\frac{-x^2}{y^2}+\frac{x}{y}$$-1$
This is clearly a homologous differential equation we can solve by substituting $x=vy$ and $\large\frac{dy}{dx}$$=v+y\large\frac{dv}{dy}$
$v+y \large\frac{dv}{dy}=\frac{-v^2y^2}{y^2}+\frac{vy}{y}-1$
=>$y.\large\frac{dv}{dy}$$=-v^2+v-1-v$
=>$y.\large\frac{dv}{dy}$$=-(v^2+1)$
Now seperating the variables we get,
$\large\frac{dv}{v^2+1}=-\frac{dy}{y}$
Step 2:
Now integrating on both sides we get,
$\int\large\frac{dv}{v^2+1}=-\int\frac{dy}{y}$
$\int\large\frac{dv}{v^2+1}$ is of the form $ \int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}\bigg(\large\frac{x}{a}\bigg)+c$
Hence $\tan ^{-1}(v)=-\log y+c$
Substituting for $v=x/y$ we get,
$\tan ^{-1}\bigg(\large\frac{x}{y}\bigg)$$=-\log y+c$
Hence the required solution is $\tan ^{-1}\bigg(\large\frac{x}{y}\bigg)$$+\log y=c$

 

answered May 9, 2013 by meena.p
 
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