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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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## 1 Answer

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• 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

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