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Solve the differebtail equation of
(x-y)^2 dx + 2xy dy =0

Sonali , 6 Years ago
Grade 12th pass
anser 1 Answers
Vinod Ramakrishnan Eswaran

Last Activity: 6 Years ago

They have given the equation as (x-y)2dx+2xydy=0
Divide with x2dx on both sides we get
\frac{(x-y)^{2}}{x^{2}}+2\frac{y}{x}\frac{dy}{dx}=0
=>(1-\frac{y}{x})^{2}+2\frac{y}{x}\frac{dy}{dx}=0
 
Now substitute y/x as v
=> y=vx
Differentiating wrt x on both sides we get 
dy/dx=v+xdv/dx      (uv rule)
Substituting in earlier equation we get
(1-v)^{2}+2v(v+x\frac{dv}{dx})=0
1-2v+3v^{2}=-2vx\frac{dv}{dx}
Arranging them homogeneously we get
\frac{dx}{x}=\frac{-2vdv}{1-2v+3v^{2}}
Integrating on both sides we get
\int \frac{dx}{x}=\int \frac{-2vdv}{1-2v+3v^{2}}+c         (where c is integration constant)
 
log x=\frac{-1}{3}\int \frac{6vdv}{1-2v+3v^{2}}+c
log x=\frac{-1}{3}\int \frac{(6v-2)dv}{1-2v+3v^{2}}+\frac{-1}{3}\int \frac{2dv}{1-2v+3v^{2}}+c
 
log x=\frac{-1}{3}log(1-2v+3v^{2})+\frac{-2}{9}\int \frac{dv}{v^{2}-\frac{2v}{3}+\frac{1}{3}}+c
log x=\frac{-1}{3}log(1-2v+3v^{2})+\frac{-2}{9}\int \frac{dv}{(v^{2}-\frac{2v}{3}+\frac{1}{9})+(\frac{1}{3}-\frac{1}{9})}+c
log x=\frac{-1}{3}log(1-2v+3v^{2})+\frac{-2}{9}\int \frac{dv}{(v-\frac{1}{3})^{2}+(\frac{2}{9})}+c
log x=\frac{-1}{3}log(1-2v+3v^{2})+\frac{-2}{9}*\frac{3}{\sqrt{2}}*\tan(\frac{3v-1}{\sqrt{2}}) +c
Substituting v=y/x we get the required answer
log x=\frac{-1}{3}log(1-2\frac{y}{x}+3(\frac{y}{x})^{2})+\frac{-\sqrt{2}}{3}*\tan(\frac{3\frac{y}{x}-1}{\sqrt{2}}) +c
 
 
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