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प्रश्न
Solve the differential equation :
`y+x dy/dx=x−y dy/dx`
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उत्तर
`y+x dy/dx=x−y dy/dx`
`x dy/dx + y dy/dx=x−y`
`⇒dy/dx=(x−y)/(x+y) ` ......(1)
`Let F(x, y) =(x−y)/(x+y)`
`F(λx, λy) = λF(x, y)`
Therefore, F(x, y) is a homogeneous function of degree zero.
Let `y=vx`
`dy/dx=v+x (dv)/dx`
Substituting the value of y and dy/dx in (1) we get,
`v + x (dv)/dx=(x−vx)/(x+vx)=(1−v)/(1+v)`
`x (dv)/dx=(1−v)/(1+v)−v=(1−v−v^2−v)/(1+v)=(1−2v−v^2)/(1+v)`
`(1+v)/(v^2+2v−1)dv=−dx/x`
Integrating both sides, we have
`1/2 log∣(y^2/x^2)+(2y)/x−1∣+log|x|=logc`
`⇒log∣(y^2/x^2)+(2y)/x−1∣+2log|x|=2logc`
`⇒log((y^2/x^2)+(2y)/x−1)(x^2)=logc^2`
`⇒((y^2+2yx−x^2)/x^2)(x^2) = c^2`
`⇒y^2+2yx−x^2=C (where C=c^2)`
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