The general solution of y² dx+ (x²-x y+y²) d y=0 is :

Correct Answer is : tan⁡−1(yx)=log⁡y+C^{-1} ≤ft( y/x ) = y + Ctan−1(xy)=logy+C

2 ^{-1} ≤ft( x/y ) + x + C = 0

≤ft( y + {x²+y² ) + y + C = 0

Correct Answer is : tan⁡−1(yx)=log⁡y+C^{-1} ≤ft( y/x ) = y + Ctan−1(xy)=logy+C

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EAMCET Engineering 2003 Solved Paper

Section: Mathematics

Question : 130 of 160

Marks: +1, -0

The general solution of

y^2 dx+ (x^2-x y+y^2) d y=0

Solution:

We have,

y^2 dx+ (x^2-x y+y^2) d y=0

\Rightarrow \frac{dy}{dx} = \frac{-y^2}{x^2-xy+y^2}

It is a homogeneous linear differential equation

\text{ Put } y = v x \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}

\therefore v + x \frac{dv}{dx} = \frac{-v^2 x^2}{x^2 - v x^2 + x^2 v^2} = \frac{-v^2}{v^2 - v + 1}

x \frac{dv}{dx} = \frac{-v^2 - v^3 + v^2 - v}{v^2 - v + 1} = \frac{-v^3 - v}{v^2 - v + 1}

\Rightarrow \frac{v^2 - v + 1}{-v^3 - v} dv = \frac{1}{x} dx

\Rightarrow \frac{-(v^2+1)+v}{v(v^2+1)} dv = \frac{1}{x} dx

\therefore -\frac{1}{v} dv + \frac{1}{v^2+1} dv = \frac{1}{x} dx

On integrating both sides, we get

\Rightarrow -\log v + \tan^{-1} v = \log x + C

\Rightarrow \tan^{-1} v = \log x v + C

\therefore \tan^{-1} \left( \frac{y}{x} \right) = \log y + C

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