The solution of the differential equation (d y)/(d x)=-((x^2+3 y^2)/(3 x^2+y^2)), y(1)=0 i…

Question

The solution of the differential equation

$\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is :

Accepted Answer

$y = vx$

$$v + x{{dv} \over {dx}} =  - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}}} \right)$$

$$x{{dv} \over {dx}} =  - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}} + v} \right)$$

$${{dv} \over {dx}} =  - \left( {{{{{(1 + v)}^3}} \over {3 + {v^2}}}} \right)$$

$$ \Rightarrow {{3 + {v^2}} \over {{{(1 + v)}^3}}} = {{ - dx} \over x}$$

$$ \Rightarrow \ln |v + 1| + {{2v} \over {{{(v + 1)}^2}}} = C - \ln |x|$$

$$x = 1,v = 0 \Rightarrow C = 0$$

$$ \Rightarrow \ln |x + y| - \ln |x| + {{2xy} \over {{{(x + y)}^2}}} =  - \ln |x|$$

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