JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 12)
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 :
$\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is :
$\log _e|x+y|+\frac{x y}{(x+y)^2}=0$
$\log _e|x+y|-\frac{x y}{(x+y)^2}=0$
$\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0$
$\log _e|x+y|-\frac{2 x y}{(x+y)^2}=0$
Explanation
$$y = vx$$
$$v + x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}}} \right)$$
$$x{{dv} \over {dx}} = - \left( {{{1 + 3{v^2}} \over {3 + {v^2}}} + v} \right)$$
$${{dv} \over {dx}} = - \left( {{{{{(1 + v)}^3}} \over {3 + {v^2}}}} \right)$$
$$ \Rightarrow {{3 + {v^2}} \over {{{(1 + v)}^3}}} = {{ - dx} \over x}$$
$$ \Rightarrow \ln |v + 1| + {{2v} \over {{{(v + 1)}^2}}} = C - \ln |x|$$
$$x = 1,v = 0 \Rightarrow C = 0$$
$$ \Rightarrow \ln |x + y| - \ln |x| + {{2xy} \over {{{(x + y)}^2}}} = - \ln |x|$$
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