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Q. If the solution of the differential equation $\frac{d y}{d x}=\frac{x^{3} + x y^{2}}{y^{3} - y x^{2}}$ is $y^{k}-x^{k}=2x^{2}y^{2}+\lambda $ (where, $\lambda $ is an arbitrary constant), then the value of $k$ is

NTA AbhyasNTA Abhyas 2022

Solution:

We have, $\frac{d y}{d x}=\frac{x^{3} + x y^{2}}{y^{3} - y x^{2}}$
Now, $\frac{y d y}{x d x}=\frac{x^{2} + y^{2}}{y^{2} - x^{2}}$
Let, $y^{2}=Y;x^{2}=X$
$\Rightarrow \frac{y d y}{x d x}=\frac{d Y}{d X}$
Hence, the equation is $\frac{d Y}{d X}=\frac{X + Y}{Y - X}$
$\Rightarrow YdY-XdX=XdY+YdX$
On integrating we get $\frac{Y^{2}}{2}-\frac{X^{2}}{2}=\displaystyle \int d \left(X Y\right) = X Y + c$
or $Y^{2}-X^{2}=2XY+\lambda $
$\Rightarrow y^{4}-x^{4}=2x^{2}y^{2}+\lambda $