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Q.
The solution of the equation $\frac{d y}{d x}+x(x+y)=$ $x^{3}(x+y)^{3}-1$ is
Differential Equations
Solution:
The given equation can be written as
$\left(\frac{d y}{d x}+1\right)+x(x+y)=x^{3}(x+y)^{3} $
$\Rightarrow \frac{d(x+y)}{d x}+x(x+y)=x^{3}(x+y)^{3}$
$\Rightarrow (x+y)^{3} \frac{d(x+y)}{d x}+x(x+y)^{-2}=x^{3}$
Let $(x+y)^{-2}=z$ so that
$-2(x+y)^{-3} \frac{d(x+y)}{d x}=\frac{d z}{d x}$
The given equation reduces to
$\frac{-1}{2} \frac{d z}{d x}+x z=x^{3}$
i.e., $ \frac{d z}{d x}-2 x z=-2 x^{3} $
I.F. $=e^{\int-2 x d x}=e^{-x 2}$
$\therefore $ The solution is,
$z \cdot e^{-x^{2}}=\int-2 x^{3} \cdot e^{-x^{2}} d x=\left(x^{2}+1\right) e^{-x^{2}}+c$
or, $\frac{1}{(x+y)^{2}}=c e^{x^{2}}+x^{2}+1$