Question

Find the particular solution of the differential equation :$y{e}^{y}dx=(y^3+2x{e}^y)dy,y\left(0\right)=1.$

Answer 1

We've $y{e}^{y}dx=(y^3+2x{e}^y)dy,y\left(0\right)=1\Rightarrow \frac{dx}{dy}=\frac{y^3+2x{e}^y}{y{e}^y}\Rightarrow \frac{dx}{dy}+(-\frac{2}{y})x=\frac{y^2}{e^y}.$
As this diff. eq. is in the form $\frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)$ so, $P\left(y\right)=-\frac{2}{y},Q\left(y\right)=\frac{y^2}{e^y}.$
Now I.F. $=e^{\int -\frac{2}{y}dy}=e^{-2logy}=\frac{1}{y^2}.$
Hence solution is, $x\left(\frac{1}{y^2}\right)=\int \frac{1}{y^2}\times \frac{y^2}{e^y}dy+C\Rightarrow \frac{x}{y^2}=-\frac{1}{e^y}+C.$
As y(0) = 1 so, $\frac{0}{1^2}=-\frac{1}{e^1}+C\Rightarrow C=\frac{1}{e}.$
Hence the required particular solution is $\frac{x}{y^2}=-\frac{1}{e^y}+\frac{1}{e}$ or, $x=y^2(e^{-1}-e^{-y})$