Question

$(1+xy)\frac{dy}{dx}+y^3=0y\left(0\right)=1$ Find k if the constant in the solution is $\frac{k}{e}$

Answer 1

$(1+xy)\frac{dy}{dx}+y^3=0\Rightarrow y^3\frac{dx}{dy}+xy=-1\Rightarrow \frac{dx}{dy}+x.\frac{1}{y^2}=-\frac{1}{y^3}$ ...(1)

Here $P=\frac{1}{y^2}\Rightarrow \int PdP=\int \frac{1}{y^2}dy=-\frac{1}{y}$

$\therefore I.F.=e^{\int pdy}=e^{-1/y}$

Multiplying (1) by $I.F.$ we get

$e^{-1/y}\frac{dx}{dy}+e^{-1/y}.x.\frac{1}{y^2}=-e^{-1/y}.\frac{1}{y^3}$

Integrating both sides, we get

$x.e^{-1/y}=\int e^{-1/y}.\left(\frac{1}{y^2}\right)(-\frac{1}{y})dy$

Putting $-\frac{1}{y}=t\Rightarrow \frac{1}{y^2}dy=dt$

$\therefore x{e}^t=\int t.e^t=(t-1)e^t+c$

Hence $x{e}^{-1/y}=(-\frac{1}{y}-1)e^{-1/y}+c$

As $y\left(0\right)=1$$\Rightarrow \frac{c}{e}=2$