Solution of the differential equation $\left(2x-10y^3\right)\frac{dy}{dx}+y=0$ is

Step1. Find the solution of the differential equation.

Given $\left(2x-10y^3\right)\frac{dy}{dx}+y=0$

The solution of the given differential equation $\frac{dy}{dx}+px=Q$ is given by:

$x\left(\mathrm{IF}\right)=\int Q\left(\mathrm{IF}\right)dy+C$

Where,$\mathrm{IF}=e^{\int pdy}$

Step 2. Reduce the equation in the form of $\frac{dy}{dx}+px=Q$

Given :$\left(2x-10y^3\right)\frac{dy}{dx}+y=0$

$\begin{array}{l}\left(2x-10y^3\right)\frac{dy}{dx}=-y\\ \Rightarrow \frac{dy}{dx}=\frac{-y}{\left(2x-10y^3\right)}\\ \Rightarrow \frac{dx}{dy}=\frac{2x-10y^3}{-y}\\ \Rightarrow \frac{dx}{dy}=-\frac{2x}{y}+10y^2\\ \Rightarrow \frac{dx}{dy}+\frac{2x}{y}=+10y^2\left[\because \frac{dy}{dx}+px=Q\right]\end{array}$

Here, $P=\frac{2}{y}$and $Q=10y^2$.

Step 3. Find the integrating factor (IF):

$\begin{array}{c}\mathrm{IF}=e^{\int \frac{2}{y}dy}\\ =e^{2\log y}\\ =e^{\log y^2}\\ =y^2\end{array}$

Step 4: Find the solution of the given differential equation.

$$\begin{gathered} x\left(\mathrm{IF}\right)=\int Q\left(\mathrm{IF}\right)dy+C \\ \begin{array}{l}\Rightarrow x\left(y^2\right)=\int 10y^2.y^{2}dy+C\\ \Rightarrow x{y}^2=\int 10y^{4}dy+C\\ \Rightarrow x{y}^2=10\int y^{4}dy+C\\ \Rightarrow x{y}^2=10y^5+C\end{array} \end{gathered}$$

Hence, the solution of the given differential equation is $x{y}^2=10y^5+C$.