Question

solve : $x\frac{dy}{dx}+y=x^2{y}^4$

Answer 1

Consider the given equation.

$x\frac{dy}{dx}+y=x^2{y}^4$

$\frac{dy}{dx}+\frac{y}{x}=x{y}^4$ …….. (1)

We know that the general equation,

$\frac{dy}{dx}+Py=Q$ ……… (2)

On comparing both equations, we get

$P=\frac{1}{x},Q=x{y}^4$

Therefore, the integrating factor

$I.F=e^{\int Pdx}$

$I.F=e^{\int \frac{1}{x}dx}$

$I.F=e^{\int \ln x}$

$I.F=x$

Therefore, the general solution

$y\times I.F=\int (x{y}^4\times I.F)dx+C$

$y\times x=\int (x{y}^4\times x)dx+C$

$yx=y^4\int x^{2}dx+C$

$yx=y^4\left(\frac{x^3}{3}\right)+C$

Hence, this is the answer.