Question

Solve the following Differential equation :
$\frac{dy}{dx}=(y-4x^2{)}^2$

Answer 1

Given,

$\frac{dy}{dx}=(y-4x^2{)}^2$

integrating on both sides, we have,

$\int \frac{dy}{dx}=\int (y-4x^2{)}^2$

$y+c=\int (y^2-8y{x}^2+16x^4)dx$

$y+c=y^{2}x-\frac{8y{x}^3}{3}+\frac{16x^5}{5}+c$

$y=y^{2}x-\frac{8y{x}^3}{3}+\frac{16x^5}{5}$