Question

Solve the given initial-value problem. the de is a bernoulli equation. x2 dy dx − 2xy = 4y4, y(1) = 1 2

Answer 1

$x^2\dfrac{\mathrm dy}{\mathrm dx}-2xy=4y^4$
$x^2y^{-4}\dfrac{\mathrm dy}{\mathrm dx}-2xy^{-3}=4$

Substitute $z=y^{-3}$, so that $\dfrac{\mathrm dz}{\mathrm dx}=-3y^{-4}\dfrac{\mathrm dy}{\mathrm dx}$. Then the ODE in terms of $z(x)$ is


$-\dfrac13x^2\dfrac{\mathrm dz}{\mathrm dx}-2xz=4$
$x^2\dfrac{\mathrm dz}{\mathrm dx}+6xz=-12$
$x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-12x^4$

$\dfrac{\mathrm d}{\mathrm dx}\left[x^6z\right]=-12x^4$
$\implies x^6z=-\dfrac{12}5x^5+C$
$\implies z=-\dfrac{12}{5x}+\dfrac C{x^6}$
$\implies\dfrac1{y^3}=-\dfrac{12}{5x}+\dfrac C{x^6}$

and you can attempt to solve for $y$ explicitly from here if you wish.
In any case, given that $y(1)=\dfrac12$ (presumably that's what you meant to write), we'd get

$\dfrac1{\left(\frac12\right)^3}=-\dfrac{12}{5\cdot1}+\dfrac C{1^6}$
$8=-\dfrac{12}5+C\implies C=\dfrac{52}5$

giving a particular solution of

$\dfrac1{y^3}=-\dfrac{12}{5x}+\dfrac{52}{5x^6}$