Question

Find the solution to the initial value problem that is a non-zero polynomial function in x. $xy'-15y=(4 x^{2} -3x+7) y^{ \frac{4}{5} }$, y(1)=0

Answer 1

Bernoulli type.

$xy'-15y=(4x^2-3x+7_y^{4/5}$
$xy^{-4/5}y'-15y^{1/5}=4x^2-3x+7$

Let $z=y^{1/5}$ so that $z'=\dfrac15y^{-4/5}$. Then the ODE becomes linear in $z$ with

$5xz'-15z=4x^2-3x+7$
$z'-\dfrac3xz=\dfrac45x-\dfrac35+\dfrac7{5x}$
$\dfrac1{x^3}z'-\dfrac3{x^4}z=\frac4{5x^2}-\dfrac3{5x^3}+\dfrac7{5x^4}$
$\left(\dfrac1{x^3}z\right)'=\frac4{5x^2}-\dfrac3{5x^3}+\dfrac7{5x^4}$
$\dfrac1{x^3}z=-\dfrac4{5x}+\dfrac3{10x^2}-\dfrac7{15x^3}+C$
$z=Cx^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}$

$y^{1/5}=Cx^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}$
$y=\left(Cx^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}\right)^5$

Given that $y(1)=0$, we have

$0=\left(C-\dfrac45+\dfrac3{10}-\dfrac7{15}\right)^5$
$\implies C=\dfrac{29}{30}$

and so the particular solution is

$y=\left(\dfrac{29}{30}x^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}\right)^5$

Feel free to expand the solution to get it in the standard polynomial form.