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^{ (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'=\frac15y^(-4/5)`. Then the ODE becomes linear in
`z` with


`5xz'-15z=4x^2-3x+7`

`z'-\frac3xz=\frac45x-\frac35+\frac7{5x}`

`\frac1{x^3}z'-\frac3{x^4}z=\frac4{5x^2}-\frac3{5x^3}+\frac7{5x^4}`

`\left(\frac1{x^3}z\right)'=\frac4{5x^2}-\frac3{5x^3}+\frac7{5x^4}`

`\frac1{x^3}z=-\frac4{5x}+\frac3{10x^2}-\frac7{15x^3}+C`

`z=Cx^3-\frac45x^2+\frac3{10}x-\frac7{15}`


`y^(1/5)=Cx^3-\frac45x^2+\frac3{10}x-\frac7{15}`

`y=\left(Cx^3-\frac45x^2+\frac3{10}x-\frac7{15}\right)^5`

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


`0=\left(C-\frac45+\frac3{10}-\frac7{15}\right)^5`

`\implies C=(29)/(30)`

and so the particular solution is


`y=\left((29)/(30)x^3-\frac45x^2+\frac3{10}x-\frac7{15}\right)^5`

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