Question

Use an appropriate substitution to solve the equation


`y'- (8)/(x) y= \frac{ y^(5) } { x^(20) }` and find the solution that satisfies y(1)=1.

Answer 1

`y'-\frac8xy=(y^5)/(x^(20))`

This ODE is in standard Bernoulli form, so we can divide through both sides by
`y^5` to get


`y^(-5)y'-\frac8xy^(-4)=\frac1{x^(20)}`

Substitute
`z=y^(-4)`, so that
`z'=-4y^(-5)y'`. Then the ODE can be written as a linear one in
`z`:


`-\frac14z'-\frac8xz=\frac1{x^(20)}`

`z'+\frac{32}xz=-\frac4{x^(20)}`

Multiplying both sides by
`x^(32)` gives


`x^(32)z'+32x^(31)z=-4x^(12)`

`(x^(32)z)'=-4x^(12)`

`x^(32)z=\displaystyle-4\int x^(12)\,\mathrm dx`

`x^(32)z=-\frac4{13}x^(13)+C`

`z=-\frac4{13x^(19)}+\frac C{x^(32)}`

Back-substitute to solve for
`y`:


`\frac1{y^4}=-\frac4{13x^(19)}+\frac C{x^(32)}`

`y^4=\frac1{Cx^(-32)-\frac4{13}x^(-19)}`

`y=\left(\frac{x^(32)}{C-\frac4{13}x^(13)}\right)^(1/4)`

`y=\frac{x^8}{(C-\frac4{13}x^(13))^(1/4)}`

Given that
`y(1)=1`, you have


`1=\frac{1^8}{(C-\frac4{13}1^(13))^(1/4)}`

`1=\frac1{(C-\frac4{13})^(1/4)}`

`1=\left(C-\frac4{13}\right)^(1/4)`

`1=C-\frac4{13}`

`C=(17)/(13)`

so that the particular solution is


`y=\frac{x^8}{((17)/(13)-\frac4{13}x^(13))^(1/4)}`