Question

Solve the initial-value problem

y' = x^4 - \frac{1}{x}y, y(1) = 1.

Answer 1

The ODE is linear:

`y'=x^4-\frac yx`

`y'+\frac yx=x^4`

Multiplying both sides by
`x` gives

`xy'+y=x^5`

Notice that the left side can be condensed as the derivative of a product:

`(xy)'=x^5`

Integrating both sides with respect to
`x` yields

`xy=\frac{x^6}6+C`

`\implies y(x)=\frac{x^5}6+\frac Cx`

Since
`y(1)=1`,

`1=\frac16+C\implies C=\frac56`

so that

`\boxed{y(x)=\frac{x^5}6+\frac5{6x}}`