Question

Solve the initial-value problem y' = x^4 - \frac{1}{x}y, y(1) = 1.

Answer 1

The ODE is linear:

$y'=x^4-\dfrac yx$

$y'+\dfrac yx=x^4$

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

$xy=\dfrac{x^6}6+C$

$\implies y(x)=\dfrac{x^5}6+\dfrac Cx$

Since $y(1)=1$,

$1=\dfrac16+C\implies C=\dfrac56$

so that

$\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}$