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

Question

asked Apr 10, 2020 17.8k views

1 Answer

Siyu Song · Answer 1 · 2020-04-15T23:47:16+0000

The ODE is linear:

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

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

Multiplying both sides by
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
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

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