Solve the ode

**sarah232** · April 17th 2009, 10:35 AM

x^3y' + 2y = x^3 + 2x y(1) = e+1

Which method would you use to solve this?

**Danny** · April 17th 2009, 11:19 AM

Originally Posted by sarah232

x^3y' + 2y = x^3 + 2x y(1) = e+1

Which method would you use to solve this?

The equation is linear so first put it in standard form

$y' + \frac{2y}{x^3} = 1 + \frac{2}{x^2}$

The integrating factor is

$\mu = e^{-\frac{1}{x^2}}$ so

$\frac{d}{dx} e^{-\frac{1}{x^2}} \cdot y = \left( 1 + \frac{2}{x^2}\right) e^{-\frac{1}{x^2}}$ which integrates giving

$e^{-\frac{1}{x^2}} \cdot y = x e^{-\frac{1}{x^2}} + c$

I think you can take it from here.

**Prove It** · April 17th 2009, 11:27 PM

Originally Posted by danny arrigo

The equation is linear so first put it in standard form

$y' + \frac{2y}{x^3} = 1 + \frac{2}{x^3}$

The integrating factor is

$\mu = e^{-\frac{1}{x^2}}$ so

$\frac{d}{dx} e^{-\frac{1}{x^2}} \cdot y = \left( 1 + \frac{2}{x^2}\right) e^{-\frac{1}{x^2}}$ which integrates giving

$e^{-\frac{1}{x^2}} \cdot y = x e^{-\frac{1}{x^2}} + c$

I think you can take it from here.

Your standard form is wrong.

If $x^3\frac{dy}{dx} + 2y = x^3 + 2x$ then

$\frac{dy}{dx} + 2x^{-3}y = 1 + 2x^{-2}$ .

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