Differential Equation

**ah-bee** · October 4th 2008, 01:27 AM

xy' + 3y = sin x/x^2 ive tried using the "general" equation for a linear first order equation but to no avail. any help would be appreciated. im stuck at the bit where i have to integrate sin(x)/x^3 btw

**nocturnal** · October 4th 2008, 01:40 AM

Why? You only need to integrate $\sin{x}$ . First, switch the DE to standard form. Then multiply the whole DE by $x^3$ , which is the integrating factor. You've probably forgotten to multiply $x^3$ in the right-hand side.

**ah-bee** · October 4th 2008, 01:49 AM

can u explain more in detail, im afraid i dont understand what u mean by "normal" form

**nocturnal** · October 4th 2008, 01:53 AM

Ack, I meant standard form. That is, the coefficient of y' should be 1. Afterwards, find the integrating factor. Are you familiar with solving first-order linear ODE's?

**ah-bee** · October 4th 2008, 01:55 AM

yes i am familiar with that and i have tried doing it, it doesnt work, can u just give it a go. im pretty sure i have done nothing wrong

**Prove It** · October 4th 2008, 02:22 AM

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

$\frac{dy}{dx} + \frac{3}{x}y = \frac{\sin{x}}{x^3}$

The Integrating Factor is $e^{\int \frac{3}{x}\, dx} = e^{3\ln x} = e^{\ln x^3} = x^3$ .

Multiply both sides by $x^3$ .

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

The left hand side is a product rule expansion of $\frac{d}{dx}(x^3y)$ .

$\frac{d}{dx}(x^3y) = \sin{x}$

$x^3y = \int \sin{x} \, dx$

$x^3y = -\cos{x} + C$

$y = -x^{-3}\cos{x} + Cx^{-3}$ .