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

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- Oct 4th 2008, 02:27 AMah-beeDifferential Equation
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

- Oct 4th 2008, 02:40 AMnocturnal
Why? You only need to integrate . First, switch the DE to standard form. Then multiply the whole DE by , which is the integrating factor. You've probably forgotten to multiply in the right-hand side.

- Oct 4th 2008, 02:49 AMah-bee
can u explain more in detail, im afraid i dont understand what u mean by "normal" form

- Oct 4th 2008, 02:53 AMnocturnal
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?

- Oct 4th 2008, 02:55 AMah-bee
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

- Oct 4th 2008, 03:22 AMProve It

The Integrating Factor is .

Multiply both sides by .

The left hand side is a product rule expansion of .

. - Oct 4th 2008, 03:28 AMbkarpuz