# Thread: Differential Equation

1. ## Differential 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

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

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

4. 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?

5. 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

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

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

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

Multiply both sides by $\displaystyle x^3$.

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

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

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

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

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

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

7. Originally Posted by ah-bee
can u explain more in detail, im afraid i dont understand what u mean by "normal" form
Normal form is for second-order equations; that is, $\displaystyle y^{\prime\prime}(t)+p(t)y(t)=0$.
It is not related for this equation.