Solving linear differential equations

[simba_]

Homework Statement

the full question is asking me to solve
dy/dx + (1/x)y = 3cos2x

i think i know what i am doing up to a point,
but for me to continue with the question i have to integrate

exp(x^-1)3cos2x

and I am not sure how to do this, once i get this part i would know how to continue...
can anyone integrate this, or does anyone know of a different approach?
(And another thing not really related to the above question, is there a good site that i can use to check my answers against when they are in the form of the original question??)

 

[snipez90]

No you did it incorrectly. The integrating factor, or whatever you call it, should be x, not 1/x.

 

[LCKurtz]

Remember your integrating factor should be

e^{\int \frac 1 x\, dx}

You have to integrate to get it.

 

[HallsofIvy]

Or (the reason for that formula LCKurtz gives)

To be "exact", the left side must be a single derivative. u(x) will be an "integrating factor if and only if
\frac{d(uy)}{dx}= u\frac{dy}{dx}+ \frac{u}{x}y

By the product rule
\frac{d(uy)}{dx}= u\frac{dy}{dx}+ \frac{du}{dx}y[/math]<br /> <br /> Comparing those, we must have<br /> \frac{du}{dx}= \frac{u}{x}<br /> which is a separable differential equation:<br /> \frac{du}{u}= \frac{dx}{x}