Question

Solve the DE 2y"+4y'+y=cos(2x)

Answer 1

Oh, yummy.

So you know that there is a particular solution to this inhomogeneous equation, call it yp; and two linearly independent solutions tot he homogeneous equation, y1 and y2.

The general solution to the equation is

y = yp + c1.y1 + c2.y2 , for constants c1 and c2

So to solve, we need to solve in two phases, in whatever order
- find the homogeneous solutions
- find the particular solution

Which of those two are you having trouble with?

Answer 2

I'm actually tempted not to tell you any more now, but direct you to watch a couple of of lectures so you know the theory here:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-11-theory-of-general-second-order-linear-homogeneous-odes/

Answer 3

Thank you so much, I think the solution to the homogeous will be Y0=exp(a)cos(bx)+exp(a)sin(-ax), x=a+/1bi

Answer 4

No. One of the homogeneous solutions is
\[y_1(x) = \exp(-(1+\sqrt{2}/2)x)\]
as you can confirm by substitution.

Answer 5

It looks like you should start here:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-9-solving-second-order-linear-odes-with-constant-coefficients/

Answer 6

OH yeah I got it, the auxiliary equation has real roots . the other solution will be y=C(exp[(-1-(sqrt(2)/2)x], then we can use the method of undertemined coeff. to gt the particular silution to th DE