Initital value problem: x" - 2x' + 2x = e^(2t) sin(t), with x(0) = 3/5, x'(0) = 1

[PhysicsCat]

Hello!

I have the following task:
" Solve the initial value problem
[math]x^{\prime \prime} - 2x^{\prime} +2x = e^{2t} \sin(t)[/math] with [math]x(0) = \frac{3}{5}[/math] and [math]x^{\prime}(0) = 1[/math]"

I solved the homogenous problem and got to the solution [imath]x_h(t) = e^t \cos(t) (c_1 + c_2)[/imath], where [imath]c_1[/imath] and [imath]c_2[/imath] are constants.
This solution does solve the homogenous problem, but now Im stuck with the inhomogeneous problem. I had the idea of using the variation of constants, but considering I have two different constants, that scared me a bit. Is there a better way to solve the inhomogeneous problem in this case?

Thank you very much and have a nice day!

 

[nasi112]

Look for a particular solution of the form:

\(\displaystyle x_p(t) = e^{2t}(A\cos t + B\sin t)\)

See my post here to continue:

Third order non-homogenous diff-eq w/ constant coefficients: y''' - y'' + y' - y = cos(x) + 2e^x

The particular problem I have in mind is the following: y''' - y'' + y' - y = cosx + 2e^x I know the general way of solving non-homogenous differential equations of higher orders with constant coefficients. First, we are supposed to define it's analogous homogenous equation, then it's...

www.freemathhelp.com

 

[PhysicsCat]

Thank you very much!