y"-2y+2y=e2t (cos(t)-3sin(t))
I don't even know what should be the form of the particular solution . Could someone at least help me approach this problem??
thanks!!
I'm assuming you made a small mistake typing the equation and actually meant this:
$\displaystyle y''-2y'+2y = e^{2t}(cos(t) - 3sin(t))$
You are trying to solve a non-homogenous second order ordinary differential equation. You can use the annihilator approach to find the simplest differential operator that will "destroy" the right hand side of your equation and turn it into 0. From your equation, alpha = 2 and beta = 1.
So the annihilator is (D^2 - 2*alpha*D + (alpha^2 + beta^2)) which simplifies to (D^2 - 4D + 5). Solving for the root should give $\displaystyle 2 +/- i$
Since the particular solution and the complementary solution must be linearly independation to form a valid linear combination, we multiple the particular solution by an x.
So the particular solution should be $\displaystyle Y_p = A*xe^{2t}cos(t) + B*xe^{2t}sin(t)$
Now solving for the coeff. A and B should be easy. Just go back to your original equation... differentiate the Yp I gave you two times, then minus 2 times the 1st deriv. of Yp, and add 2 times the original particular solution. Then do some simple algebra to the undetermined coefficient.