2nd order nonhomogeneous differential equations?

Question

How to you find the general solution to this equation? y'' + y = 3sin(2t) + tcos(2t). Thanks a lot. The answer by the way is y = (c1)cos(t) + (c2)sin(t) - 1/3*t*cos(2t) - (5/9)sin(2t).

Thanks blondie, I'm in college like most successful people in this world. Why did you answer this question anyway?

Answer 1

This is not too hard; it's a linear constant coefficient ODE.

First find the general solution to the corresponding homogeneous ODE: y" + y = 0
This has characteristic equation r^2 + 1 = 0 <=> r = ± i, so the solutions are cos t and sin t. (In general if there are complex conjugate solutions r = a ± ib then the solutions to the ODE are e^a cos bt and e^a sin bt.)
So the complementary solution is y = A cos t + B sin t.

Now we need to find a particular solution. The function we're after is 3 sin 2t + t cos 2t, so we expect it will be of the form
y_p = C cos 2t + D sin 2t + E t cos 2t + F t sin 2t
None of these terms are present in the complementary solution, so we'll go with this.
y_p' = - 2C sin 2t + 2D cos 2t + E cos 2t - 2E t sin 2t + F sin 2t + 2F t cos 2t
= (F-2C) sin 2t + (2D+E) cos 2t - 2E t sin 2t + 2F t cos 2t
y_p" = (2F-4C) cos 2t + (-4D-2E) sin 2t - 2E sin 2t - 4E t cos 2t + 2F cos 2t - 4F t sin 2t
= (4F-4C) cos 2t - (4D+4E) sin 2t - 4E t cos 2t - 4F t sin 2t
So y_p" + y_p = [(4F-4C) cos 2t - (4D+4E) sin 2t - 4E t cos 2t - 4F t sin 2t] + [C cos 2t + D sin 2t + E t cos 2t + F t sin 2t]
= (4F-3C) cos 2t - (3D+4E) sin 2t- 3E t cos t - 3F t sin 2t
which must identically equal 3 sin 2t + t cos 2t. Equating coefficients gives us
4F - 3C = 0
3D + 4E = -3
3E = -1
3F = 0
Hence F = C = 0, E = -1/3 and 3D - 4/3 = -3 => D = -5/9.
So y_p = (-5/9) sin 2t - (1/3) t cos 2t.

So the general solution is the sum of the complementary and particular solutions, i.e.
y = A cos t + B sin t - (5/9) sin 2t - (1/3) t cos 2t.