How to find the General and particular solution of an ODE

Question

Find a particular solution and the general solution to the ODE:

1. y" + 2y' + y = 3t^(2) + 5e^(2t)
2. y" - y' - 2y = 4sin(3t)
3. y" - y' - 12y = 3te^(2t)

Answer 1

Hi! I solved the equation number 2, and I think that you can solve 1 and 3 following the same idea.

I am assuming that you know how to calculate the corresponding homogeneous equation y" - y' - 2y = 0. Let me know if you don't.

The general solution is given by y(t) = y_c(t) + y_p(t), where y_c(t) is the solution for the corresponding homogeneous and y_p(t) is the particular solution that satisfies the nonhomogeneous equation.

Working with the nonhomogeneous equation:

y" - y' - 2y = 4sin(3t)

A possible solution is given by the combination of the trigonometric functions sin(3t) and cos(3t)

y_p(t) = Asin(3t) + Bcos(3t)

y'_p(t) = 3Acos(3t) - 3Bsin(3t)

y"_p(t) = -9Asin(3t) - 9Bcos(3t)

Replacing them on the equation:

-9Asin(3t) - 9Bcos(3t) - 3Acos(3t) + 3Bsin(3t) - 2Asin(3t) - 2Bcos(3t) = 4sin(3t)

(-9A +3B-2A) sin(3t) + (-9B -3A-2B) cos(3t) = 4sin(3t)

So we get:

-9A + 3B - 2A = 4

-9B - 3A - 2B = 0

Solving the linear system, we find that A = -22/65 and B = 6/65

Putting those values on the solution y_p(t):

y_p(t) = -22/65 sin(3t)+ 6/65 cos(3t)

The general solution is:

y(t) = c₁e^2t + c₂e^-t - 22/65 sin(3t) + 6/65 cos(3t)

where y_c(t) = c₁e^2t + c₂e^-t is the solution of the corresponding homogeneous equation and c₁ and c₂ are other constants to be determined.