Find the general solution?

Question

Find the general solution of y"-2y'-3y=3e^2t.

Answer: c₁e^3t+c₂e^-t-e^2t

 

Answer 1

Your answer should be C₁e^3t +C₂e^-t -3e^2t. In order to have the answer shown above, your equation was suppposed to be

                                                       v'' -2y' - 3y = e^2t

but not 3 e^2t. Now let's check.

The general solution y = y₀ + y₁

where y₀ - solution of the homogeneous equation y'' - 2y' -3y = 0, and

y₁  is a particular solution of the oirginal equation. Plug e^kt into the homogeneous equation to find k.

You will come upo with the equation: k² -2k - 3 = 0   or

                 (k+1)(k-3) = 0  with  solutions: k=-1 and k = 3

Thus, you have

                     y₀ = C₁ e^3t + C₂ e^-t

For a particular solution you can notice, that y₁ = C e^2t since the right side of the equation

contains the function e^2t. Plug y₁ into the equation to find C. You will have

                                          y' = 2C e^2t and y" = 4C e^2t

Canceling both sides of the equation by e^2t you will obtain                       

                             4C - 2C -3C = 3  or C = -3.

Thus, we have for the general solution

                                               y = C₁ e^3t + C₂ e^-t - 3e^2t

Your solution, shown in the tex,t is true for the equation

                                                y" - 2y' - 3y = e^2t

as it is seen from the current calculations.