Question

Question #95164

Answer 1

`(d^2y)/(dt^2)− y = 2e^t qquad star`

The general solution (`y_g`) is made up of the complementary (`y_c`) and particular (`y_p`) solutions.

The complementary solution is the solution to homogeneous DE:

`(d^2y)/(dt^2)− y = 0`

This has characteristic equation `lambda^2 - 1 = 0`, `lambda = pm 1`

And so `y_c = A e^t + B e^(-t)`

For the particular solution, we might think that `y_g = alpha e^(t)` is the way to go but that term is already in the complementary solution and so we look instead at `y_p = alpha t e^t`

So:

`y_p = alpha t e^t`

`y_p' = alpha e^t + alpha t e^t = alpha e^t(1 + t )`

`y_p'' = alpha e^t (1 + t ) + alpha e^t = alpha e^t (2+t)`

Plugging that into `star` yields:

`alpha e^t (2+t) - alpha t e^t = 2 e^(t) implies alpha = 1`

So the general solution is:

`y_g = A e^t + B e^(-t) + te^t`

Answer 2

`y(t)=te^t+c_1e^t+c_2e^(-t)`

Explanation:

First, solve for `(d^2y)/dt^2-y=0` to obtain the solution of the associated homogeneous equation. This gives `c_1e^t+c_2e^(-t)`.

Now try to solve `(d^2y)/dt^2-y=2e^t`. Guess that a particular solution `y` is of the form `Ae^t`, where `A` is a constant. Thus, `Ae^t-Ae^t=2e^t`, which is clearly not possible. Modify our guess to `Ate^t`. Thus, `A(te^t+e^t)+Ae^t-Ate^t=2e^t`, or `A=1`. This means that `te^t` is one of the solutions. Now, using the solutions of the associated homogenous equation, all solutions `y(t)` of this differential equation will satisfy `y(t)-te^t=c_1e^t+c_2e^(-t)`. Thus, all solutions satisfy `y(t)=te^t+c_1e^t+c_2e^(-t)`.