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Here is a collection of differential equations, I suggest trying to find the correct guess for the particular solution $y_p$ before moving onto the answers.

1. $y’’ + 2y’ + y = e^{4t}$
2. $y’’ + 6y = \cos(3t)$
3. $y’’ + 7y’ + 6y = e^{-t}$
4. $y’’ + 3y’ + 2y = t^2 +5$
5. $y’‘+5y’ + 4y = t^2e^t$
6. $y’‘-4y = te^{2t}$
7. $y’‘+2y’+y = e^{-t} + e^t$
8. $y’‘+10 y’ + 16y = e^t + \cos(2t)+ t$

Solutions

1

\[y'' + 2y' + y = e^{4t}\]

Since the homogeneous solution is \(y_h = (c_1+c_2t) e^{-t}.\)

Our normal guess for the particular solution is $y_p = Ae^{4t}$. Since there are no terms which overlap with the homogeneous solutions we do not have to include any correction (which is the multiplication by $t$).

\[y_p = Ae^{4t}.\]

2

\[y'' + 6y = \cos(3t)\]

The homogeneous solution is

\[y_h = c_1\cos(\sqrt{6}t) + c_2\sin(\sqrt{6}t).\]

The normal guess for the particular solution is $y_p = A\cos(3t)+B\sin(3t)$, which does not have any term appearing in the homogeneous solution. So there is no correction again. Thus

\[y_p = A\cos(3t) + B\sin(3t).\]

3

\[y'' + 7y' + 6y = e^{-t}\]

The homogeneous solution is \(y_h = c_1e^{-6t}+c_2e^{-t}\)

Our normal guess for the particular solution is $y_p = Ae^{-t}$ which does appear in the homogeneous solution exactly once. We therefore have to include the correction by multiplying the $t$: \(y_p = A te^{-t}.\)

4

\[y'' + 3y' + 2y = t^2 +5\]

The homogeneous solution is \(y_h = c_1 e^{-2t} + c_2 e^{-t}.\)

The guess for the particular solution is \(y_p = A t^2 + Bt + C,\) and we don’t have to have any correction.

5

\[y''+5y' + 4y = t^2e^t\]

The homogeneous solution is \(y_h = c_1e^{-t} + c_2e^{-4t}.\)

The particular solution is

\(y_p = (At^2 + Bt + C) e^{t},\) and none of these terms appear in the homogeneous solution and so there is no correction need as well.

6

\[y''-4y = te^{2t}\]

The homogeneous solution is

\[y_h = c_1 e^{2t} + c_2e^{-2t}.\]

The normal guess for the particular solution would be \(y_p = (At+B)e^{2t}.\)

However, the $Be^{2t}$ term overlaps with the homogeneous solution. Now if we just multiply the $Be^{2t}$ term by $t$ then \(y_p = (At+Bt)e^{2t} = (A+B)te^{2t},\) which can be simplified by setting $C = A+B$. This idea turns out to be wrong!

The correct thing to do is to multiply everything involving $t^ne^{2t}$ for some $n = 0,1,\dotsm$ by $t$. That means; \(y_p = t(At+B)e^{2t}\) is the correct guess.

7

\[y''+2y'+y = e^{-t} + e^t\]

The homogeneous solution is \(y_h = (c_1+c_2 t) e^{t}.\)

The normal particular solution is \(y_p = Ae^{-t} + B e^t.\)

The $e^t$ term appears in the homogeneous solution twice! In order to correct the guess for the particular solution we have to multiply the $t^n e^{t}$ terms by $t^2$: \(y_p = Ae^{-t} + t^2 B e^t.\)

8

\[y''+10 y' + 16y = e^t + \cos(2t)+ t\]

The homogeneous solution is \(y_h = c_1 e^{-2t} + c_2 e^{-8t}.\)

The guess for the particular solution is

\[y_p = A e^t + B \cos(2t) + C\sin(2t) + Dt +E,\]

and does not need any correction terms.