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Solve $y''-2y'-3y=3te^{-t}$

Attempt: now the solution has two parts. Namely, $y=y_h+y_p$ where $y_h$ is the solution of the homogeneous equation and $y_p$ is the solution of the nonhomogeneous equation. $y_h$ is easy to find.


I am struggling with finding $y_p$. The way I set it up is:


Since $Be^{-t}$ is also a solution for $y_h$ I multiply RHS of $y_p$ by $t$ and get:


Then, I find derivatives of $y_p$ and plug them into differential equation, simplify, and get:


Now, if this is correct how would I solve for A and B?

It is not homework. Thank you for your help.

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If that's correct, there is no solution as you can't have $-2A=0$ and $-4A=-3$. But I don't think it's correct. However you do it, it's very messy. – Mike Feb 21 '13 at 23:12
Agreeing with @Mike. You must have made a mistake in differentiating $y_p$ or in algebra, as the method is correct. – Gerry Myerson Feb 21 '13 at 23:21
If you know how to use integrating factors, you can solve the equation directly. Try the substituion $z=y'+y$. – Mike Feb 21 '13 at 23:37
up vote 3 down vote accepted



For $y''-2y'-3y=3te^{-t}$, we get:

$$(2 A - 4B - 8At)e^{-t} = 3t e^{-t}$$

So, equating sides, we get: $$-8A = 3 \rightarrow A = -\frac{3}{8}$$

and we have:

$$2A - 4B = 0 \rightarrow B = \frac{1}{2} A \rightarrow B = -\frac{3}{16}$$

This agrees with WA


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Very good answer. Thanks. – Koba Feb 22 '13 at 5:06
@Dostre: You are very welcome. – Amzoti Feb 22 '13 at 5:31
Great answer = +1, nice feedback $+ \to \infty$ – amWhy May 1 '13 at 0:58
@amWhy: Thanks, I like when things are that clear, although it does not always happen! :-) – Amzoti May 1 '13 at 0:59

You definitely made a mistake, that's what Mathematica shows

In[27]:= ClearAll[A, B]
y = (A*t^2 + B*t)*E^-t;
y' = D[y, t];
y'' = D[y', t];
S = Simplify[y'' - 2 y' - 3 y];

In[64]:= S

Out[64]= E^-t (-4 B + A (2 - 8 t))

In[65]:= sol = Solve[-4 q + 2 p == 0 && -8 p == 3, {p, q}];

In[63]:= Simplify[S /. {A -> p /. sol[[1]], B -> q /. sol[[1]]}]

Out[63]= 3 E^-t t
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You are on the right track as far as a guess for $y_p$; something went wrong with your algebra. If we guess that $y_p = (b+c t)e^{-t}$, then plugging this into the diff eq'n gives us, after much simplification:

$$(-4 b+2 c -8 c t) e^{-t}$$

Equating this with your RHS gives $c=-3/8$ and $b=-3/16$. Plugging this specific form back into the equation produces $3 t e^{-t}$, as expected.

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For getting the particular solution of this inhomogeneous equation, you guess an expression for it with two undetermined coefficients A and B, however you may neglect some conditions. There is a systematic method to cope with those kinds of inhomogenous equation, which the general solutions of its associated homogeneous equations are known.

As you posted it: the associated general solutions of its homogeneous equations are: $$y_h=c_1e^{3t}+c_2e^{−t}$$

now supposing the particular solutions is of the form: $$y_p=v_1(t)e^{3t} + v_2(t)e^{-t} $$

The Wronskian of this homogeneous equ. is: $$W=\begin{vmatrix} e^{3t} & e^{-t} \\ (e^{3t})' & (e^{-t})' \end{vmatrix}=-4e^{2t}$$

denote the inhomogeneous term as $Q(t)$: $3te^{-t}=Q(t)$

the systematic method states that :(note the integral constants are not important.) $$ v_1(t) = \int \frac{-y_2(t)Q(t)}{W} \mathrm{d}t =-\frac{3}{64}e^{-4t}(4t+1) \\ v_2(t)=\int \frac{y_1(t)Q(t)}{W} \mathrm{d}t = -\frac{3}{8}t^2$$

Therefore one of the particular solutions is : $$y_p=-\frac{3}{64}e^{-t}(8t^2+4t+1)$$

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This systematic method is called The Method of Variation of Parameters. – Zoe Rowa Feb 22 '13 at 4:12
More, for those kinds of inhomogeneous equation which inhomogeneous term is $P_n(t)e^{at}$, you should guess the particular solution is of the form :$\hat{P}_{n+1}(t)e^{-t}$, $P_n(t),\hat{P}_{n+1}$ is a polynomial of order n or n+1. – Zoe Rowa Feb 22 '13 at 4:16
I have not learned the method of variation of parameters but thank you for introduction. – Koba Feb 22 '13 at 5:07

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