I'm trying to solve the following Initial value problem using the method of undetermined coefficients, but I keep getting the wrong answer. Any pointers?
$y''-9y=20e^{2t} - 81\quad\quad y(0)=10\quad y'(0)=17$
So I start by finding $y_c(x)$:
$m^2 -9 = 0 \implies m=\pm3$, then
$y_c(x)=c_1e^{3x}+c_2e^{-3x}$
For the undetermined coefficients part, I look at $20e^{2t}-18$ to get $Ae^{2t}$, and then to find $A$ I plug it into the original equation to get$$4Ae^{2t}-9(Ae^{2t})=20e^{2t}-81$$ And end up with $A = 81e^{-2t}/5 -4$
I could go on, but at this point I'm pretty sure I've done somthing wrong. I'm pretty sure $A$ isn't supposed to be this ugly.
$\endgroup$2 Answers
$\begingroup$For the particular solution try $y_p = Ae^{2t} + B$ substitute it into the DE.
$$ 4Ae^{2t} - 9Ae^{2t} - 9B = 20e^{2t} - 81$$ equate coefficients $$ -5A = 20 \, , \, 9B = 81$$
$$ A = -4 \, , \, B = 9$$
$$ y(x)=c_1e^{3t}+c_2e^{-3t} -4e^{2t} + 9$$
$$ y(0) = 10 = c_1 + c_2 -4 + 9$$ $$ c_1 + c_2 = 5$$
$$ y'(0) = 17 = 3c_1 -3c_2 -8$$ $$ c_1 - c_2 = \frac{25}{3}$$
$$ c_1 = \frac{20}{3} \, , \, c_2 = \frac{-5}{3}$$
$$ y(x)=\frac{20}{3} e^{3t}- \frac{5}{3}e^{-3t} -4e^{2t} + 9$$
$\endgroup$ 3 $\begingroup$You are correct up until the point of applying the undetermined coefficient strategy. Notice that the right hand side of your initial differential equation is a linear combination of e^(2t) and 1. Once you add the constant 1 to your partial solutions and then add another undetermined coefficient B, I think you will be able to solve this problem. (After this you should get A = -4 and B = 9).
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