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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.

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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 $$ -5A = 20 \, , \, 9B = 81$$

$$ A = -4 \, , \, B = 9$$

$$ y(x)=c_1e^{3t}+c_2e^{-3t} -4e^{2t} + 9$$

$$ y(0) = 10 = c_1 + c_2 -4 + 9$$ $$ c_1 + c_2 = 5$$

$$ y'(0) = 17 = 3c_1 -3c_2 -8$$ $$ c_1 - c_2 = \frac{25}{3}$$

$$ c_1 = \frac{20}{3} \, , \, c_2 = \frac{-5}{3}$$

$$ y(x)=\frac{20}{3} e^{3t}- \frac{5}{3}e^{-3t} -4e^{2t} + 9$$

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  • $\begingroup$ Thanks! I knew I was missing something, this makes more sense $\endgroup$ – kojak Nov 7 '16 at 1:20
  • $\begingroup$ I'm getting 20/3 and 5/3 for c_1 and c_2. I've checked and your answer is right, but what am I missing? $\endgroup$ – kojak Nov 7 '16 at 1:42
  • $\begingroup$ You're right. I made a sign error. I've corrected it and checked it on wolfram wolframalpha.com/input/… $\endgroup$ – arthur Nov 7 '16 at 2:35
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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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