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**Term Test 1 / Re: TT1 Problem 3 (afternoon)**

« **on:**October 16, 2018, 09:36:14 AM »

Since the first post did not type out the solution, I will type out my solution and fix my mistake in the picture.

First, we need to find the homogeneous solution

$r^2-5r-8=0$

$(r+4)(r-2)=0$

$r_1=-4,r_2=2$

$y_c(t) = c_1e^{-4t}+c_2e^{2t}$

Secondly, we need to find particular solution of $y" +2y' -8y = -30e^{2t}$

$y_p(t) =Ate^{2t}, y'_p(t) = Ae^{2t} +2Ate^{2t}, y"_p(t) = 4Ae^{2t}+4Ate^t$

$4Ae^{2t}+4Ate^{2t}+2Ae^{2t}+4Ate^{2t}-8Ate^{2t}=-30e^{2t}$

$6Ae^{2t} = -30e^{2t}$

A=-5

Thus, $y_p(t) =-5te^{2t}$

Thirdly, we need to find particular solution of $y" +2y' -8y = 24e^{-2t}$

$y_p(t) =Bte^{-2t}, y'_p(t) = -2Be^{-2t}, y"_p(t) = 4Be^{-2t}$

plug in back to the equation we get,

$4Be^{-2t}-4Be^{-2t} - 8Be^{-2t}=24e^{2t}$

Thus, B=-3

Thus, $y_p(t) =-3e^{-2t}$

Therefore, $y(t) = c_1e^{-4t} +c_2e^{2t}-5te^{2t}-3e^{-2t}$

(b) $y(0) = 0$ so that $c_1+c_2 -3= 0$

$y'(0) =0$ so that $-4c_1+2c_2 +1= 0$

Therefore $c_1=\frac{7}{6}, c_2 = \frac{11}{6}$ and $y(t) = \frac{7}{6}e^{-4t} +\frac{11}{6}e^{2t}-5te^{2t}-3e^{-2t}$

First, we need to find the homogeneous solution

$r^2-5r-8=0$

$(r+4)(r-2)=0$

$r_1=-4,r_2=2$

$y_c(t) = c_1e^{-4t}+c_2e^{2t}$

Secondly, we need to find particular solution of $y" +2y' -8y = -30e^{2t}$

$y_p(t) =Ate^{2t}, y'_p(t) = Ae^{2t} +2Ate^{2t}, y"_p(t) = 4Ae^{2t}+4Ate^t$

$4Ae^{2t}+4Ate^{2t}+2Ae^{2t}+4Ate^{2t}-8Ate^{2t}=-30e^{2t}$

$6Ae^{2t} = -30e^{2t}$

A=-5

Thus, $y_p(t) =-5te^{2t}$

Thirdly, we need to find particular solution of $y" +2y' -8y = 24e^{-2t}$

$y_p(t) =Bte^{-2t}, y'_p(t) = -2Be^{-2t}, y"_p(t) = 4Be^{-2t}$

plug in back to the equation we get,

$4Be^{-2t}-4Be^{-2t} - 8Be^{-2t}=24e^{2t}$

Thus, B=-3

Thus, $y_p(t) =-3e^{-2t}$

Therefore, $y(t) = c_1e^{-4t} +c_2e^{2t}-5te^{2t}-3e^{-2t}$

(b) $y(0) = 0$ so that $c_1+c_2 -3= 0$

$y'(0) =0$ so that $-4c_1+2c_2 +1= 0$

Therefore $c_1=\frac{7}{6}, c_2 = \frac{11}{6}$ and $y(t) = \frac{7}{6}e^{-4t} +\frac{11}{6}e^{2t}-5te^{2t}-3e^{-2t}$