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$y'' + 4y' + 5y = t^2$

So I solve for $r^2 + 4r + 5 = 0$ returning $r = 2 \pm 2i$. So $y_c = e^{-2t}(C_1\cos(2t) + C_2\sin(2t)$. For $y_p(t)$ I pick $At^2$. So $2A + 8At + 5At^2 = t^2$. I have been stuck on finding a particular solution for 30 minutes now.

My first question: Is there any particular method for finding $y_p(t)$?

Secondly, how to find $y_p(t)$ here?

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You should use method of undetermined coefficients. Take $$y_{p}(t)=at^2+bt+c$$ then $$ y'_{p}(t)=2at+b $$

$$ y''_{p}(t)=2a $$

and substitue it in differential equation $$2a+8at+4b+5at^2+5bt+5c=t^2$$ from equality of polynomials we have $5a=1,8a+5b=0,2a+4b+5c=0$ from here we can find $a=\frac{1}{5},b=\frac{-8}{25},c=\frac{22}{125}$. So $$ y_{p}(t)=\frac{1}{5}t^2+\frac{-8}{25}t+\frac{22}{125} $$

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Hint: try a quadratic polynomial

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First question: Look for these in your book...

(1) method of undetermined coefficients

(2) variation of parameters

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I haven't solved one using this method in awhile, but I think to get this to work, one needs more room to maneuver. Try

$y_p(t) = At^2 + Bt + C, \tag{1}$

then

$y_p' = 2At + B, \tag{2}$

$y_p'' = 2A, \tag{3}$

all of which, when inserted into

$y'' + 4y' + 5y = t^2 \tag{4}$

lead to

$2A + 4(2At + B) + 5(At^2 + Bt + C) = t^2, \tag{5}$

which in turn leads to the three equations

$2A + 4B + 5C = 0, \tag{6}$

$8A + 5B = 0, \tag{7}$

$5A = 1, \tag{8}$

so that

$A = \frac{1}{5}, \tag{9}$

$B = - \frac{8}{25}, \tag{10}$

$C = \frac{22}{125}, \tag{11}$

whence

$y_p(t) = \frac{1}{5}t^2 - \frac{8}{25}t + \frac{22}{125}. \tag{12}$

That should work if the arithmetic is right!

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