Question

Using the variation of parameters (Diffeq):
y''+4y'+4y=t^(-2)*e^(-2t)
I've found the roots, the Wronskian , and set up for my particular solution leaving me with:
-e^(-2t)*ln(abs(t))+e^(-2t)
However the answer key only has
-e^(-2t)*ln(abs(t))
I cant figure out how to get rid of that last term

Answer 1

\[y''+4y'+4y=0\]
gives the characteristic equation
\[r^2+4r+4=(r+2)^2=0\]
which gives homogeneous solutions \(y_1=C_1e^{-2t}\) and \(C_2te^{-2t}\). Checking for linear independence:
\[\begin{align*}W(y_1,y_2)&=\begin{vmatrix}e^{-2t}&te^{-2t}\\-2e^{-2t}&(1-2t)e^{-2t}\end{vmatrix}\\\\
&=(1-2t)e^{-4t}+2te^{-4t}\\\\
&=e^{-4t}\\\\
&\not=0
\end{align*}\]
Factors given by VoP:
\[\begin{align*}u_1&=-\int\frac{y_2t^{-2}e^{-2t}}{W(y_1,y_2)}~dt\\\\
&=-\int\frac{dt}{t}\\\\
&=-\ln|t|
\end{align*}\]
\[\begin{align*}u_2&=\int\frac{y_1t^{-2}e^{-2t}}{W(y_1,y_2)}~dt\\\\
&=\int\frac{dt}{t^2}\\\\
&=-\frac{1}{t}
\end{align*}\]
which means the nonhomogeneous solution is
\[\begin{align*}y_p&=u_1y_1+u_2y_2\\\\
&=-e^{-2t}\ln|t|-\frac{e^{-2t}}{t}
\end{align*}\]

Answer 2

Oops, hold on, the final solution is indeed the same as yours. I meant to write
\[y_p=-e^{-2t}\ln|t|-e^{-2t}\]
The reason it disappears is because of the homogeneous solution, \(y_1=Ce^{-2t}\). The solution \(u_2y_2=e^{-2t}\) gets absorbed into \(y_1\).

Answer 3

In other words,
\[Ce^{-2t}+e^{-2t}=(C+1)e^{-2t}\implies Ce^{-2t}\]

Answer 4

Basically, I can assume that because \[e ^{-2t}\] is present in all parts, I can just assign a constant to that and because it is the general solution just disregard it? I remember my professor mentioning that now that I see it again, but quite honestly both his accent and handwriting are to hard to understand to get much out of it.