- #1
PatsyTy
- 30
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Homework Statement
Solve the following differential equation and compare computer solutions
\begin{equation*}
4y''+12y'+9=0
\end{equation*}
Homework Equations
None
The Attempt at a Solution
[/B]
First of all looking at this equation, even though it is in the section where we learned about D.Es with zero right hand side, I don't believe this is the case since the equation is equal to
\begin{equation*}
4y''+12y'=-9
\end{equation*}
Which is fine, I know how to solve D.E's using the annihilator method (to an extent). I'm sure I am missing something simply in my solution but I just can't get it.
I start out by writing this out in the operator form
\begin{equation*}
4D(D+3)y=-9
\end{equation*}
Solving the homogenous equation with roots ##0## and ##-3## which gives me the solutions to the complementary equation
\begin{equation*}
y_g=c_1+c_2e^{-3x}
\end{equation*}
I then use the operator ##D## to annihilate the constant ##-9## on the RHS to help me solve for the particular solution
\begin{equation*}
4D^2(D+3)y=0
\end{equation*}
I already know the solution for the term ##(D+3)##, and can find the solution for the term ##d^2y## as follows
\begin{equation*}
D^2y=DDy=0 \\
\end{equation*}
integrate both sides to solve this D.E
\begin{equation*}
DDy=0 \rightarrow \int{\frac{d}{dx} \Big(\frac{dy}{dx}dx\Big)}=\int{0dx} \rightarrow \frac{dy}{dx}=c_3
\end{equation*}
Integrate everything again
\begin{equation*}
\int{ \frac{d}{dx}(y)dx}=\int{c_1 dx} \rightarrow y= c_3x+c_4
\end{equation*}
So from my understanding I take the solutions for the terms ##D^2## and ##D+3##, differentiate them and sub them back into the original D.E ##4y''+12y'+9=0##. I also know that the terms from ##D+3## are part of the general solution and should be annihilated so I don't need to keep them (I did just to make sure)
\begin{equation*}
y=c_1+c_2e^{-3x}+c_3+c_4x \\
y' = -3c_2e^{-3x}+c_4\\
y''=9c_2e^{-3x}
\end{equation*}
Sub into the D.E
\begin{equation*}
4(9c_2e^{-3x})+12(-3c_2e^{-3x}+c_4)+9=0\\
36c_2e^{-ex}-36c_2e^{-3x}+12c_4+9=0\\
c_4=-\frac{3}{4}
\end{equation*}
This is where I get confused, the general solution is the complementary solution plus the particular solution, however in the particular solution I have only been able to solve for ##c_4##, does this mean I can throw away ##c_1##,##c_2## and ##c_3## of the particular solution? This would mean my particular solution is a constant ##-3/4##. In this case my general solution would be
\begin{equation*}
y=y_c+y_p \\
y=c_1+c_2e^{-3x}-\frac{3}{4}x
\end{equation*}
The issue is my textbook says the answer is
\begin{equation*}
y=(c_1+c_2x)e^{-3x/2}
\end{equation*}
I am unsure where I went wrong to get this answer. I suspect the text authors used a different technique than the annihilator method but I can't see how since ##g(x)## (the RHS) does not equal zero. Any help would be greatly appreciated. I am struggling with the D.E portion of my mathematical physics course because we only spent 4 classes on D.Es and I haven't worked with them before in any of my calc classes.
Thanks again!
Edit: Error in my second last equation ##y=c_1+c_2e^{-3x}-\frac{3}{4}##, I have edited it to be ##y=c_1+c_2e^{-3x}-\frac{3}{4}x##
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