Question

solve the initial value problem using laplace transform w''(t)-2w'(t)+w=6t-2 when w(-1)=3, w'(-1)=7
I know you have to use y(t)=w(t-1), but i don't know if i have is right! my solution that i found was
w(t)= 10+2t-4e^(t+1)+5te^(t+1)

Answer 1

\[w''(t)-2w'(t)+w=6t-2\]
Let \(x=t+1\), so \(t=x-1\).
\[w''(x-1)-2w'(x-1)+w=6(x-1)-2\]
Let \(w(t)=w(x-1)=y(x)\), so that \(w'(t)=y'(x)\) and \(w''(t)=y''(x)\).
\[y''(x)-2y'(x)+y=6x-8\]
with initial conditions
\[w(-1)=y(-1+1)=y(0)=3\\
w'(-1)=y'(-1+1)=y'(0)=7\]
Taking the LT, you have
\[\begin{align*}
\mathcal{L}\{y''(x)-2y'(x)+y\}&=\mathcal{L}\{6x-8\}\\
(s^2\mathcal{L}\{y\}-sy(0)-y'(0))-2(s\mathcal{L}\{y\}-y(0))+\mathcal{L}\{y\}&=\frac{6}{s^2}-\frac{8}{s}\\
(s^2-2s+1)\mathcal{L}\{y\}-3s-1&=\frac{6}{s^2}-\frac{8}{s}\\
\mathcal{L}\{y\}&=\frac{6-8s+s^2+3s^3}{s^2(s^2-2s+1)}
\end{align*}\]
Is this what you did?

Answer 2

@SithsAndGiggles I don't understand why we have to convert to y ? Why don't we solve directly from the original problem?

Answer 3

See example 4 in the link: http://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx

Answer 4

I got it, thank you very much :)

Answer 5

Sorry, slight mistake:
\(x-1=t\)

Then \(w(t)=w(x-1)=y(x)=y(t+1)\), which gives \(w(-1)=y(0)\).

Answer 6

yes i have that work! but, when i break up the denominator into partial fractions to simplify, i must be doing something wrong..

Answer 7

when i break into partial fractions i get:
=\[\frac{ 4 }{ s} +\frac{ 6 }{ s ^{2} }-\frac{ 1 }{ s-1 }+\frac{ 2}{ (s-1)^2 }\]

Answer 8

is that right? my final answer is currently, w(t)= 12+6t-e^(t+1) +2te^(t+1)

Answer 9

Hmm, that doesn't look completely correct...

Here's the partial fraction decomposition:
\[\frac{3s^3+s^2-8s+6}{s^2(s-1)^2}=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{s-1}+\frac{d}{(s-1)^2}\]
which gives
\[3s^3+s^2-8s+6=as(s-1)^2+b(s-1)^2+cs^2(s-1)+ds^2\]
or
\[3s^3+s^2-8s+6=(a+c)s^3+(-2a+b-c+d)s^2+(a-2b)s+b\]
So you have
\[\begin{cases}a+c=3\\-2a+b-c+d=1\\a-2b=-8\\b=6\end{cases}~~\Rightarrow~~a=4,~b=6,~c=-1,~d=2\]
So your decomposition is right - I just wanted to check for myself.

Now, you want to get the following:
\[y(x)=\mathcal{L}^{-1}\left\{\frac{4}{s}+\frac{6}{s^2}-\frac{1}{s-1}+\frac{2}{(s-1)^2}\right\}\]
Taking the inverse transforms, you have
\[y(x)=4+6x-e^x+2xe^x\]
Back-substituting with \(x=t+1\), you ahve
\[\begin{align*}y(t+1)&=w(t)\\
&=4+6(t+1)-e^{t+1}+2(t+1)e^{t+1}\\
&=4+6t+6-e^{t+1}+2te^{t+1}+2e^{t+1}\\
&=10+6t+e^{t+1}+2te^{t+1}\end{align*}\]
You weren't too far off.

Answer 10

thank you very much! I understand it now, great instructions!

Answer 11

yw