Question

Check my answer!
Using Laplace Transform.Solve the following differential equation with given
condition(D^2-2D+1)x=e^t,with x=2,Dx=-1,at t=0.

i got \[x= 2 e^t+3t e^t+(e^t t^2)/2 \]

Answer 1

Tell me if there is any mistake in any step...
and whether the final answer is correct or not.

Answer 2

Partial fractions part is correct. wolf confirmed it
but the part before that...i am not sure

Answer 3

I haven't scanned the entire document, but I see a slight error where you have the transform of \(-2Dx\), which should be
\[-2\left(sX-x(0)\right)=-2sX+2x(0)=-2sX+2(2)\]
if I'm correctly understanding that the initial condition is \(x(0)=2\) - the notation is unfamiliar territory for me.

Answer 4

oh :(
i missed the parenthesis!
that changes everything :O

Answer 5

You could always check your final answer with a different method, or with WA:
http://www.wolframalpha.com/input/?i=x%27%27-2x%27%2Bx%3De%5Et%2C+x%280%29%3D2%2C+x%27%280%29%3D-1

Your answer isn't actually that far off!

Answer 6

\((2s^2-7s+6) \)
is that the new numerator ???

Answer 7

i didn't know how to put initial conditions in WA

Answer 8

I didn't know it was possible myself until fairly recently

Answer 9

thats not correct numerator...

Answer 10

Let's see:
\[\begin{align*}s^2F(s)-2s+1-2\left(sF(s)-2\right)+F(s)&=\frac{1}{s-1}\\
\left(s^2-2s+1\right)F(s)&=\frac{1}{s-1}+2s-5\\
\left(s^2-2s+1\right)F(s)&=\frac{1}{s-1}+\frac{2s(s-1)}{s-1}-\frac{5(s-1)}{s-1}\\
\left(s^2-2s+1\right)F(s)&=\frac{2s^2-7s+6}{s-1}\\
F(s)&=\frac{2s^2-7s+6}{(s-1)^3}
\end{align*}\]
I'm getting the same numerator, unless I made a mistake.

Answer 11

WA agrees, it gives the right solution:
http://www.wolframalpha.com/input/?i=InverseLaplaceTransform%5B%282s%5E2-7s%2B6%29%2F%28s-1%29%5E3%2Cs%2Ct%5D

Answer 12

Thanks again! :)

Answer 13

You're welcome!

Answer 14

@hartnn
At first it appeared that your initial solution was correct.
Then I noticed that the second term sign should have been a negative.

Refer to the Mathematica 9 solution attached.

Answer 15

thanks robtobey :)
SithsAndGiggles pointed that out earlier and i corrected it :)