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Transform the differential equation to find a nontrivial solution such that x(0) = 0.
$\displaystyle tx''+(t-2)x'+x=0 $
What I have so far after the laplace transform is:
$\displaystyle -2sX-s^2X'-X'-2X-2sX+X=0$
$\displaystyle \frac {X'}{X} = - \frac {4s +1}{s^2+1} $
Then taking the integral, I have $\displaystyle lnX = -2ln(s^2+1) - tan^{-1}s $
Try e to the (what you had)
The X is found, and then you backtrack to x.
I have no time now but will take a closer look later.Quote:
Originally Posted by tttcomrader
Just for the record, Mathematica doesn't like either as a closed form solution. (That doesn't always mean that there isn't one, though.)Quote:
Originally Posted by mr fantastic
-Dan
what you got is incorrect! if, as your notation, $\displaystyle \mathcal{L}(x(t))=X(s),$ then since we are given that $\displaystyle x(0)=0,$ we will get:Quote:
Originally Posted by tttcomrader
$\displaystyle \mathcal{L}(tx''(t))=-s^2 X'(s) - 2sX(s), \ \ \ \mathcal{L}(tx'(t))=-sX'(s)-X(s), \ \ \ \mathcal{L}(x'(t))=sX(s).$ thus the transformation of your
differential equation is: $\displaystyle -(s^2+s)X'(s) - 4sX(s)=0.$ that will give you $\displaystyle X(s)=\frac{C}{(s+1)^4},$ where $\displaystyle C$ is any constant.
so the general solution is: $\displaystyle x(t)=kt^3e^{-t},$ where $\displaystyle k$ is any constant. Q.E.D.
