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$y'' -4y' = -4te^{2t}, y(0)=0, y'(0)=1$ If you take laplace Transform of all terms, isolate L(y), I got $$L(y) = \frac{1}{(p-2)^2} + \frac{-2}{(p-2)^2 -4}$$

Then, taking inverse Laplace, you get $$y(t) = te^{2t} - e^{2t}sinh(2t)$$

But the solution is just $te^{2t}$. WHat am I doing wrong

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  • $\begingroup$ GIven solution is just $te^{2t}$, initial conditions y(0) = 0, y'(0)=1 $\endgroup$
    – MinYoung Kim
    Feb 7, 2019 at 5:10
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    $\begingroup$ At the outset, your solution is not correct, because $y'(0)=-1$ $\endgroup$
    – Shubham Johri
    Feb 7, 2019 at 7:08

3 Answers 3

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What are the initial conditions? What is the given solution?

Your $-e^{2t}\sinh(2t)$ is a solution to the homogeneous equation $y''-4y'=0$, and can also be written as $\frac12e^{4t}-\frac12$. The homogeneous equation has solutions $ae^{4t}+b$ for constants $a$ and $b$. If there are no initial conditions given, we take all of them, while if there are specific initial conditions we'll need specific values - which can be found by applying appropriate care in the Laplace transform equation.

Given initial conditions $y(0)=c_0$, $y'(0)=c_1$, the Laplace transform equation becomes $$(p^2L(p) - pc_0 - c_1) - 4(pL(p) - c_0) = \frac{-4}{(p-2)^2}$$ Solve that for $L(p)$, and you'll have the exact solution with initial conditions. If your form has hyperbolic functions $\cosh$ or $\sinh$ in it, remember that those can be converted into exponentials.

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  • $\begingroup$ So the initial condition was $y(0) = 0, y'(0) = 1$, which I've already taken into account, and yes I know sinh can be expressed as combination of exponentials. The answer is just $te^{2t}$ $\endgroup$
    – MinYoung Kim
    Feb 7, 2019 at 5:10
  • $\begingroup$ You've already taken those initial conditions into account? I doubt that. Calculating... official solution confirmed. $\endgroup$
    – jmerry
    Feb 7, 2019 at 5:18
  • $\begingroup$ I plugged in co=0, c1=1?after taking laplace of each term $\endgroup$
    – MinYoung Kim
    Feb 7, 2019 at 6:09
  • $\begingroup$ See the equation in my post - the initial conditions come in by modifying the Laplace transforms of the derivative terms. If you ignore them until after that, you'll need to solve a system - take your particular solution, add the general solution of the homogeneous equation, then match $y(0)$ and $y'(0)$. $\endgroup$
    – jmerry
    Feb 7, 2019 at 6:29
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    $\begingroup$ Checking my scratch paper ... yes. I think that was just a typo, as I used the correct version calculating the solution. $\endgroup$
    – jmerry
    Feb 7, 2019 at 8:52
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Your solution works for the initial conditions $y(0)=0,y'(0)=-1$, not the ones given in the question.

$\mathcal L[y'']-4\mathcal L[y']=[s^2Y(s)-sy(0)-y'(0)]-4[sY(s)-y(0)]\\=(s^2-4s)Y(s)\color{red}{-1}=\dfrac{-4}{(s-2)^2}\\\therefore Y(s)=\dfrac1{s^2-4s}-\dfrac4{(s^2-4s)(s-2)^2}=\dfrac1{(s-2)^2}\\\therefore y(t)=te^{2t}$

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First you should determine whether you want to use one or two sided Laplace transform. This problem can be solved easily using two sided Laplace transform in which case the initial conditions should not be applied. By definition$$F(s)=\int_{-\infty}^{\infty}f(t)e^{-st}dt$$and$$f'(t)\iff sF(s)\\f''(t)\iff s^2F(s)$$also $$e^{at}\iff \delta(s-a)$$therefore $$te^{at}\iff -\delta'(s-a)$$by substitution we obtain $$(s^2-4s)Y(s)=-4\delta'(s-2)\iff Y(s)={-4\delta'(s-2)\over s^2-4s}=-\delta'(s-2)$$therefore $$y(t)=te^{2t}$$

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  • $\begingroup$ Hm we've kind of went over using convolution integral to solve, without actually calculating the laplace transform on RHS, but I"m not that comfortable with it to use it. And the problem is in a section of chapter where delta function is not yet introduced so I think I'll stick to laplace T. both sides.. $\endgroup$
    – MinYoung Kim
    Feb 7, 2019 at 8:22
  • $\begingroup$ Of course for solving that using Laplace transform, you should use $\delta$ since the functions of form $e^{at}$ are the eigenfunctions of this transform and they lead to such a $delta$-shape LT. $\endgroup$
    – Mostafa Ayaz
    Feb 7, 2019 at 8:41

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