I am trying to solve the following ordinary differential equation, where $y$ is a function of $t$:


My attempt was to use the Laplace Transform. The following formulae were assumed, where $\mathcal{L}(y)$ is the Laplace transform of a function $y$ defined by $$\mathcal{L}(y(t))=\int_0^{\infty}e^{-st}y(t)dt=Y(s):$$

  • $\mathcal{L}(ty'')=-s^2Y'(s)-2sY(s)+y(0)$;
  • $\mathcal{L}(ty')=-sY'(s)-Y(s)$;
  • $\mathcal{L}(ty)=-Y'(s)$;
  • $\mathcal{L}(y')=sY(s)-y(0)$.

After applying $\mathcal{L}$ to both sides of $ty''+(2+4t)y'+(4+4t)y=0$ and simplifying, I obtain $$Y(s)=\frac{y(0)}{s+2}+C$$ and here $C=0$ because $\displaystyle\lim_{s\to\infty}Y(s)=0$. Then applying the inverse Laplace Transform to $Y(s)=\dfrac{y(0)}{s+2},$ we obtain the solution $$y(t)=y(0)e^{-2t}.$$

But the general solution to the original differential equation should be $$y(t)=y(0)e^{-2t}+C_1e^{-2t}t^{-1}.$$

My Question:

Why is the term $C_1e^{-2t}t^{-1}$ missing in my solution? Somehow I failed to find any mistake in my process, but the general solution is incomplete (although applying variation of parameters may make it complete, but my question is about the Laplace Transform method).


$$ty''+(2+4t)y'+(4+4t)y=0.$$ The general solution is $\quad y(t)=c_1e^{-2t}+c_2\frac{e^{-2t}}{t}$

$\begin{cases} \mathcal{L}(e^{-2t})=\frac{1}{s+2}\\ \mathcal{L} (\frac{e^{-2t}}{t})=\infty \end{cases} \quad\implies\quad \mathcal{L}(y(t))$ doesn't exists if $c_2\neq 0$

The Laplace method works only if the function solution of the ODE has a Laplace transform.

You correctly obtained $\quad Y(s)=\frac{y(0)}{s+2}+C$

But your assumption $\quad C=0$ because $\displaystyle\lim_{s\to\infty}Y(s)=0\quad$ is true only if $Y(s)$ exists, that is if the searched function has a Laplace transform. This is true for the solution which has a Laplace transform. It is false for the solution which has no Laplace transform.

So, assuming $C=0$ is equivalent to setting $c_2=0$.

In the case of we are compeled to do this, we must be aware that the Laplace transform method will lead to some solution, but not to all solutions. The missing solution has to be searched with a different approach (for example a change of function, $u(t)=ty(t)$ in the present case).

  • $\begingroup$ I would like to say a big thank you for your answer! It clears all my questions about this. $\endgroup$ – Zuriel Jul 18 '17 at 12:52
  • $\begingroup$ Just out of curiosity, how would you solve this differential equation? $\endgroup$ – Zuriel Jul 18 '17 at 12:54
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    $\begingroup$ The change of variable $y(t)=\frac{u(t)}{t}$ transforms the initial ODE into : $$u''+4u'+4u=0$$ $\endgroup$ – JJacquelin Jul 18 '17 at 13:36
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    $\begingroup$ Another way (without guess of form of function) : The Laplace transform method gave a particular solution : $y_p(t)=e^{-2t}$. it is known that changing of function $ y(t)=f(t)y_p(t)$ reduces the second order ODE to a first order ODE where $f(t)$ is the unknown. $\endgroup$ – JJacquelin Jul 18 '17 at 13:48

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