Laplace transform of integral constant From the book First course on DE, there's the problem $2y''+ty'-2y=10, y(0)=0, y'(0)=0$. Implying laplace transform to both sides give a linear homogenuous DE $Y'(s)-(2s-\frac{3}{s})Y(s)=-\frac{10}{s^2}$ and it gives $Y=\frac{5}{s^3}+C\frac{e^{s^2}}{s^3}$, but I'm stuck here. The answer is $\frac{5}{2}t^2$, which means that $c=0$. Now I can't find the reason why $c=0$. Can anyone help???
 A: Apply the exponential order property of fuctions.
Definition: A function $f$ is said to be of exponential order $c$ if there exist constants $c,M>0,T>0$ such that $|f(t)|≤Me^{ct}$ for all $t>T$.
In order for $f(t)$ to have a Laplace Transform then in a race between $|f(t)|$ and $e^{ct}$ as $t\to\infty$ then $e^{ct}$ must approach its limit first, i.e. $\lim_{t\to\infty}\frac{f(t)}{e^{ct}}=0.$
Theorem: If $f$ is piecewise continuous on $[0,\infty)$ and of exponential order $c$, then $F(s)=L[f(t)]$ exists for $s>c$ and $\lim_{s\to\infty}F(s)=0.$
Therefore
$$\lim_{s\to\infty}Y(s)=0$$
implies
$$\lim_{s\to\infty}\frac{5}{s^3}+C\frac{e^{s^2}}{s^3}=0$$
which is true if and only if $C=0$. Thus
$$Y(s)=\frac{5}{s^3}$$
A: We have 
$~\mathcal L\{y(t)\}=\bar y(p)~$
$\mathcal L\{y'(t)\}=p~\bar y(p)-y(0)=p~\bar y(p)~~~~$ (as it is given that $~y(0)=0~$)
$\mathcal L\{y''(t)\}=p^2~\bar y(p)-p~y(0)-y'(0)=p^2~\bar y(p)~~~~$(as it is given that $~y(0)=0,~y'(0)=0$)
$\mathcal L\{ty'(t)\}=-\dfrac{d}{dp}\{p~\bar y(p)\}=-\bar y(p)-p~\dfrac{d\bar y}{dp}~$
Now given equation is 
$$2y''+ty'-2y=10$$
Taking Laplace transform both side we have 
$$2~\mathcal L\{y''\}+\mathcal L\{ty'\}-2\mathcal L\{y\}=\mathcal L\{10\}$$
$$\implies 2p^2\bar y(p)-\bar y(p)-p~\dfrac{d\bar y}{dp}-2\bar y(p)=\dfrac{10}{p}$$
$$\implies \dfrac{d\bar y}{dp}+\left(\dfrac{3}{p}-2p\right)\bar y(p)=-\dfrac{10}{p^2}\tag1$$
Which is a first order linear differential equation.
Integrating factor (I.F.) $$~=e^{\int\left(\frac{3}{p}-2p\right)dp}=e^{3\ln p-p^2}=p^3e^{-p^2}~$$ 
Multiplying both side of $(1)$ by the I.F. we have
$$p^3e^{-p^2}\dfrac{d\bar y}{dp}+e^{-p^2}(3p^2-2p^4)\bar y(p)=-10pe^{-p^2}$$
$$\implies \dfrac{d}{dp}\left\{p^3e^{-p^2}\bar y\right\}=-10pe^{-p^2}$$
Integrating we have,
$$p^3e^{-p^2}\bar y=-10\int pe^{-p^2}$$
$$\implies p^3e^{-p^2}\bar y=5e^{-p^2}+c$$
$$\implies \bar y=\dfrac{1}{p^3}\left(c~e^{p^2}+5\right)\tag2$$where $~c~$ is integrating constant.
Now by the definition $~\bar y(p)=\int^{\infty}_0 y(t)e^{-pt}dt~,~$which$~~\to 0~$ if $~p\to\infty~$. So from $(2)$, $~c=0~$.
Therefore  $$ \bar y=\dfrac{5}{p^3}$$
So 
$~y(t)=\mathcal L^{-1}\{\bar y(p)\}~$
$~~~~~~=5~\mathcal L^{-1}\left\{\dfrac{1}{p^3}\right\}$
$~~~~~~=5\dfrac{t^2}{2}$
A: However I reached the answer using that this is the IVP!
Note that the lVP uniqueness condition satifies to this problem. Now if we plug in the initial condition, then whatever $L^{-1} (\frac{e^{s^2}}{s^3})$ is, if it exists then $C=0$ since $\frac{5}{2}t^2$ is a solution for every single t. I think the approach of calculating $L^{-1} (\frac{e^{s^2}}{s^3})$ is harder.
