Solve equation taking Laplace transforms

Question

I'm solving this equation using Laplace transform

$$ Y''(t) + (t+1)Y'(t) + tY(t) = 0 $$

and I know that $Y(0)=1$ and $Y`(0)=-1$, so I start solving it taking Laplace transform

$$ \mathcal{L}\{Y''(t)\} + \mathcal{L}\{tY'(t)\} + \mathcal{L}\{Y'(t)\} + \mathcal{L}\{tY(t)\} $$

calculating the transforms,

$$ \mathcal{L}\{Y(t)\} = f(s) $$ $$ \mathcal{L}\{Y'(t)\} = sf(s)-1 $$ $$ \mathcal{L}\{Y''(t)\} = s^2f(s)-s+1 $$ $$ \mathcal{L}\{tY(t)\} = -f'(s) $$ $$ \mathcal{L}\{tY'(t)\} = -f(s)-sf'(s)$$

I get

$$ s^2f(s) + sf(s) - f(s) -s -sf'(s) - f'(s) = 0 $$

what I have to do now?

Answer 1

Try solving the ODE! We have

$\frac{s^2+s-1}{s+1}f(s) - f'(s) = \frac{s}{s+1}$

Now you can apply the standard method: factorize the LHS into $(-g(s)f(s))'/g(s)$ where $g(s) = \exp(-\int \frac{s^2+s+1}{s+1})$ and then integrate up and solve. From experience it looks like the solution should be something like

$f(s) = \frac{A}{s+1}$

Plugging this in gives $A=1$ so this is indeed the correct solution

Now just invert this to find $Y(t) = \exp(-t)$

Answer 2

The general solution from the differential equation \begin{align} \left[ \partial_{t}^{2} + (t+1) \partial_{t} + t \right] y(t) = 0 \end{align} is \begin{align} y(t) &= c_{1} e^{-t^{2}/2} H_{-1}\left( \frac{t-1}{\sqrt{2}} \right) + c_{2} e^{1/2 -t} \\ &= e^{-t} \left[ A \,\, erf\left(\frac{t-1}{\sqrt{2}}\right) + B \right]. \end{align} Now the Laplace transform of the equation is as follows. \begin{align} L\{ \left[ \partial_{t}^{2} + (t+1) \partial_{t} + t \right] y(t) \} = 0 \end{align} which becomes \begin{align} (s+1) \partial_{s} y(s) - (s^{2} + s -1) y(s) = -y(0)(s+1) - y^{'}(0) \end{align} or \begin{align} \partial_{s} y(s) - \left( s - \frac{1}{s+1} \right) y(s) &= -y(0) - \frac{y^{'}(0)}{s+1} \\ \frac{1}{s+1} \, e^{s^{2}/2} \partial_{s} \left[ (s+1) e^{-s^{2}/2} y(s) \right] &= -y(0) - \frac{y^{'}(0)}{s+1} \\ \partial_{s} \left[ (s+1) e^{-s^{2}/2} y(s) \right] &= y(0)(s+1) \, e^{-s^{2}/2} - y^{'}(0) e^{-s^{2}/2} \\ (s+1) e^{-s^{2}/2} y(s) &= y(0) e^{-s^{2}/2} - (y(0) + y^{'}(0)) \int^{s} e^{-u^{2}/2} du \end{align} which gives \begin{align} y(s) &= \frac{y(0)}{s+1} - \sqrt{ \frac{\pi}{2} }(y(0) + y^{'}(0)) \, \frac{1}{s+1} \, e^{-s^{2}/2} \, erf\left(\frac{s}{\sqrt{2}}\right). \end{align} Inverting the transform yields \begin{align} y(t) &= y(0) e^{-t} - \sqrt{ \frac{\pi}{2} }(y(0) + y^{'}(0)) L^{-1}\{\frac{1}{s+1} \, e^{-s^{2}/2} \, erf\left(\frac{s}{\sqrt{2}}\right) \} \\ &= y(0) e^{-t} - \sqrt{ \frac{\pi}{2} }(y(0) + y^{'}(0)) \, e^{-t^{2}/2} \,\,\cdot \frac{1}{2\pi i} \int_{\gamma-i \infty}^{\gamma + i \infty} e^{(s+t)^{2}/2} \, erf(s+t/\sqrt{2}) \frac{ds}{s+1} \\ &= y(0) e^{-t} - \sqrt{ \frac{\pi}{2} }(y(0) + y^{'}(0)) \, e^{-t^{2}/2} \,\, e^{(t-1)^{2}/2} \, erf(t-1/\sqrt{2}) \\ y(t) &= e^{-t} \left[ y(0) - \sqrt{ \frac{\pi}{2} }(y(0) + y^{'}(0)) \,\, erf\left(\frac{t-1}{\sqrt{2}}\right) \right]. \end{align}

Since the initial conditions are $y(0) =1$ and $y^{'}(0) = -1$ then the solution is \begin{align} y(t) = e^{-t}. \end{align}