Show that $Y(s)$ satisfies $(1+s^2)Y'(s) + sY(s) = 0$ for $ty'' + y' + ty = 0$

Question

I first approached the problem by finding the Laplace transform of $ty'' + y' + ty = 0$, such that:

$ts^2Y(s) - tsy(0) - ty'(0) + sY(s) - y(0) + tY(s) = 0. $

I solved for $Y(s)$:

$Y(s) = \frac{tsy(0) + ty'(0) + y(0)}{ts^2 + t + 2}$

But this is where I am stuck, as finding $Y'(s)$ by hand seems like a waste of time.

Thank you in advanced, any input helps!

Answer 1

Using standard Laplace properties from tables, \begin{align*}% %TCIMACRO{\tciLaplace}% %BeginExpansion \mathcal{L}% %EndExpansion \left( ty\left( t\right) \right) & =-Y^{\prime}\left( s\right) \\% %TCIMACRO{\tciLaplace}% %BeginExpansion \mathcal{L}% %EndExpansion \left( y^{\prime}\left( t\right) \right) & =sY\left( s\right) \\% %TCIMACRO{\tciLaplace}% %BeginExpansion \mathcal{L}% %EndExpansion \left( ty^{\prime}\left( t\right) \right) & =-Y\left( s\right) -sY^{\prime}\left( s\right) \\% %TCIMACRO{\tciLaplace}% %BeginExpansion \mathcal{L}% %EndExpansion \left( ty^{\prime\prime}\left( t\right) \right) & =-2sY\left( s\right) -s^{2}Y^{\prime}\left( s\right) \end{align*}

Then \begin{align*}% %TCIMACRO{\tciLaplace}% %BeginExpansion \mathcal{L}% %EndExpansion \left( ty^{\prime\prime}+y^{\prime}+ty\right) & =0\\ -2sY\left( s\right) -s^{2}Y^{\prime}\left( s\right) +sY\left( s\right) -Y^{\prime}\left( s\right) & =0\\ Y^{\prime}\left( s\right) \left( -1-s^{2}\right) +Y\left( s\right) \left( -2s+s\right) & =0\\ -Y^{\prime}\left( s\right) \left( 1+s^{2}\right) -sY\left( s\right) & =0\\ Y^{\prime}\left( s\right) \left( 1+s^{2}\right) +sY\left( s\right) & =0 \end{align*}