Finding an inverse Laplace Transform for a function - solving IVPs with Laplace

[VinnyCee]

Homework Statement

Use Laplace Transforms to solve the following initial value problems

a. t\,y''\, - \,t\,y'\, + \,y\; = \;2\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-1

b. y''\,+\,2\,y'\,-3\,y\;=\;\delta(t\,-\,1)\,-\,\delta(t\,-\,2)\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-2

Homework Equations

Laplace Transforms

The Attempt at a Solution

PART A

- \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y'\left( 0 \right)} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}

- \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,2\,s\, + \,1} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,2} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}

\begin{array}{l}<br /> - \left[ {2\;s\;Y\left( s \right)\; + \;s^2 \;Y&#039;\left( s \right)\; - 2} \right]\; + \;\left[ {Y\left( s \right)\; + \;s\;Y&#039;\left( s \right)} \right]\; + \;Y\left( s \right)\; = \;\frac{2}{s} \\ <br /> Y&#039;\left( s \right)\;\left( { - s^2 \; + \;s} \right)\; + \;Y\left( s \right)\;\left( { - 2\;s\; + \;2} \right)\; = \;\frac{2}{s}\;\;\;\; \to \;\;\;\;Y&#039;\; + \;Y\;\left( {\frac{{2\; - \;2\;s}}{{ - s^2 \; + \;s}}} \right)\; = \; - \frac{2}{{s^3 \; - \;s^2 }} \\ <br /> \end{array}

\begin{array}{l}<br /> \mu \left( s \right)\; = \;e^{\int {\frac{2}{s}\;ds} } \; = \;s^2 \;\;\;\; \to \;\;\;\;\frac{d}{{ds}}\left\{ {\mu \left( s \right)\;Y\left( s \right)} \right\}\; = \;\mu \left( s \right)\;Q\left( s \right) \\ <br /> \mu \left( s \right)\;Y\left( s \right)\; = \;\int { - \frac{2}{{s\; - \;1}}} \;ds\;\;\;\; \to \;\;\;\;Y\left( s \right)\; = \;\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }} \\ <br /> \end{array}

How do I do an inverse transform for

\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }}PART B

\begin{array}{l}<br /> \left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y&#039;\left( 0 \right)} \right]\, + \,2\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, - \,3\,Y\left( s \right)\; = \;e^{ - s} \; - \;e^{ - 2\,s} \\ <br /> Y\,\left( s \right)\left( {s^2 \; + \;2\,s\; - \;3} \right)\; - \;2\,s\; - \;2\; = \;e^{ - s} \; - \;e^{ - 2\,s} \;\;\;\; \to \;\;\;\;Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{s^2 \; + \;2\,s\; - \;3}}\end{array}

Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{\left( {s\; - \;1} \right)\,\left( {s\; + \;3} \right)}}

How would I go about the partial fraction expansion of the last expression?

 

[Kummer]

f(t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s)e^{st} dt

 

[HallsofIvy]

Homework Statement

Use Laplace Transforms to solve the following initial value problems

a. t\,y&#039;&#039;\, - \,t\,y&#039;\, + \,y\; = \;2\;\;\;y(0)\;=\;2\;\;\;y&#039;(0)\;=\;-1

b. y&#039;&#039;\,+\,2\,y&#039;\,-3\,y\;=\;\delta(t\,-\,1)\,-\,\delta(t\,-\,2)\;\;\;y(0)\;=\;2\;\;\;y&#039;(0)\;=\;-2

Homework Equations

Laplace Transforms

The Attempt at a Solution

PART A

- \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y&#039;\left( 0 \right)} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}

- \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,2\,s\, + \,1} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,2} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}

These are meaningless. The Laplace transform of a derivative does not involve a derivative. You seem to be writing "d/dx" of the Laplace transform of the derivative. If that is true you do not want the "d/dx" in the expression.

\begin{array}{l}<br /> - \left[ {2\;s\;Y\left( s \right)\; + \;s^2 \;Y&#039;\left( s \right)\; - 2} \right]\; + \;\left[ {Y\left( s \right)\; + \;s\;Y&#039;\left( s \right)} \right]\; + \;Y\left( s \right)\; = \;\frac{2}{s} \\ <br /> Y&#039;\left( s \right)\;\left( { - s^2 \; + \;s} \right)\; + \;Y\left( s \right)\;\left( { - 2\;s\; + \;2} \right)\; = \;\frac{2}{s}\;\;\;\; \to \;\;\;\;Y&#039;\; + \;Y\;\left( {\frac{{2\; - \;2\;s}}{{ - s^2 \; + \;s}}} \right)\; = \; - \frac{2}{{s^3 \; - \;s^2 }} \\ <br /> \end{array}

\begin{array}{l}<br /> \mu \left( s \right)\; = \;e^{\int {\frac{2}{s}\;ds} } \; = \;s^2 \;\;\;\; \to \;\;\;\;\frac{d}{{ds}}\left\{ {\mu \left( s \right)\;Y\left( s \right)} \right\}\; = \;\mu \left( s \right)\;Q\left( s \right) \\ <br /> \mu \left( s \right)\;Y\left( s \right)\; = \;\int { - \frac{2}{{s\; - \;1}}} \;ds\;\;\;\; \to \;\;\;\;Y\left( s \right)\; = \;\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }} \\ <br /> \end{array}

How do I do an inverse transform for

\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }}


PART B

\begin{array}{l}<br /> \left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y&#039;\left( 0 \right)} \right]\, + \,2\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, - \,3\,Y\left( s \right)\; = \;e^{ - s} \; - \;e^{ - 2\,s} \\ <br /> Y\,\left( s \right)\left( {s^2 \; + \;2\,s\; - \;3} \right)\; - \;2\,s\; - \;2\; = \;e^{ - s} \; - \;e^{ - 2\,s} \;\;\;\; \to \;\;\;\;Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{s^2 \; + \;2\,s\; - \;3}}\end{array}

Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{\left( {s\; - \;1} \right)\,\left( {s\; + \;3} \right)}}

How would I go about the partial fraction expansion of the last expression?