Activities for Laplace Transforms

Section 5.10 Activities for Laplace Transforms

Subsection 5.10.1 Using Laplace Transforms to Solve DEs

Activity 5.10.1.

Solve this DE using Laplace transforms.

\begin{align*} y^{\prime \prime} + 4y^\prime + 4y \amp = \sin (3t)\\ y(0) \amp = 0\\ y^\prime (0) \amp = 0 \end{align*}

Solution.

First I apply the Laplace transform, produce an algebraic equation for \(Y\) and solve for \(Y\text{.}\)

\begin{align*} \calL \left\{ y^{\prime \prime} + 4y^\prime + 4y \right\} \amp = \calL \left\{ \sin(3t) \right\} \\ (s^2Y - sy(0) - y^\prime(0)) + 4(sY - y(0)) + 4Y \amp = \frac{3}{s^2 + 9} \\ (s^2 + 4s + 4)Y \amp = \frac{3}{s^2 + 9} \\ Y \amp = \frac{3}{(s^2+9)(s+2)^2} \end{align*}

I use partial fractions to split up the rational function.

\begin{align*} Y \amp = \frac{-3}{169} \frac{4s+5}{s^2+9} + \frac{12}{169} \frac{1}{s+2} + \frac{3}{13} \frac{1}{(s+2)^2} \end{align*}

I need to split up the first fractions into two forms that fit the inverse Laplace transform.

\begin{align*} Y \amp = \frac{-12}{169} \frac{s}{s^2+9} + \frac{-5}{169} \frac{3}{s^2+9} + \frac{12}{169} \frac{1}{s+2} + \frac{3}{13} \frac{1}{(s+2)^2} \end{align*}

I apply the inverse Laplace transform.

\begin{align*} \calL^{-1} \{Y\} \amp = \calL^{-1} \left\{ \frac{-12}{169} \frac{s}{s^2+9} + \frac{-5}{169} \frac{3}{s^2+9} + \frac{12}{169} \frac{1}{s+2} + \frac{3}{13} \frac{1}{(s+2)^2} \right\}\\ y \amp = \frac{-12}{169} \cos 3t - \frac{5}{169} \sin (3t) + \frac{12}{169} e^{-2t} + \frac{3}{13} te^{-2t} \end{align*}

Activity 5.10.2.

Solve this DE using Laplace transforms.

\begin{align*} y^{\prime \prime} + 2y^\prime + 12y \amp = 7 u_5(t)\\ y(0) \amp = 0\\ y^\prime (0) \amp = -6 \end{align*}

Solution.

First I apply the Laplace transform, produce an algebraic equation for \(Y\) and solve for \(Y\text{.}\)

\begin{align*} \calL \left\{ y^{\prime \prime} + 2y^\prime + 12y \right\} \amp = \calL \left\{ 7u_5(t) \right\} \\ (s^2Y - sy(0) - y^\prime(0)) \amp +2 (sY - y(0)) + 12Y = 7\frac{e^{-5s}}{s} \\ (s^2 + 2s + 12)Y + 6 \amp = 7\frac{e^{-5s}}{s} \\ (s^2 + 2s + 12)Y \amp = 7\frac{e^{-5s}}{s} - 6 \\ Y \amp = 7\frac{e^{-5s}}{s((s+1)^2+11)} - 6 \frac{1}{(s+1)^2+11} \end{align*}

I use partial fractions to split up the first rational function.

\begin{align*} Y \amp = -\frac{7}{12}\frac{e^{-5s}(s+2)}{(s+1)^2+11} + \frac{7}{12}\frac{e^{-5s}}{s} - 6 \frac{1}{(s+1)^2+11} \end{align*}

I need to make some adjustments so that the numerators and constants fit the pattern of inverse Laplace transforms.

\begin{align*} Y \amp = -\frac{7}{12}\frac{e^{-5s}(s+1)}{(s+1)^2+11} - \frac{7}{12}\frac{e^{-5s}}{(s+1)^2+11} + \frac{7}{12}\frac{e^{-5s}}{s} - 6 \frac{1}{\sqrt{11}}\frac{\sqrt{11}}{(s+1)^2+11}\\ Y \amp = -\frac{7}{12}\frac{e^{-5s}(s+1)}{(s+1)^2+11} - \frac{7}{12\sqrt{11}}\frac{e^{-5s}\sqrt{11}}{(s+1)^2+11} + \frac{7}{12}\frac{e^{-5s}}{s} - 6 \frac{1}{\sqrt{11}}\frac{\sqrt{11}}{(s+1)^2+11} \end{align*}

I apply the inverse Laplace transform.

\begin{align*} y \amp = \frac{-7}{12} u_5(t) e^{-(t-5)} \cos (\sqrt{11}(t-5)) - \frac{7}{12\sqrt{11}} u_5(t) e^{-(t-5)} \sin (\sqrt{11}(t-5)) \\ \amp + \frac{7}{12} u_5(t) - \frac{6}{\sqrt{11}} e^{-t} \sin (\sqrt{11}t) \end{align*}

Activity 5.10.3.

Solve this DE using Laplace transforms.

\begin{align*} y^{\prime \prime} + 2y^\prime + 9y \amp = u_3(t) \sin (2(t-3)) + u_6(t) \cos (2(t-6)) \\ y(0) \amp = 0\\ y^\prime (0) \amp = 0 \end{align*}

Solution.

First I apply the Laplace transform, produce an algebraic equation for \(Y\) and solve for \(Y\text{.}\)

\begin{align*} \calL \left\{ y^{\prime \prime} + 2y^\prime + 9y \right\} \amp = \calL \left\{ u_3(t) \sin (2(t-3)) + u_6(t) \cos(2(t-6)) \right\} \\ (s^2Y - sy(0) - y^\prime(0)) \amp + 2 (sY - y(0)) + 9Y = e^{-3s} \frac{2}{s^2 + 4} + e^{-6s} \frac{s}{s^2+4}\\ (s^2 + 2s + 9)Y \amp = e^{-3s} \frac{2}{s^2 + 4} + e^{-6s} \frac{s}{s^2+4}\\ Y \amp = e^{-3s} \frac{2}{(s^2 + 4)((s+1)^2+8)} + e^{-6s} \frac{s}{(s^2+4)((s+1)^2+8)} \end{align*}

I use partial fractions to split up the rational function.

\begin{align*} Y \amp = e^{-3s} \frac{2}{41} \frac{2s-1}{(s+1)^2+8} - e^{-3s} \frac{2}{41} \frac{2s-5}{s^2 + 4} - e^{-6s} \frac{1}{41} \frac{5s+18}{(s+1)^2+8} + e^{-6s} \frac{1}{41} \frac{5s+8}{s^2+4} \end{align*}

I need to make some adjustments so that the numerators and constants fit the pattern of inverse Laplace transforms. This leads to an annoying proliferation of terms, but it is necessary.

\begin{align*} Y \amp = e^{-3s} \frac{2}{41} \frac{2s+2}{(s+1)^2+8} - e^{-3s} \frac{2}{41} \frac{3}{(s+1)^2+8} - e^{-3s} \frac{2}{41} \frac{2s}{s^2 + 4} \\ \amp + e^{-3s} \frac{2}{41} \frac{5}{s^2 + 4} - e^{-6s} \frac{1}{41} \frac{5s+5}{(s+1)^2+8} - e^{-6s} \frac{1}{41} \frac{13}{(s+1)^2+8}\\ \amp + e^{-6s} \frac{1}{41} \frac{5s}{s^2+4} + e^{-6s} \frac{1}{41} \frac{8}{s^2+4}\\ Y \amp = e^{-3s} \frac{4}{41} \frac{s+1}{(s+1)^2+8} - e^{-3s} \frac{6}{41\sqrt{8}} \frac{\sqrt{8}}{(s+1)^2+8} - e^{-3s} \frac{4}{41} \frac{s}{s^2 + 4} \\ \amp + e^{-3s} \frac{5}{41} \frac{2}{s^2 + 4} - e^{-6s} \frac{5}{41} \frac{s+1}{(s+1)^2+8} - e^{-6s} \frac{13}{41\sqrt{8}} \frac{\sqrt{8}}{(s+1)^2+8}\\ \amp + e^{-6s} \frac{5}{41} \frac{s}{s^2+4} + e^{-6s} \frac{4}{41} \frac{2}{s^2+4} \end{align*}

I apply the inverse Laplace transform.

\begin{align*} y \amp = \frac{4}{41} u_3(t) e^{-(t-3)} \cos(\sqrt{8}(t-3)) - \frac{6}{41\sqrt{8}} u_3(t) e^{-(t-3)} \sin(\sqrt{8}(t-3))\\ \amp - \frac{4}{41} u_3(t) \cos(2(t-3)) + \frac{5}{41} u_3(t) \sin(2(t-3))\\ \amp \frac{5}{41} u_6(t) e^{-(t-6)} \cos (\sqrt{8}(t-6)) - \frac{13}{41\sqrt{8}} u_6(t) e^{-(t-6)} \sin (\sqrt{8}(t-6))\\ \amp + \frac{5}{41} u_6(t) \cos(2(t-6)) + \frac{4}{41} u_6(t) \sin(2(t-6)) \end{align*}

Activity 5.10.4.

Solve this DE using Laplace transforms.

\begin{align*} y^{\prime \prime} + 6y^\prime + 5y \amp = t^2 + 1 + \delta_4(t) \\ y(0) \amp = 2\\ y^\prime (0) \amp = 0 \end{align*}

Solution.

First I apply the Laplace transform, produce an algebraic equation for \(Y\) and solve for \(Y\text{.}\)

\begin{align*} \calL \left\{ y^{\prime \prime} + 6y^\prime + 5y \right\} \amp = \calL \left\{ t^2 + 1 + \delta_4(t) \right\} \\ (s^2Y - sy(0) - y^\prime(0)) \amp + 6(sY - y(0)) + 5Y = \frac{2}{s^3} + \frac{1}{s} + e^{-4s} \\ (s^2 + 6s + 5)Y - 2s - 12 \amp = \frac{2}{s^3} + \frac{1}{s} + e^{-4s}\\ (s+5)(s+1)Y \amp = \frac{2}{s^3} + \frac{1}{s} + e^{-4s} + 2s + 12\\ Y \amp = \frac{1}{(s+5)(s+1)} \left( \frac{2}{s^3} + \frac{1}{s} + e^{-4s} + 2s + 12 \right) \\ Y \amp = \frac{2}{s^3(s+5)(s+1)} + \frac{1}{s(s+5)(s+1)} + \frac{e^{-4s}}{(s+5)(s+1)}\\ \amp + \frac{2s}{(s+5)(s+1)} + \frac{10}{(s+5)(s+1)} \end{align*}

I use partial fractions to split up the rational function.

\begin{align*} Y \amp = \frac{2}{5} \frac{1}{s^3} - \frac{12}{25} \frac{1}{s^2} - \frac{1}{2} \frac{1}{s+1} + \frac{1}{250} \frac{1}{s+5} - \frac{62}{125} \frac{1}{s} - \frac{1}{4} \frac{1}{s+1} + \frac{1}{20} \frac{1}{s+5} + \frac{1}{5} \frac{1}{s}\\ \amp + \frac{1}{4} \frac{e^{-4s}}{s+1} - \frac{1}{4} \frac{e^{-4s}}{s+5} + \frac{5}{2} \frac{1}{s+5} - \frac{1}{2} \frac{1}{s+1} + 3 \frac{1}{s+1} - 3 \frac{1}{s+5} \end{align*}

I group like terms before going the inverse transform.

\begin{align*} Y \amp = \frac{1}{4} \frac{e^{-4s}}{s+1} - \frac{1}{4} \frac{e^{-4s}}{s+5} + \frac{2}{5} \frac{1}{s^3} - \frac{12}{25} \frac{1}{s^2} - \frac{62}{125} \frac{1}{s} + \frac{7}{4} \frac{1}{s+1} - \frac{223}{500} \frac{1}{s+5} \end{align*}

I apply the inverse Laplace transform.

\begin{align*} y \amp = \frac{1}{4} u_4(t) e^{-(t-4)} - \frac{1}{4} u_4(t) e^{-5(t-4)} - \frac{1}{5} t^2 - \frac{12}{25} t - \frac{62}{125} + \frac{7}{4} e^{-t} - \frac{223}{500} e^{-5t} \end{align*}

Prev Top Next