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Solution to these problems are provided by STEM Jock.

\begin{equation*} F(s)=\frac{2}{s^{2}+3 s-4} \end{equation*}

Factor the denominator.

\begin{aligned} F(s) &=\frac{2}{s^{2}+3 s-4} \\ &=\frac{2}{(s+4)(s-1)} \\ &=\frac{2 / 5}{s-1}-\frac{2 / 5}{s+4} \end{aligned}

Take the inverse Laplace transform now to get \(f(t)\).

\begin{align*} f(t)=\frac{2}{5} e^{t}-\frac{2}{5} e^{-4 t} \end{align*}

\begin{align*} y^{\prime \prime}-2 y^{\prime}+4 y=0 ; \quad y(0)=2, \quad y^{\prime}(0)=0 \end{align*}

Because the ODE is linear, the Laplace transform can be applied to solve it. The Laplace transform of a function \(y(t)\) is defined here as \begin{align*} Y(s)=\mathcal{L}\{y(t)\}=\int_{0}^{\infty} e^{-s t} y(t) d t \end{align*}

Consequently, the first and second derivatives transform as follows.

\begin{aligned} \mathcal{L}\left\{\frac{d y}{d t}\right\} &=s Y(s)-y(0) \\ \mathcal{L}\left\{\frac{d^{2} y}{d t^{2}}\right\} &=s^{2} Y(s)-s y(0)-y^{\prime}(0) \end{aligned}

Apply the Laplace transform to both sides of the ODE.

\begin{align*} \mathcal{L}\left\{y^{\prime \prime}-2 y^{\prime}+4 y\right\}=\mathcal{L}\{0\} \end{align*}

Use the fact that the transform is a linear operator.

\begin{align*} \begin{array}{c} \mathcal{L}\left\{y^{\prime \prime}\right\}-2 \mathcal{L}\left\{y^{\prime}\right\}+4 \mathcal{L}\{y\}=0 \\ {\left[s^{2} Y(s)-s y(0)-y^{\prime}(0)\right]-2[s Y(s)-y(0)]+4 Y(s)=0} \end{array} \end{align*}

Plug in the initial conditions, \(y(0)=2\) and \(y^{\prime}(0)=0\).

\begin{align*} \left[s^{2} Y(s)-2 s\right]-2[s Y(s)-2]+4 Y(s)=0 \end{align*}

As a result of applying the Laplace transform, the ODE has reduced to an algebraic equation for Solution \(Y\) , the transformed solution.

\begin{equation*} \begin{array}{c} s^{2} Y(s)-2 s Y(s)+4 Y(s)-2 s+4=0 \\ \left(s^{2}-2 s+4\right) Y(s)=2 s-4 \end{array} \end{equation*}

\begin{aligned} Y(s) &=\frac{2 s-4}{s^{2}-2 s+4} \\ &=\frac{2 s-4}{s^{2}-2 s+1+4-1} \\ &=\frac{2 s-4}{(s-1)^{2}+3} \\ &=\frac{2 s-2-4+2}{(s-1)^{2}+3} \\ &=\frac{2(s-1)-2}{(s-1)^{2}+3} \\ &=2 \frac{s-1}{(s-1)^{2}+3}-\frac{2}{(s-1)^{2}+3} \\ &=2 \frac{s-1}{(s-1)^{2}+3}-\frac{2}{\sqrt{3}} \frac{\sqrt{3}}{(s-1)^{2}+3} \end{aligned}

Take the inverse Laplace transform of \(Y(s)\) now to recover \(y(t)\).

\begin{aligned} y(t) &=\mathcal{L}^{-1}\{Y(s)\} \\ &=\mathcal{L}^{-1}\left\{2 \frac{s-1}{(s-1)^{2}+3}-\frac{2}{\sqrt{3}} \frac{\sqrt{3}}{(s-1)^{2}+3}\right\} \\ &=2 \mathcal{L}^{-1}\left\{\frac{s-1}{(s-1)^{2}+3}\right\}-\frac{2}{\sqrt{3}} \mathcal{L}^{-1}\left\{\frac{\sqrt{3}}{(s-1)^{2}+3}\right\} \\ &=2 e^{t} \cos \sqrt{3} t-\frac{2}{\sqrt{3}} e^{t} \sin \sqrt{3} t \end{aligned}

\begin{equation*} y^{\prime \prime}+y=\left\{\begin{array}{ll} t, & 0 \leq t<1 \\ 2-t, & 1 \leq t<2, \\ 0, & 2 \leq t<\infty \end{array} \quad y(0)=0, \quad y^{\prime}(0)=0\right. \end{equation*}

Let \(f(t)\) represent the piecewise function on the right side.

\begin{align*} y^{\prime \prime}+y=f(t)=\left\{\begin{array}{ll} t, & 0 \leq t<1 \\ 2-t, & 1 \leq t<2 \\ 0, & 2 \leq t<\infty \end{array}\right. \end{align*}

Because this ODE is linear, the Laplace transform can be applied to solve it. The Laplace transform of a function \(y(t)\) is defined here as

\begin{align*} Y(s)=\mathcal{L}\{y(t)\}=\int_{0}^{\infty} e^{-s t} y(t) d t . \end{align*}

Consequently, the first and second derivatives transform as follows.

\begin{aligned} \mathcal{L}\left\{\frac{d y}{d t}\right\} &=s Y(s)-y(0) \\ \mathcal{L}\left\{\frac{d^{2} y}{d t^{2}}\right\} &=s^{2} Y(s)-s y(0)-y^{\prime}(0) \end{aligned}

Apply the Laplace transform to both sides of the ODE.

\begin{align*} \mathcal{L}\left\{y^{\prime \prime}+y\right\}=\mathcal{L}\{f(t)\} \end{align*}

Use the fact that the transform is a linear operator.

\begin{align*} \begin{array}{c} \mathcal{L}\left\{y^{\prime \prime}\right\}+\mathcal{L}\{y\}=\mathcal{L}\{f(t)\} \\ {\left[s^{2} Y(s)-s y(0)-y^{\prime}(0)\right]+Y(s)=\int_{0}^{\infty} e^{-s t} f(t) d t} \end{array} \end{align*}

Plug in the initial conditions, \(y(0)=0\) and \(y^{\prime}(0)=0\), and \(f(t)\).

Divide both sides by \(s^{2}+1\).

\begin{aligned} Y(s) &=\frac{1}{s^{2}\left(s^{2}+1\right)}+\frac{e^{-2 s}}{s^{2}\left(s^{2}+1\right)}-\frac{2 e^{-s}}{s^{2}\left(s^{2}+1\right)} \\ &=\frac{1}{s^{2}}-\frac{1}{s^{2}+1}+\left(\frac{1}{s^{2}}-\frac{1}{s^{2}+1}\right) e^{-2 s}-2\left(\frac{1}{s^{2}}-\frac{1}{s^{2}+1}\right) e^{-s} \end{aligned}

Take the inverse Laplace transform of \(Y(s)\) now to recover \(y(t)\). Note that \(H(t)\) is the Heaviside function, which is defined to be 1 if \(t>0\) and 0 if \(t<0\).

\begin{equation} f(t)=t e^{a t} \end{equation}

The Laplace transform of a function \(f(t)\) is defined here as

\begin{align*} F(s)=\mathcal{L}\{f(t)\}=\int_{0}^{\infty} e^{-s t} f(t) d t \end{align*}

Substitute the given function and evaluate the integral.

\begin{aligned} F(s) &=\int_{0}^{\infty} e^{-s t} t e^{a t} d t \\ &=\int_{0}^{\infty}\left(-\frac{\partial}{\partial s} e^{-s t}\right) e^{a t} d t \\ &=-\frac{d}{d s} \int_{0}^{\infty} e^{-s t} e^{a t} d t \\ &=-\frac{d}{d s} \int_{0}^{\infty} e^{(a-s) t} d t \\ &=-\frac{d}{d s}\left[\left.\frac{1}{a-s} e^{(a-s) t}\right|_{0} ^{\infty}\right] \\ &=-\frac{d}{d s}\left(\frac{1}{s-a}\right) \\ &=-\left[-\frac{1}{(s-a)^{2}}\right] \\ &=\frac{1}{(s-a)^{2}} \end{aligned}

\begin{equation*} f(t)=\left\{\begin{array}{ll} 1, & 0 \leq t<2 \\ e^{-(t-2)}, & t \geq 2 \end{array}\right. \end{equation*}

Write \(f(t)\) in terms of the Heaviside function, \(H(t)\), which is defined to be 1 if \(t>0\) and 0 if \(t<0\).

\begin{aligned} f(t) &=1[H(t)-H(t-2)]+e^{-(t-2)} H(t-2) \\ &=H(t)+\left[e^{-(t-2)}-1\right] H(t-2) \\ &=u_{0}(t)+\left[e^{-(t-2)}-1\right] u_{2}(t) \end{aligned}

\begin{equation*} F(s)=\frac{3 !}{(s-2)^{4}} \end{equation*}

Apply the two transforms, \begin{align*} \mathcal{L}\left\{t^{n}\right\}=\frac{n !}{s^{n+1}} \quad \text { and } \quad \mathcal{L}\left\{e^{c t} f(t)\right\}=F(s-c), \end{align*} together to solve this problem.

\begin{aligned} f(t) &=\mathcal{L}^{-1}\{F(s)\} \\ &=\mathcal{L}^{-1}\left\{\frac{3 !}{(s-2)^{4}}\right\} \\ &=t^{3} e^{2 t} \end{aligned}

\begin{equation*} F(s)=\frac{e^{-s}+e^{-2 s}-e^{-3 s}-e^{-4 s}}{s} \end{equation*}

Apply the two transforms, \begin{align*} \mathcal{L}\left\{t^{n}\right\}=\frac{n !}{s^{n+1}} \quad \text { and } \quad \mathcal{L}\{f(t-c) H(t-c)\}=F(s) e^{-c s} \end{align*} together to solve this problem.

\begin{aligned} f(t) &=\mathcal{L}^{-1}\{F(s)\} \\ &=\mathcal{L}^{-1}\left\{\frac{e^{-s}+e^{-2 s}-e^{-3 s}-e^{-4 s}}{s}\right\} \\ &=\mathcal{L}^{-1}\left\{\frac{1}{s} e^{-s}\right\}+\mathcal{L}^{-1}\left\{\frac{1}{s} e^{-2 s}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s} e^{-3 s}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s} e^{-4 s}\right\} \\ &=(t-1)^{0} H(t-1)+(t-2)^{0} H(t-2)-(t-3)^{0} H(t-3)-(t-4)^{0} H(t-4) \\ &=H(t-1)+H(t-2)-H(t-3)-H(t-4) \\ &=u_{1}(t)+u_{2}(t)-u_{3}(t)-u_{4}(t) \end{aligned}

\begin{equation*} F(s)=\frac{1}{9 s^{2}-12 s+3} \end{equation*}

Observe that the denominator can be written in terms of \(3 s\).

\begin{align*} F(s)=\frac{1}{(3 s)^{2}-4(3 s)+3} \end{align*}

Factor the denominator.

\begin{align*} F(s)=\frac{1}{[(3 s)-1][(3 s)-3]} \end{align*}

Partially decompose the fraction.

\begin{align*} F(s)=\frac{-\frac{1}{2}}{(3 s)-1}+\frac{\frac{1}{2}}{(3 s)-3} \end{align*}

Apply the two transforms,

\begin{align*} \mathcal{L}\left\{e^{a t}\right\}=\frac{1}{s-a} \quad \text { and } \quad F(k s)=\mathcal{L}\left\{\frac{1}{k} f\left(\frac{t}{k}\right)\right\}, \end{align*}

together to get \(f(t)\)

\begin{aligned} f(t) &=\mathcal{L}^{-1}\{F(s)\} \\ &=-\frac{1}{2}\left(\frac{1}{3} e^{t / 3}\right)+\frac{1}{2}\left(\frac{1}{3} e^{3 t / 3}\right) \\ &=-\frac{1}{6} e^{t / 3}+\frac{1}{6} e^{t} \\ &=\frac{1}{6}\left(e^{t}-e^{t / 3}\right) \end{aligned}

\begin{equation*} y^{\prime \prime}+y=f(t) ; \quad y(0)=0, \quad y^{\prime}(0)=1 ; \quad f(t)=\left\{\begin{array}{ll} 1, & 0 \leq t<3 \pi \\ 0, & 3 \pi \leq t<\infty \end{array}\right. \end{equation*}

Because the ODE is linear, the Laplace transform can be applied to solve it. The Laplace transform of a function \(y(t)\) is defined here as

\begin{align*} Y(s)=\mathcal{L}\{y(t)\}=\int_{0}^{\infty} e^{-s t} y(t) d t \end{align*}

Consequently, the first and second derivatives transform as follows. Apply the Laplace transform to both sides of the ODE.

\begin{align*} \mathcal{L}\left\{y^{\prime \prime}+y\right\}=\mathcal{L}\{f(t)\} \end{align*}

Use the fact that the transform is a linear operator.

\begin{align*} \begin{array}{c} \mathcal{L}\left\{y^{\prime \prime}\right\}+\mathcal{L}\{y\}=\mathcal{L}\{f(t)\} \\ {\left[s^{2} Y(s)-s y(0)-y^{\prime}(0)\right]+Y(s)=\int_{0}^{3 \pi} e^{-s t}(1) d t+\int_{3 \pi}^{\infty} e^{-s t}(0) d t} \end{array} \end{align*}

Plug in the initial conditions, \(y(0)=0\) and \(y^{\prime}(0)=1\). \begin{align*} \left[s^{2} Y(s)-1\right]+Y(s)=\int_{0}^{3 \pi} e^{-s t} d t \end{align*}

As a result of applying the Laplace transform, the ODE has reduced to an algebraic equation for \(Y\), the transformed solution.

\begin{align*} \begin{array}{c} \left(s^{2}+1\right) Y(s)-1=\left.\left(-\frac{1}{s} e^{-s t}\right)\right|_{0} ^{3 \pi} \\ \left(s^{2}+1\right) Y(s)=\frac{1}{s}-\frac{1}{s} e^{-3 \pi s}+1 \\ Y(s)=\frac{1}{s\left(s^{2}+1\right)}-\frac{1}{s\left(s^{2}+1\right)} e^{-3 \pi s}+\frac{1}{s^{2}+1} \\ =\left(\frac{1}{s}-\frac{s}{s^{2}+1}\right)-\left(\frac{1}{s}-\frac{s}{s^{2}+1}\right) e^{-3 \pi s}+\frac{1}{s^{2}+1} \end{array} \end{align*}

Take the inverse Laplace transform of \(Y(s)\) now to get \(y(t)\).

\begin{aligned} y(t) &=\mathcal{L}^{-1}\{Y(s)\} \\ &=\mathcal{L}^{-1}\left\{\left(\frac{1}{s}-\frac{s}{s^{2}+1}\right)-\left(\frac{1}{s}-\frac{s}{s^{2}+1}\right) e^{-3 \pi s}+\frac{1}{s^{2}+1}\right\} \\ &=(1-\cos t)-[1-\cos (t-3 \pi)] H(t-3 \pi)+\sin t \\ &=1+\sin t-\cos t-[1-\cos (t-\pi)] H(t-3 \pi) \\ &=1+\sin t-\cos t-(1+\cos t) H(t-3 \pi) \\ &=1+\sin t-\cos t-(1+\cos t) u_{3 \pi}(t) \end{aligned}

Evaluate the inverse Laplace transforms.

In order to write \(Y(s)\) in terms of known transforms, use partial fraction decomposition.

\begin{align*} \frac{1}{s\left(s^{2}+2 s+2\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+2} \end{align*}

Multiply both sides by \(s\left(s^{2}+2 s+2\right)\).

\begin{align*} 1=A\left(s^{2}+2 s+2\right)+(B s+C) s \end{align*} Plug in three random values of \(s\) to get a system of three equations for \(A, B\), and \(C .\)

\begin{align*} \begin{array}{ll} s=0: & 1=2 A \\ s=1: & 1=5 A+B+C \\ s=2: & 1=10 A+4 B+2 C \end{array} \end{align*}

Solving this system yields \(A=1 / 2, B=-1 / 2\), and \(C=-1\).

Complete the square in the denominators.

Make it so that \(s+1\) appears in the numerators.

Take the inverse Laplace transform of \(Y(s)\) now to get \(y(t)\).

\begin{align*} y(t)=\mathcal{L}^{-1}\{Y(s)\} \end{align*}

Therefore,