3.12 HW 11

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3.12.1 chapter 15, problem 2.11

Problem Find the inverse transform of the function $F\left(p\right)=\frac{3p+2}{3p^2+5p-2}$

Solution

Need to simplify the above expression to some expressions which are shown in the table on page 636.

$$\begin{array}{rl}F(p)& =\frac{3p+2}{3p^2+5p-2}\\ \text{(1)}& & =\frac{2}{(3p-1)(p+2)}+\frac{3p}{(3p-1)(p+2)}\end{array}$$

Expanding in partial fractions. For the first term in (1):

$$\begin{array}{rl}\frac{1}{(3p-1)(p+2)}& =\frac{A}{(3p-1)}+\frac{B}{(p+2)}=\frac{A(p+2)+B(3p-1)}{(3p-1)(p+2)}\\ A(p+2)+B(3p-1)& =1\\ Ap+2A+3Bp-B& =1\end{array}$$

Hence $2A-B=1$ and $(A+3B)=0$ which gives $A=\frac{1}{2}+\frac{B}{2}$. Therefore $\frac{1}{2}+\frac{B}{2}+3B=0$ $$  or $\frac{1+B+6B}{2}=0$ or $1+7B=0$ or $B=\frac{-1}{7}$. Hence $A=\frac{1}{2}+\frac{\frac{-1}{7}}{2}=\frac{1}{2}-\frac{1}{14}=\frac{7-1}{14}=\frac{6}{14}$.

Now the first term in (1) can be written as $2(\frac{A}{(3p-1)}+\frac{B}{(p+2)})$ or $2(\frac{\frac{6}{14}}{(3p-1)}+\frac{\frac{-1}{7}}{(p+2)})$ or

$$\begin{array}{}\text{(2)}& \frac{6}{7(3p-1)}-\frac{2}{7(p+2)}\end{array}$$

Doing partial fraction on the second term in (1) which is $\frac{3p}{(3p-1)(p+2)}$ gives

$$\begin{array}{rl}\frac{p}{(3p-1)(p+2)}& =\frac{A}{(3p-1)}+\frac{B}{(p+2)}=\frac{A(p+2)+B(3p-1)}{(3p-1)(p+2)}\\ A(p+2)+B(3p-1)& =p\\ Ap+2A+3Bp-B& =p\end{array}$$

Hence $2A-B=0$ and $(A+3B)=1$, therefore $A=\frac{B}{2}$. Hence $(\frac{B}{2}+3B)=1$ or $\frac{B+6B}{2}=1$ or $7B=2$ or $B=\frac{2}{7}$. Therefore $A=\frac{B}{2}=\frac{2}{14}$. Hence $\frac{3p}{(3p-1)(p+2)}=3(\frac{A}{(3p-1)}+\frac{B}{(p+2)})=3(\frac{\frac{2}{14}}{(3p-1)}+\frac{\frac{2}{7}}{(p+2)})$ or

$$\begin{array}{}\text{(3)}& \frac{3}{7(3p-1)}+\frac{6}{7(p+2)}\end{array}$$

Combining (2) and (3) gives

$$\begin{array}{rl}F(p)& =\frac{6}{7(3p-1)}-\frac{2}{7(p+2)}+\frac{3}{7(3p-1)}+\frac{6}{7(p+2)}\\ & =\frac{6}{7}\frac{1}{(3p-1)}-\frac{2}{7}\frac{1}{(p+2)}+\frac{2}{7}\frac{1}{(3p-1)}+\frac{6}{7}\frac{1}{(p+2)}\\ & =\frac{4}{7}\frac{1}{(p+2)}+\frac{9}{7}\frac{1}{(3p-1)}\\ & =\frac{4}{7}\frac{1}{(p+2)}+\frac{9}{21}\frac{1}{(p-\frac{1}{3})}\end{array}$$

Now we can use the table to find the inverse transform. Use property L2, which says $$\mathcal{L}\left(e^{-at}\right)=\frac{1}{p+a}$$ Hence , setting $a=2,$ gives $\mathcal{L}\left(e^{-2t}\right)=\frac{1}{p+2}$ and setting $a=-\frac{1}{3}$ gives $\mathcal{L}{\left(e^{-\frac{1}{3}t}\right)}^{\frac{1}{3}t}=\frac{1}{p-\frac{1}{3}}$. Hence $f(t)=\frac{4}{7}{e}^{-2t}+\frac{9}{21}{e}^{\frac{1}{3}t}$ and the inverse Laplace transform is $$f(t)=\frac{4}{7}{e}^{-2t}+\frac{3}{7}{e}^{\frac{1}{3}t}$$

3.12.2 chapter 15, problem 2.17

Problem Use L32 and L11 to obtain $\mathcal{L}(t^2\sin at)$

Solution

$$\begin{array}{}\text{(L32)}& \mathcal{L}(t^{n}f(t))& ={(-1)}^n\frac{d^{n}F(p)}{d{p}^n}\text{(L11)}& \mathcal{L}(t\sin at)& =\frac{2ap}{{(p^2+a^2)}^2}\end{array}$$

we set $f(t)=t\sin at$ then we can write using L32

$$\begin{array}{}\text{(1)}& \mathcal{L}(tf(t))=(-1)\frac{d\mathcal{L}(f(t))}{dp}\end{array}$$

But $\mathcal{L}(f(t))=\mathcal{L}(t\sin at)$  $=\frac{2ap}{{(p^2+a^2)}^2}$from table L11 (1) becomes

$$\begin{array}{rl}\mathcal{L}(tf(t))& =-\frac{d}{dp}\left(\frac{2ap}{{(p^2+a^2)}^2}\right)\\ \mathcal{L}(t\times t\sin at)& =-(p\frac{-2\times 2a}{{(p^2+a^2)}^3}\times 2p+\frac{2a}{{(p^2+a^2)}^2}\times 1)\\ \mathcal{L}(t^2\sin at)& =\frac{8a{p}^2}{{(p^2+a^2)}^3}-\frac{2a}{{(p^2+a^2)}^2}\\ & =\frac{8a{p}^2-2a(p^2+a^2)}{{(p^2+a^2)}^3}\\ & =\frac{a(8p^2-2p^2-2a^2)}{{(p^2+a^2)}^3}\\ & =\frac{a(6p^2-2a^2)}{{(p^2+a^2)}^3}\end{array}$$

Or

$$\mathcal{L}(t^2\sin at)=\frac{6a{p}^2-2a^3}{{(p^2+a^2)}^3}$$

3.12.3 chapter 15, problem 2.18

Problem Use L31 to derive L21

Solution

$$\begin{array}{}\text{(L31)}& \mathcal{L}\left(\frac{f(t)}{t}\right)& ={\int }_{u=p}^{\mathrm{\infty }}F(u)du\text{(L21)}& \mathcal{L}\left(\frac{e^{-at}-e^{-bt}}{t}\right)& =\ln \frac{p+b}{p+a}\end{array}$$

we set $f(t)=e^{-at}-e^{-bt}$ then we can write using L31

$$\begin{array}{rl}\mathcal{L}\left(\frac{f(t)}{t}\right)& ={\int }_{u=p}^{\mathrm{\infty }}F\left(u\right)du\\ & ={\int }_{u=p}^{\mathrm{\infty }}\mathcal{L}[f(t)]du\end{array}$$

but $\mathcal{L}[f(t)]=\mathcal{L}(e^{-at}-e^{-bt})=\mathcal{L}\left(e^{-at}\right)-\mathcal{L}\left(e^{-bt}\right)=\frac{1}{p+a}-\frac{1}{p+b}$  By using L2. But since we are using $u$ in place of $p$ in integral, we need to call $p=u$. Hence

$$\begin{array}{rl}\mathcal{L}\left(\frac{f(t)}{t}\right)& ={\int }_{u=p}^{\mathrm{\infty }}(\frac{1}{u+a}-\frac{1}{u+b})du\\ & ={\int }_{u=p}^{\mathrm{\infty }}\frac{1}{u+a}du-{\int }_{u=p}^{\mathrm{\infty }}\frac{1}{u+b}du\\ & ={[\ln (u+a)]}_p^{\mathrm{\infty }}-{[\ln (u+b)]}_p^{\mathrm{\infty }}\\ & =[\ln (\mathrm{\infty }+a)-\ln (p+a)]-[\ln (\mathrm{\infty }+b)-\ln (p+b)]\\ & =\ln \left(\mathrm{\infty }\right)-\ln (p+a)-\ln \left(\mathrm{\infty }\right)+\ln (p+b)\\ & =\ln (p+b)-\ln (p+a)\end{array}$$

but $\ln A-\ln B=\ln \frac{A}{B}$, therefore

$$\begin{array}{rl}\mathcal{L}\left(\frac{f(t)}{t}\right)& =\ln (p+b)-\ln (p+a)\\ \mathcal{L}\left(\frac{e^{-at}-e^{-bt}}{t}\right)& =\ln \left(\frac{p+b}{p+a}\right)\end{array}$$

3.12.4 chapter 15, problem 2.2

Problem Use relation L2 to find L7 and L8 in laplace table.

Solution

$$\begin{array}{}\text{(L2)}& \mathcal{L}\left(e^{-at}\right)=\frac{1}{p+a}\end{array}$$

for $\mathrm{\Re }(p+a)>0$

$$\begin{array}{}\text{(L7)}& \mathcal{L}\frac{e^{-at}-e^{-bt}}{b-a}=\frac{1}{(p+a)(p+b)}\end{array}$$

$$\begin{array}{}\text{(L8)}& \mathcal{L}\frac{a{e}^{-at}-b{e}^{-bt}}{a-b}=\frac{p}{(p+a)(p+b)}\end{array}$$

For $\mathrm{\Re }(p+a)>0$ and $\mathrm{\Re }(p+b)>0$. Where $\mathcal{L}$$f(t)$ is the laplace transform of $f(t)$ defined as $\mathcal{L}$$f(t)=F(p)={\int }_0^{\mathrm{\infty }}{e}^{-pt}f\left(t\right)dt.$

From the linearity property of the $\mathcal{L}$ operator, expand the LHS of L7, we get

$$\mathcal{L}\frac{e^{-at}-e^{-bt}}{b-a}=\frac{1}{b-a}\mathcal{L}(e^{-at}-e^{-bt})=\frac{1}{b-a}(\mathcal{L}\left(e^{-at}\right)-\mathcal{L}\left(e^{-bt}\right))$$

Now applying L2 gives for $\mathrm{\Re }(p+a)>0$ and $\mathrm{\Re }(p+b)>0$

$$\begin{array}{rl}\mathcal{L}\frac{e^{-at}-e^{-bt}}{b-a}& =\frac{1}{b-a}(\frac{1}{p+a}-\frac{1}{p+b})\\ & =\frac{1}{b-a}\left(\frac{p+b-(p+a)}{(p+a)(p+b)}\right)\\ & =\frac{1}{b-a}\left(\frac{b-a}{(p+a)(p+b)}\right)\\ & =\frac{1}{(p+a)(p+b)}\end{array}$$

For Which is L7 as required to show. Similarly for L8, expand the LHS of L8 we get for $\mathrm{\Re }(p+a)>0$ and $\mathrm{\Re }(p+b)>0$

$$\begin{array}{rl}\mathcal{L}\frac{a{e}^{-at}-b{e}^{-bt}}{a-b}& =\frac{1}{a-b}\mathcal{L}(a{e}^{-at}-b{e}^{-bt})\\ & =\frac{1}{a-b}(\mathcal{L}\left(e^{-at}\right)-\mathcal{L}\left(e^{-bt}\right))\\ & =\frac{1}{a-b}(a\mathcal{L}\left(e^{-at}\right)-b\mathcal{L}\left(e^{-bt}\right))\\ & =\frac{1}{a-b}(a\frac{1}{p+a}-b\frac{1}{p+b})\\ & =\frac{1}{a-b}\left(\frac{a(p+b)-b(p+a)}{(p+a)(p+b)}\right)\\ & =\frac{1}{a-b}\left(\frac{ap+ab-bp-ba}{(p+a)(p+b)}\right)\\ & =\frac{1}{a-b}\left(\frac{ap-bp}{(p+a)(p+b)}\right)\\ & =\frac{1}{a-b}\left(\frac{p(a-b)}{(p+a)(p+b)}\right)\\ & =\frac{p}{(p+a)(p+b)}\end{array}$$

Which is L8.

3.12.5 chapter 15, problem 2.21

Problem Use L29 and L11 to obtain $\mathcal{L}(t{e}^{-at}\sin bt)$

Solution

$$\begin{array}{}\text{(L29)}& \mathcal{L}(e^{-at}f(t))& =F(p+a)\text{(L11)}& \mathcal{L}(t\sin at)& =\frac{2ap}{{(p^2+a^2)}^2}\end{array}$$

Then from L11 we get

$$\mathcal{L}(t\sin bt)=\frac{2bp}{{(p^2+b^2)}^2}$$

Now, let $p=(p+a)$ then from L29, the above becomes

$$\mathcal{L}(e^{-at}t\sin bt)=\frac{2b(p+a)}{{({(p+a)}^2+b^2)}^2}$$

3.12.6 chapter 15, problem 2.22

Problem similar to problem 2.21, Use L29 and L12 to obtain $\mathcal{L}(t{e}^{-at}\cos bt)$

Solution

$$\begin{array}{}\text{(L29)}& \mathcal{L}(e^{-at}f(t))& =F(p+a)\text{(L12)}& \mathcal{L}(t\cos at)& =\frac{p^2-a^2}{{(p^2+a^2)}^2}\end{array}$$

then from L12 we get

$$\mathcal{L}(t\cos bt)=\frac{p^2-b^2}{{(p^2+b^2)}^2}$$

Now, let $p=(p+a)$ then from L29, the above becomes

$$\mathcal{L}(e^{-at}t\cos bt)=\frac{{(p+a)}^2-b^2}{{({(p+a)}^2+b^2)}^2}$$

3.12.7 chapter 15, problem 2.23

Problem use result obtained in problem 2.21 and 2.22 to find inverse transform for $\frac{p^2+2p-1}{{(p^2+4p+5)}^2}$

Solution

Recall, from 2.21 we showed that $\mathcal{L}(t{e}^{-at}\sin bt)=\frac{2b(p+a)}{{({(p+a)}^2+b^2)}^2}$ and from 2.22 $\mathcal{L}(t{e}^{-at}\cos bt)=\frac{{(p+a)}^2-b^2}{{({(p+a)}^2+b^2)}^2}$ Hence$$\begin{array}{rl}\mathcal{L}(t{e}^{-at}\cos bt)-\mathcal{L}(t{e}^{-at}\sin bt)& =\frac{{(p+a)}^2-b^2}{{({(p+a)}^2+b^2)}^2}-\frac{2b(p+a)}{{({(p+a)}^2+b^2)}^2}\\ & =\frac{{(p+a)}^2+b^2-2b(p+a)}{{({(p+a)}^2+b^2)}^2}\end{array}$$

Now Let $a=2$ and let $b=1$ we get

$$\begin{array}{rl}\mathcal{L}(t{e}^{-2t}\cos t)-\mathcal{L}(t{e}^{-2t}\sin t)& =\frac{{(p+2)}^2-1^2-2(p+2)}{{({(p+2)}^2+1^2)}^2}\\ \mathcal{L}(t{e}^{-2t}\cos t-t{e}^{-2t}\sin t)& =\frac{p^2+4+4p-1-2p-4}{{({(p+2)}^2+1^2)}^2}\\ & =\frac{p^2+2p-1}{{(p^2+4p+5)}^2}\end{array}$$

Hence inverse transform of $\frac{p^2+2p-1}{{(p^2+2p+5)}^2}$ is $t{e}^{-2t}\cos t-t{e}^{-2t}\sin t$ or $t{e}^{-2t}(\cos t-\sin t)$

3.12.8 chapter 15, problem 2.3

Problem Using either relation L2 or L3 and L4, verify L9 and L10 in laplace table.

Solution

$$\begin{array}{}\text{(L2)}& \mathcal{L}\left(e^{-at}\right)=\frac{1}{p+a}\mathrm{Re}(p+a)>0\end{array}$$

$$\begin{array}{}\text{(L3)}& \mathcal{L}\sin at=\frac{a}{p^2+a^2}\mathrm{Re}(p)>\left|\mathrm{Im}a\right|\end{array}$$

$$\begin{array}{}\text{(L4)}& \mathcal{L}\cos at=\frac{p}{p^2+a^2}\mathrm{Re}(p)>\left|\mathrm{Im}a\right|\end{array}$$

$$\begin{array}{}\text{(L9)}& \mathcal{L}\sinh at=\frac{a}{p^2-a^2}\mathrm{Re}(p)>|\mathrm{Re}(a)|\end{array}$$

$$\begin{array}{}\text{(L10)}& \mathcal{L}\cosh at=\frac{p}{p^2-a^2}\mathrm{Re}(p)>|\mathrm{Re}(a)|\end{array}$$

Where $\mathcal{L}(f(t))$ is the laplace transform of $f\left(t\right)$ defined as $\mathcal{L}(f(t))=Y(s)={\int }_0^{\mathrm{\infty }}{e}^{-pt}f(t)dt$. To derive L9, use the relation that

$$i\sinh (x)=\sin \left(ix\right)$$

Hence, using L3, we get

$$\begin{array}{rl}\mathcal{L}(i\sinh at)& =\mathcal{L}(\sin iat)\\ i\mathcal{L}(\sinh at)& =\frac{ia}{p^2+{\left(ia\right)}^2}\mathrm{Re}(p)>\left|\mathrm{Im}a\right|\\ \mathcal{L}(\sinh at)& =\frac{a}{p^2-a^2}\mathrm{Re}(p)>|\mathrm{Re}(a)|\end{array}$$

which is L9. To find L10, use the relation

$$\cosh (x)=\cos \left(ix\right)$$

And using L4, we get

$$\begin{array}{rl}\mathcal{L}(\cosh \left(at\right))& =\mathcal{L}(\cos \left(iat\right))\\ \mathcal{L}(\cosh at)& =\frac{p}{p^2+{\left(ia\right)}^2}\mathrm{Re}(p)>\left|\mathrm{Im}a\right|\\ \mathcal{L}(\sinh at)& =\frac{p}{p^2-a^2}\mathrm{Re}(p)>|\mathrm{Re}(a)|\end{array}$$

Which is L10.

3.12.9 chapter 15, problem 2.4

Problem by differentiating the appropriate formulas w.r.t. ’a’, verify L12

Solution

L12 is

$$\begin{array}{}\text{(L12)}& \mathcal{L}(t\cos t)=\frac{p^2-a^2}{{(p^2+a^2)}^2}\mathrm{Re}\left(p\right)>\left|\mathrm{Im}a\right|\end{array}$$

Where $\mathcal{L}(f(t))$ is the laplace transform of $f\left(t\right)$ defined as$\mathcal{L}(f(t))=F(p)={\int }_0^{\mathrm{\infty }}{e}^{-pt}f(t)dt$. To derive this, I start with L3, which says

$$\mathcal{L}(\sin at)=\frac{a}{p^2+a^2}\mathrm{Re}(p)>\left|\mathrm{Im}a\right|$$

The above can be rewritten in the full definition of the transform to make it easier to see

$${\int }_0^{\mathrm{\infty }}{e}^{-pt}\sin atdt=\frac{a}{p^2+a^2}$$

Taking derivative of both sides w.r.t. $a$ gives

$$\begin{array}{rl}\frac{d}{da}[{\int }_0^{\mathrm{\infty }}{e}^{-pt}\sin atdt]& =\frac{d}{da}\left[\frac{a}{p^2+a^2}\right]\\ {\int }_0^{\mathrm{\infty }}\frac{d}{da}[e^{-pt}\sin at]dt& =a\left(\frac{-2a}{{(p^2+a^2)}^2}\right)+\frac{1}{p^2+a^2}\times 1\\ {\int }_0^{\mathrm{\infty }}{e}^{-pt}t\cos atdt& =\frac{-2a^2}{{(p^2+a^2)}^2}+\frac{1}{p^2+a^2}\\ \mathcal{L}(t\cos at)& =\frac{-2a^2+(p^2+a^2)}{{(p^2+a^2)}^2}\\ & =\frac{p^2-a^2}{{(p^2+a^2)}^2}\end{array}$$

Which is L12

3.12.10 chapter 15, problem 2.5

Problem by integrating the appropriate formulas w.r.t. ’a’, verify L19

Solution L19 is

$$\begin{array}{}\text{(L19)}& \mathcal{L}\left(\frac{\sin at}{t}\right)=\arctan \frac{a}{p}\mathrm{Re}(p)>\left|\mathrm{Im}a\right|\end{array}$$

Where $\mathcal{L}(f(t))$ is the laplace transform of $f\left(t\right)$ defined as $\mathcal{L}(f(t))=F(p)={\int }_0^{\mathrm{\infty }}{e}^{-pt}f(t)dt$. To derive this, we start with L4, which says

$$\mathcal{L}(\cos at)=\frac{p}{p^2+a^2}\mathrm{Re}(p)>\left|\mathrm{Im}a\right|$$

The above can be rewritten in the full definition of the transform to make it easier to see

$${\int }_0^{\mathrm{\infty }}{e}^{-pt}\cos atdt=\frac{p}{p^2+a^2}$$

Integrating both sides w.r.t. $a$ gives

$$\begin{array}{rl}\int ({\int }_{t=0}^{t=\mathrm{\infty }}{e}^{-pt}\cos atdt)da& =\int \frac{p}{p^2+a^2}da\\ {\int }_{t=0}^{t=\mathrm{\infty }}(\int e^{-pt}\cos atda)dt& =\arctan \frac{a}{p}+K\\ {\int }_{t=0}^{t=\mathrm{\infty }}{e}^{-pt}(\int \cos atda)dt& =\arctan \frac{a}{p}+K\end{array}$$

Now, since $\int \cos atda=\frac{\sin at}{t}+k$ and choose zero for values of the K’s, the constants of integration gives

$${\int }_0^{\mathrm{\infty }}{e}^{-pt}\left(\frac{\sin at}{t}\right)dt=\arctan \frac{a}{p}$$

Hence

$$\mathcal{L}\left(\frac{\sin at}{t}\right)=\arctan \frac{a}{p}$$

Which is L19

3.12.11 chapter 15, problem 2.9

Problem Find the inverse transform of the function $F\left(p\right)=\frac{5-2p}{p^2+p-2}$

Solution

Need to simplify the above expression to some expressions which are shown in the table on page 636.

$$\begin{array}{rl}F(p)& =\frac{5-2p}{p^2+p-2}\\ & =\frac{5}{p^2+p-2}-\frac{2p}{p^2+p-2}\\ \text{(1)}& & =\frac{5}{(p-1)(p+2)}-\frac{2p}{(p-1)(p+2)}\end{array}$$

From table, L7, we see that $\mathcal{L}\left(\frac{e^{-at}-e^{-bt}}{b-a}\right)=F(p)=\frac{1}{(p+a)(p+b)}$. Setting $a=-1,b=2$ we get the first expression in (1), that is

$$\begin{array}{}\text{(2)}& \mathcal{L}\left(\frac{e^t-e^{-2t}}{3}\right)=\frac{1}{(p-1)(p+2)}\end{array}$$

also from table, we see that L8 is $\mathcal{L}\left(\frac{a{e}^{-at}-b{e}^{-bt}}{a-b}\right)=F(p)=\frac{p}{(p+a)(p+b)}$. Hence, letting $a=-1,b=2$ we get the second expression in (1), that is

$$\begin{array}{}\text{(3)}& \mathcal{L}\left(\frac{-e^{+t}-2e^{-2t}}{-3}\right)=\frac{p}{(p-1)(p+2)}\end{array}$$

So combine (2) and (3) we get

$$5\mathcal{L}\left(\frac{e^{+t}-e^{-2t}}{2+1}\right)-2\mathcal{L}\left(\frac{-e^{+t}-2e^{-2t}}{-1-2}\right)=\frac{5}{(p-1)(p+2)}-\frac{2p}{(p-1)(p+2)}$$

which is (1). Hence

$$\begin{array}{rl}f(t)& =5\frac{e^{+t}-e^{-2t}}{3}-2\frac{-e^{+t}-2e^{-2t}}{-3}\\ & =5\frac{e^{+t}-e^{-2t}}{3}+2\frac{-e^{+t}-2e^{-2t}}{3}\\ & =\frac{5e^{+t}-5e^{-2t}-2e^{+t}-4e^{-2t}}{3}\\ & =\frac{3e^t-9e^{-2t}}{3}\\ & =e^t-3e^{-2t}\end{array}$$

Hence the inverse transform of $F(p)$ is $e^t-3e^{-2t}$

3.12.12 chapter 15, problem 3.11

Problem Use laplace transform to solve $y^{\prime \prime }-4y=4e^{2t}$ ,$y_0=0,y_0^{\prime }=1$

Solution

Let $\mathcal{L}(y(t))=Y(p)$, and taking the laplace transform of both sides, noting first that $\mathcal{L}\left(y^{\prime \prime }\right)=p^{2}Y-p{y}_0-y_0^{\prime }$ then we get

$$\begin{array}{rl}\mathcal{L}\left(y^{\prime \prime }\right)-4\mathcal{L}(y(t))& =\mathcal{L}\left(4e^{2t}\right)\\ (p^{2}Y-p{y}_0-y_0^{\prime })-4Y& =4\mathcal{L}\left(e^{2t}\right)\end{array}$$

L2 from table on page 636 : $\mathcal{L}\left(e^{-at}\right)=\frac{1}{p+a}$, hence $\mathcal{L}\left(e^{2t}\right)=\frac{1}{p-2}$, hence, after applying boundary conditions, we get

$$\begin{array}{rl}(p^{2}Y-1)-4Y& =4\frac{1}{p-2}\\ Y(p^2-4)-1& =4\frac{1}{p-2}\\ Y& =4\frac{1}{(p^2-4)(p-2)}+\frac{1}{(p^2-4)}\\ Y& =4\frac{1}{(p-2)(p+2)(p-2)}+\frac{1}{(p^2-4)}\\ Y& =4\frac{1}{(p+2){(p-2)}^2}+\frac{1}{(p^2-4)}\end{array}$$

Doing partial fractions, repeated roots, gives

$$\begin{array}{rl}\frac{1}{(p+2){(p-2)}^2}& =\frac{A}{(p+2)}+\frac{B}{(p-2)}+\frac{C}{{(p-2)}^2}\\ 1& =A{(p-2)}^2+B(p+2)(p-2)+C(p+2)\\ 1& =A(p^2-4p+4)+B(p^2-4)+C(p+2)\\ 1& =p^2(A+B)+p(-4A+C)+4A-4B+2C\end{array}$$

Hence

$$\begin{array}{rl}A+B& =0\\ -4A+C& =0\\ 4A-4B+2C& =1\end{array}$$

hence $A=-B$, then $-4B-4B+2C=1$ or $-8B+2C=1$ or $C=\frac{1+8B}{2}$, therefore $4B+\frac{1+8B}{2}=0$ then $B=-\frac{1}{16}$, then $A=\frac{1}{16},C=\frac{1+8(-\frac{1}{16})}{2}=\frac{1-\left(\frac{1}{2}\right)}{2}=\frac{1}{4}$. Hence

$$\begin{array}{rl}Y& =4\frac{1}{(p+2){(p-2)}^2}+\frac{1}{(p^2-4)}\\ & =4(\frac{A}{(p+2)}+\frac{B}{(p-2)}+\frac{C}{{(p-2)}^2})+\frac{1}{(p^2-4)}\\ & =4(\frac{\frac{1}{16}}{(p+2)}+\frac{-\frac{1}{16}}{(p-2)}+\frac{\frac{1}{4}}{{(p-2)}^2})+\frac{1}{(p^2-4)}\\ & =\frac{1}{4}\frac{1}{(p+2)}-\frac{1}{4}\frac{1}{(p-2)}+\frac{1}{{(p-2)}^2}+\frac{1}{(p^2-4)}\\ & =\frac{1}{4}\frac{1}{(p+2)}-\frac{1}{4}\frac{1}{(p-2)}+\frac{1}{{(p-2)}^2}+(\frac{1}{4}\frac{1}{(p-2)}-\frac{1}{4}\frac{1}{(p+2)})\\ & =\frac{1}{{(p-2)}^2}\end{array}$$

$\frac{1}{{(p-2)}^2}\to$ using L6, we have $\mathcal{L}\left(t^k{e}^{-at}\right)=\frac{k!}{{(p+a)}^{k+1}}$, here $a=-2,k=1$ and $\frac{1}{{(p-2)}^2}\to t{e}^{2t}$. Hence ,

$$f(t)=t{e}^{2t}$$

3.12.13 chapter 15, problem 3.24

Problem Use laplace transform to solve $y^{\prime \prime }-2y^{\prime }+y=2\cos t$ ,$y_0=5,y_0^{\prime }=-2$

Solution

Let $\mathcal{L}(y(t))=Y(p)$, Take the laplace transform of both sides, noting first that $\mathcal{L}\left(y^{\prime \prime }\right)=p^{2}Y-p{y}_0-y_0^{\prime }$ , $\mathcal{L}\left(y^{\prime }\right)=pY-y_0$then we get

$$\begin{array}{rl}\mathcal{L}\left(y^{\prime \prime }\right)-2\mathcal{L}\left(y^{\prime }\right)+\mathcal{L}(y(t))& =\mathcal{L}(2\cos t)\\ (p^{2}Y-p{y}_0-y_0^{\prime })-2(pY-y_0)+Y& =2\frac{p}{p^2+1}\end{array}$$

$$\begin{array}{rl}(p^{2}Y-5p+2)-2(pY-5)+Y& =2\frac{p}{p^2+1}\\ Y(p^2-2p+1)+12-5p& =2\frac{p}{p^2+1}\\ Y(p^2-2p+1)& =\frac{2p}{p^2+1}-12+5p\\ Y& =\frac{\frac{2p}{p^2+1}-12+5p}{(p^2-2p+1)}\\ Y& =\frac{2p}{(p^2+1)(p^2-2p+1)}+\frac{5p-12}{(p^2-2p+1)}\\ Y& =\frac{2p}{(p^2+1){(p-1)}^2}+\frac{5p-12}{{(p-1)}^2}\end{array}$$

Doing partial fractions $\frac{2p}{{(p-1)}^2(p^2+1)}=\frac{A}{(p-1)}+\frac{B}{{(p-1)}^2}+\frac{Cp+D}{(p^2+1)}$. Solving, we get $A=0,B=1,C=0,D=-1$ Hence

$$\begin{array}{rl}Y& =\frac{1}{{(p-1)}^2}-\frac{1}{(p^2+1)}+\frac{5p-12}{{(p-1)}^2}\\ & =\frac{1}{{(p-1)}^2}-\frac{1}{(p^2+1)}+\frac{5p}{{(p-1)}^2}-12\frac{1}{{(p-1)}^2}\\ & =\frac{-11}{{(p-1)}^2}-\frac{1}{(p^2+1)}+\frac{5p}{{(p-1)}^2}\end{array}$$

Hence , using table, we get inverse laplace transform $\frac{1}{{(p-1)}^2}\to t{e}^t$ and $\frac{1}{(p^2+1)}\to \sin t$ and $\frac{p}{{(p-1)}^2}\to e^t+t{e}^t$, hence

$$\begin{array}{rl}f(t)& =-11t{e}^t-\sin t+5(e^t+t{e}^t)\\ & =-11t{e}^t-\sin t+5e^t+5t{e}^t\\ & =5e^t-6t{e}^t-\sin t\end{array}$$

3.12.14 chapter 15, problem 3.25

Problem Use laplace transform to solve $y^{\prime \prime }+4y^{\prime }+5y=2e^{-2t}\cos t,$ ,$y_0=0,y_0^{\prime }=3$

Solution

Let $\mathcal{L}(y(t))=Y(p)$, Taking the laplace transform of both sides, noting first that $\mathcal{L}(y^{\prime \prime }(t))=p^{2}Y-p{y}_0-y_0^{\prime }$ , $\mathcal{L}(y^{\prime }(t))=pY-y_0$then we get

$$\begin{array}{rl}\mathcal{L}(y^{\prime \prime }(t))+4\mathcal{L}(y^{\prime }(t))+5\mathcal{L}(y(t))& =\mathcal{L}(2e^{-2t}\cos t)\\ (p^{2}Y-p{y}_0-y_0^{\prime })+4(pY-y_0)+5Y& =2\mathcal{L}(e^{-2t}\cos t)\end{array}$$

Applying initial conditions

$$(p^{2}Y-3)+4\left(pY\right)+5Y=2\mathcal{L}(e^{-2t}\cos t)$$

From table using L14 $\mathcal{L}(e^{-2t}\cos t)=\frac{p+2}{{(p+2)}^2+1}$. Hence

$$\begin{array}{rl}(p^{2}Y-3)+4\left(pY\right)+5Y& =2\frac{p+2}{{(p+2)}^2+1}\\ Y(p^2+4p+5)-3& =2\frac{p+2}{{(p+2)}^2+1}\\ Y& =\frac{2\frac{p+2}{{(p+2)}^2+1}+3}{(p^2+4p+5)}\\ Y& =2\frac{p+2}{({(p+2)}^2+1)(p^2+4p+5)}+3\frac{1}{(p^2+4p+5)}\end{array}$$

but $(p^2+4p+5)={(p+2)}^2+1$, therefore

$$\begin{array}{rl}Y& =2\frac{p+2}{{({(p+2)}^2+1)}^2}+\frac{3}{{(p+2)}^2+1}\\ & =2\frac{p+2}{({(p+2)}^2+1)({(p+2)}^2+1)}+\frac{3}{{(p+2)}^2+1}\end{array}$$

But inverse transform of $\frac{p+2}{{({(p+2)}^2+1)}^2}=\frac{1}{2}t{e}^{-2t}\sin t$ and inverse transform of $\frac{1}{{(p+2)}^2+1}=e^{-2t}\sin t$. Hence

$$\begin{array}{rl}f(t)& =2(\frac{1}{2}t{e}^{-2t}\sin t)+3(e^{-2t}\sin t)\\ & =t{e}^{-2t}\sin t+3e^{-2t}\sin t\\ & =(t+3)e^{-2t}\sin t\end{array}$$

3.12.15 chapter 15, problem 3.29

Problem Use laplace transform to solve $y^{\prime }+z^{\prime }-2y=1,$ $y_0=z_0=1,z-y^{\prime }=t$

Solution

Taking laplace transform of both equations, then we get 2 equations in $Y$ and $Z$, then solve for them. Let $\mathcal{L}(y(t))=Y(p)$, $\mathcal{L}(z(t))=Z(p)$

$$\begin{array}{rl}\mathcal{L}(y^{\prime }(t))+\mathcal{L}(z^{\prime }(t))-2\mathcal{L}(y(t))& =\mathcal{L}(1)\\ \mathcal{L}(z(t))-\mathcal{L}(y^{\prime }(t))& =\mathcal{L}(t)\end{array}$$

Then we get

$$\begin{array}{rl}(pY-y_0)+(pZ-z_0)-2Y& =\frac{1}{p}\\ Z-(pY-y_0)& =\frac{1}{p^2}\end{array}$$

Putting initial conditions gives

$$\begin{array}{rl}(pY-1)+(pZ-1)-2Y& =\frac{1}{p}\\ Z-(pY-1)& =\frac{1}{p^2}\end{array}$$

$$\begin{array}{}\text{(1)}& Y(p-2)+pZ-2& =\frac{1}{p}\text{(2)}& Z-pY+1& =\frac{1}{p^2}\end{array}$$

Solving for $Y,$From (1), $Z=\frac{\frac{1}{p}+2-Y(p-2)}{p}$ , and substituting into (2) gives

$$\begin{array}{rl}\frac{\frac{1}{p}+2-Y(p-2)}{p}-pY+1& =\frac{1}{p^2}\\ \frac{1}{p}+2-Y(p-2)-p^{2}Y+p& =\frac{1}{p}\\ 2-Y(p-2)-p^{2}Y& =-p\\ Y(-p+2-p^2)& =-p-2\\ Y& =\frac{-p-2}{(-p+2-p^2)}\\ Y& =-\frac{p+2}{(-p+1)(p+2)}\\ Y& =-\frac{1}{(-p+1)}\end{array}$$

Hence $Y=\frac{1}{p-1}$ so from L2

$$y(t)=e^t$$

Now, that we have $Y,$ we solve for $Z.$ From (2)

$$\begin{array}{rl}Z-pY+1& =\frac{1}{p^2}\\ Z-p\left(\frac{1}{p-1}\right)+1& =\frac{1}{p^2}\\ Z& =\frac{1}{p^2}-1+\frac{p}{p-1}\\ Z& =\frac{p-1-p^2(p-1)+p^3}{p^2(p-1)}\\ Z& =\frac{p-1+p^2}{p^2(p-1)}\\ Z& =\frac{p-1+p^2}{p^2(p-1)}\end{array}$$

Doing partial fraction on the above, we get $Z=\frac{1}{p^2}+\frac{1}{p-1}$, Hence

$$z(t)=t+e^t$$

3.12.16 chapter 15, problem 3.30

Problem Use laplace transform to solve $y^{\prime }+2z=1,$ $y_0=0,2y-z^{\prime }=2t,z_0=1$

Solution

Take laplace transform of both equations, then we get 2 equations in $Y$ and $Z$, then solve for them.

Let $\mathcal{L}(y(t))=Y(p)$, $\mathcal{L}(z(t))=Z(p)$

$$\begin{array}{rl}\mathcal{L}(y^{\prime }(t))+2Z& =\mathcal{L}(1)\\ 2Y-\mathcal{L}(z^{\prime }(t))& =\mathcal{L}\left(2t\right)\end{array}$$

$$\begin{array}{rl}pY-y_0+2Z& =\mathcal{L}(1)\\ 2Y-(pZ-z_0)& =\mathcal{L}\left(2t\right)\end{array}$$

Then we get, by putting $z_0=1,y_0=0$

$$\begin{array}{}\text{(1)}& pY+2Z& =\frac{1}{p}\text{(2)}& 2Y-(pZ-1)& =\frac{2}{p^2}\end{array}$$

Obtain Z from first equation and sub into the second to solve for $Y$, $Z=\frac{\frac{1}{p}-pY}{2}$, Hence

$$\begin{array}{rl}2Y-(p\left(\frac{\frac{1}{p}-pY}{2}\right)-1)& =\frac{2}{p^2}\\ 2Y-\frac{p}{2}(\frac{1}{p}-pY)+1& =\frac{2}{p^2}\\ 2Y-\frac{p}{2}\left(\frac{1-p^{2}Y}{p}\right)& =\frac{2}{p^2}-1\\ 2Y-\frac{1}{2}(1-p^{2}Y)& =\frac{2-p^2}{p^2}\\ 2Y-\frac{1}{2}+\frac{1}{2}{p}^{2}Y& =\frac{2-p^2}{p^2}\\ Y(2+\frac{p^2}{2})& =\frac{2-p^2}{p^2}+\frac{1}{2}\\ Y& =\frac{\frac{4-p^2}{2p^2}}{\left(\frac{4+p^2}{2}\right)}\\ Y& =\frac{4-p^2}{p^2(4+p^2)}\end{array}$$

Hence , using partial fraction gives

$$\begin{array}{rl}Y& =\frac{4+p^2}{p^2(4+p^2)}\\ & =\frac{Ap+B}{p^2}+\frac{Cp+D}{(4+p^2)}\end{array}$$

Then

$$$$\begin{array}{rl}(Ap+B)(4+p^2)+(Cp+D)p^2& =4-p^2\\ 4Ap+A{p}^3+4B+B{p}^2+C{p}^3+D{p}^2& =4-p^2\\ p^3(A+C)+p^2(B+D)+p\left(4A\right)+4B& =4-p^2\end{array}$$

Hence, $4B=4\to B=1$ and $4A=0\to A=0$ and $B+D=-1$ and $A+C=0\to C=0$, therefore $D=-1-B\to D=-2$. Hence

$$\begin{array}{rl}Y& =\frac{Ap+B}{p^2}+\frac{Cp+D}{(4+p^2)}\\ & =\frac{1}{p^2}-\frac{2}{(4+p^2)}\end{array}$$

Using tables for inverse transform gives

$$y(t)=t-\sin 2t$$

Now, to find $z(t)$, subtituting value we found for Y into equation (1) above.

$$\begin{array}{rl}pY+2Z& =\frac{1}{p}\\ p(\frac{1}{p^2}-\frac{2}{(4+p^2)})+2Z& =\frac{1}{p}\\ \frac{1}{p}-\frac{2p}{(4+p^2)}+2Z& =\frac{1}{p}\\ Z& =\frac{p}{(4+p^2)}\\ Z& =\frac{p}{(4+p^2)}\end{array}$$

From tables, using L4

$$z(t)=\cos 2t$$

3.12.17 chapter 15, problem 3.4

Problem Use laplace transform to solve $y^{\prime \prime }+y=\sin t$ ,$y_0=1,y_0^{\prime }=0$

Solution

Let $\mathcal{L}(y(t))=Y(p)$, Take the laplace transform of both sides, noting first that $\mathcal{L}(y^{\prime \prime }(t))=p^{2}Y-p{y}_0-y_0^{\prime }$ then we get

$$\begin{array}{rl}\mathcal{L}(y^{\prime \prime }(t))+\mathcal{L}(y(t))& =\mathcal{L}(\sin t)\\ (p^{2}Y-p{y}_0-y_0^{\prime })+Y& =\frac{1}{p^2+1}\end{array}$$

Where I used L3 from table on page 636 which says that $\mathcal{L}(\sin at)=\frac{a}{p^2+a^2}$. Now solving for $Y,$ noting that $y_0=1$ and $y_0^{\prime }=0$ gives

$$\begin{array}{rl}(p^{2}Y-p)+Y& =\frac{1}{p^2+1}\\ Y(p^2+1)-p& =\frac{1}{p^2+1}\\ Y(p^2+1)& =\frac{1}{p^2+1}+p\\ \text{(1)}& Y& =\frac{1}{{(p^2+1)}^2}+\frac{p}{(p^2+1)}\end{array}$$

From table using L12, $\frac{p}{(p^2+1)}\to \cos t$ and using L17, $\frac{1}{{(p^2+1)}^2}\to \frac{1}{2}(\sin t-t\cos t)$. Hence, putting these together into (1) gives

$$f(t)=\cos t+\frac{1}{2}(\sin t-t\cos t)$$

This is the particular solution to the ODE.

3.12.18 chapter 15, problem 3.6

Problem Use laplace transform to solve $y^{\prime \prime }-6y^{\prime }+9y=t{e}^{3t}$ ,$y_0=0,y_0^{\prime }=5$

Solution

Let $\mathcal{L}(y(t))=Y(p)$, Take the laplace transform of both sides, noting first that $\mathcal{L}(y^{\prime \prime }(t))=p^{2}Y-p{y}_0-y_0^{\prime }$ and $\mathcal{L}(y^{\prime }(t))=pY-y_0$ then we get

$$\begin{array}{rl}\mathcal{L}(y^{\prime \prime }(t))-6\mathcal{L}(y^{\prime }(t))+9\mathcal{L}(y(t))& =\mathcal{L}\left(t{e}^{3t}\right)\\ (p^{2}Y-p{y}_0-y_0^{\prime })-6(pY-y_0)+9Y& =\mathcal{L}\left(t{e}^{3t}\right)\end{array}$$

I use L6 from table on page 636 which says that $\mathcal{L}\left(t^k{e}^{-at}\right)=\frac{k!}{{(p+a)}^{k+1}},$ hence for $k=1,a=-3$, we get $\mathcal{L}\left(t^k{e}^{-at}\right)=\frac{1}{{(p-3)}^2}$

$$(p^{2}Y-p{y}_0-y_0^{\prime })-6(pY-y_0)+9Y=\frac{1}{{(p-3)}^2}$$

Applying boundary conditions gives

$$\begin{array}{rl}(p^{2}Y-5)-6\left(pY\right)+9Y& =\frac{1}{{(p-3)}^2}\\ Y(p^2-6p+9)-5& =\frac{1}{{(p-3)}^2}\\ Y(p^2-6p+9)& =\frac{1}{{(p-3)}^2}+5\\ Y& =\frac{1}{(p^2-6p+9){(p-3)}^2}+\frac{5}{(p^2-6p+9)}\\ Y& =\frac{1}{{(p-3)}^2{(p-3)}^2}+\frac{5}{{(p-3)}^2}\\ \text{(1)}& Y& =\frac{1}{{(p-3)}^4}+\frac{5}{{(p-3)}^2}\end{array}$$

Now using table, from L6, $\frac{1}{{(p-3)}^2},$ let $a=-3,k=1$ hence $\frac{1}{{(p-3)}^2}\to t{e}^{3t}$ And using L6 again, $\frac{1}{{(p-3)}^4},$ let $k=3,a=-3$ then $\frac{6}{{(p-3)}^4}\to t^3{e}^{3t}$, therefore (1) becomes

$$\begin{array}{rl}f(t)& =\frac{1}{6}{t}^3{e}^{3t}+5t{e}^{3t}\\ & =e^{3t}(\frac{1}{6}{t}^3+5t)\end{array}$$

3.12.19 chapter 15, problem 3.8

Problem Use laplace transform to solve $y^{\prime \prime }+16y=8\cos 4t$ ,$y_0=0,y_0^{\prime }=0$

Solution

Let $\mathcal{L}(y(t))=Y(p)$, Taking the laplace transform of both sides, noting first that $\mathcal{L}(y^{\prime \prime }(t))=p^{2}Y-p{y}_0-y_0^{\prime }$ results in

$$\begin{array}{rl}\mathcal{L}(y^{\prime \prime }(t))+16\mathcal{L}(y(t))& =\mathcal{L}(8\cos 4t)\\ (p^{2}Y-p{y}_0-y_0^{\prime })+16Y& =8\mathcal{L}(\cos 4t)\\ Y(p^2+16)& =8\frac{p}{p^2+16}\end{array}$$

I used L6 from table on page 636 : $\mathcal{L}(\cos at)=\frac{p}{p^2+a^2}$

$$Y=8\frac{p}{{(p^2+16)}^2}$$

Now looking at L11, which says $\mathcal{L}(t\sin at)=\frac{2ap}{{(p^2+a^2)}^2}$, hence letting $a=4$ gives the solution

$$f(t)=t\sin 4t$$