4.1.1: För funktionen $LaTeX: e^{t^2}$ divergerar integralen för Laplacetransformen för alla värden på LaTeX: s .

4.1.2: Svar: $LaTeX: \frac{1-e^{-2\pi s}}{1+s^2}$ . Ledning: $LaTeX: \mathcal{L}[f](s)= \int_0^{2\pi}e^{-st}f(t)dt+\cancelto{0}{\int_{2\pi}^\infty e^{-st}f(t)dt}=\{\text{partiellintegrera 2 ggr.\}}=\left[-e^{-st}\cos t\right]_0^{2\pi}-\cancelto{0}{\left[se^{-st}\sin t\right]_0^{2\pi}}-s^2\int_0^{2\pi}e^{-st}\sin t dt.$ Lös sedan ut $LaTeX: \int_0^{2\pi}e^{-st}\sin t dt$ från ekvationen.

4.1.3: Bevis: Låt integranden i Laplanceintegralen heta LaTeX: i . Då har vi $LaTeX: \mathcal{L}[f](s)=\int_0^\infty idt=\int_0^pidt+\int_p^{2p}idt+\int_{2p}^{3p}idt+...=\sum_{n=0}^\infty\int_{np}^{(n+1)p}idt.$ Utför nu en variabelsusbtitution från LaTeX: t till $LaTeX: \tau$ enligt $LaTeX: \tau=t-np$ . Då får vi $LaTeX: \mathcal{L}[f](s)=\sum_{n=0}^\infty\int_{0}^{p}e^{-s\tau-snp}f({\tau+np})d\tau.$ Använd nu att LaTeX: f är LaTeX: p- periodisk, dvs. $LaTeX: f(\tau+np)=f(\tau)$ , och flytta ut den konstanta termen $LaTeX: e^{-snp}$ ur integralen. Vi noterar att integralen ej längre beror på LaTeX: n , och kan skriva att $LaTeX: \mathcal{L}[f](s)=\int_{0}^{p}e^{-s\tau}f({\tau})d\tau\cdot\sum_{n=0}^\infty (e^{-sp})^n.$ Den senare termen är summan av en oändlig geometrisk serie som konvergerar till $LaTeX: \frac{1}{1-e^{-sp}}$ . Vi är klara. Not: Andra lösningssätt finns här.

4.1.4: Bevis: Använd resultatet från den föregående uppgiften för att erhålla $LaTeX: \mathcal{L}[f](s)=\frac{1}{1-e^{-4s}}\int_0^4e^{-st}f(t)dt=\frac{1}{1-e^{-4s}}\left(\int_0^2e^{-st}dt-\int_2^4e^{-st}dt\right)$ . Beräkning ger $LaTeX: \frac{1}{s}\frac{e^{-4s}-2e^{-2s}+1}{1-e^{-4s}}$ . Multiplicera nu kvoten med $LaTeX: \frac{e^{2s}}{e^{2s}}$ och notera att den kan skrivas $LaTeX: \frac{1 }{s }\frac{e^{-2s}-2+e^{2s} }{e^{2s}-{e^-2s} }=\frac{1}{s}\frac{(e^s-e^{-s})^\cancel{2}}{(e^s+e^{-s})\cancel{(e^s-e^{-s})}}=\frac{\tanh s}{s}$ .

4.2.1a: $LaTeX: \frac{s+1}{4+\left(s+1\right)^2}$ ,

4.2.1b: $LaTeX: \mathcal{L}[e^{-4t}\cosh(2t)]=\mathcal{L}[\cosh(2t)](s+4)=\frac{s+4}{(s+4)^2-4}$

4.2.1c: $LaTeX: \frac{24 }{ s^5}+\frac{ 4}{ s^3}+\frac{ 1}{ s}$ (Ledning: Använd linearitetsegenskaper),

4.2.1d: $LaTeX: \mathcal{L}[3\cosh(t)-4\sinh(5t)]=\frac{3s}{s^2-1}-\frac{20}{s^2-25}$

4.2.1e: $LaTeX: \left(-1\right)^n\frac{d^n}{ds^{^n}}\left(\frac{1}{s^2+1}\right)$ (Formel 7 på Laplacebladet)

4.2.2: $LaTeX: \frac{d}{ds}\mathcal{L}[f](s)=-\mathcal{L}[tf(t)](s)=-\mathcal{L}[\sin(t)](s)=-\frac{1}{1+s^2}$ . First order differential equation $LaTeX: Y^{\prime}\left(s\right)=-\frac{1}{1+s^2}$ with initial condition $LaTeX: Y\left(0\right)=\int_0^{\infty}\frac{\sin(t)}{t}\;dt=\frac{\pi}{2}$ has solution $LaTeX: Y\left(s\right)=-\arctan(s)+\frac{\pi}{2}$ .

4.2.3: We know that $LaTeX: \frac{1}{s^{2}-a^{2}} = \frac{1}{a} \frac{a}{s^{2}-a^{2}}$ , and since we know $LaTeX: \mathcal{L}[\sinh(at))]=\frac{a}{s^2-a^2}$ , we can use formula 16: $LaTeX: \frac{1}{a} \frac{d^n}{ds^n}\mathcal{L}[\sinh(at)]=\mathcal{L}[\frac{1}{a} (-t)^n\sinh(at)]$ , i.e $LaTeX: f(t) = \frac{1}{a} (-t)^n\sinh(at)$

4.2.4a: Use the formula $LaTeX: \sin^2(t)=-\frac{1}{2}\cos(2t)+\frac{1}{2}$ . Then $LaTeX: \mathcal{L}[\sin^2(t)]=-\frac{1}{2}\mathcal{L}[\cos(2t)]+\frac{1}{2}\mathcal{L}[1]=-\frac{s}{2(s^2+4)}+\frac{1}{2s}=\frac{2}{s(s^2+4)}$

4.2.4b: Use the "double angle formula": $LaTeX: \cos(\omega t+\theta)=\cos(\theta)\cos(\omega t)-\sin(\theta)\sin(\omega t)$ . Then $LaTeX: \mathcal{L}[A\cos(\omega t+\theta)]=A\cos(\theta)\mathcal{L}[\cos(\omega t)]-A\sin(\theta)\mathcal{L}[\sin(\omega t)]$

4.2.4c: $LaTeX: \mathcal{L}[\cos(at)\cosh(at)]=\frac{1}{2}\mathcal{L}[\cos(at)](s-a)+\frac{1}{2}\mathcal{L}[\cos(at)](s+a)=\frac{1}{2}\left(\frac{s-a}{(s-a)^2+a^2}+\frac{s+a}{(s+a)^2+a^2}\right)$ then simplify.

4.2.4d: $LaTeX: \mathcal{L}[t^2](s-2)-5\mathcal{L}[t](s-2)+6\mathcal{L}[1](s-2)=\frac{2}{(s-2)^3}-\frac{5}{(s-2)^2}+\frac{6}{s-2}$

4.2.5a: $LaTeX: \int_0^\infty te^{-2t}\cos(t)\; dt=\mathcal{L}[t\cos(t)](2)=-\left.\frac{d}{ds}\frac{s}{s^2+1}\right|_{s=2}=\left.\frac{s^2-1}{(s^2+1)^2}\right|_{s=2}=\frac{3}{25}$

4.2.5b: $LaTeX: \int_0^\infty t^3e^{-t}\sin(t)\; dt=\mathcal{L}[t^3\sin(t)](1)=-\left.\frac{d^3}{ds^3}\frac{1}{s^2+1}\right|_{s=1}=\left.\frac{24s(s^2-1)}{(s^2+1)^4}\right|_{s=1}=0$

4.2.5c: $LaTeX: \int_0^\infty t^4e^{-t}\; dt=\mathcal{L}[t^4](1)=\left.\frac{24}{s^5}\right|_{s=1}=24$

4.2.5d: $LaTeX: \int_0^\infty t^6e^{-3t}\; dt=\mathcal{L}[t^6](3)=\left.\frac{6!}{s^7}\right|_{s=3}=\frac{80}{243}$

4.3.1a: $LaTeX: f\left(t\right)=3e^{2t}\cos\left(\sqrt{14}t\right)-\frac{8}{\sqrt{14}}e^{2t}\sin\left(\sqrt{14}t\right)$

4.3.1b: $LaTeX: f\left(t\right)=t-\sin(t)$

4.3.1c: $LaTeX: f\left(t\right)=2e^{\frac{3t}{2}}\sinh\left(\frac{t}{2}\right)$

4.3.1.d: $LaTeX: \frac{s^2}{(s^2+4)^2}=-\frac{s}{2}\frac{d}{ds}(\frac{1}{s^2+4})=-\frac{s}{4}\frac{d}{ds}\mathcal{L}[\sin(2t)]=\frac{1}{4}\mathcal{L}[\frac{d}{dt}(t\sin(2t))]=\mathcal{L}[\frac{1}{4}(\sin(2t)+2t\cos(2t))]$

4.3.1e: $LaTeX: f\left(t\right)=\frac{1}{4}t\cdot\sin\left(2t\right)$

4.3.1f: Set $LaTeX: \ln(1+\frac{1}{s^2})=\mathcal{L}[f]$ , then $LaTeX: \mathcal{L}[tf]=-\frac{d}{ds}\mathcal{L}[f]=\frac{2}{s(1+s^2)}=\frac{2s}{s^2+1}-\frac{2}{s}=\mathcal{L}[2\cos(t)-2]\implies f(t)=\frac{2\cos(t)-2}{t}$

4.3.1g: $LaTeX: \mathcal{L}[f]=\frac{1}{(s^2+4)^2}=-\frac{1}{4s}\frac{d}{ds}\frac{2}{s^2+4}=\frac{1}{4s}\mathcal{L}[t\sin(2t)]\implies f'(t)=\frac{1}{4}t\sin(2t)\implies f(t)=\frac{1}{16}\sin(2t)-\frac{1}{8}t\cos{(2t)}$

4.3.1h: $LaTeX: \mathcal{L}^{-1}\left[\frac{s^3}{s^4-16}\right]=\mathcal{L}^{-1}\left[\frac{1}{2}\frac{s}{s^2+4}+\frac{1}{2}\frac{s}{s^2-4}\right]=\frac{1}{2}\cos(2t)+\frac{1}{2}\cosh(2t)$

4.3.1i: $LaTeX: \mathcal{L}^{-1}\left[\frac{1}{s^4+1}\right]=\frac{1}{2\sqrt{i}}\mathcal{L}^{-1}\left[\frac{\sqrt{i}}{s^2+i}-\frac{\sqrt{i}}{s^2-i}\right]=\frac{1}{2\sqrt{i}}\left(\sin(\sqrt{i}t)-\sinh(\sqrt{i}t)\right)$

4.3.1j: $LaTeX: \mathcal{L}^{-1}\left[\frac{2s^2+s-10}{(s-4)(s^2+2s+2)}\right]=\mathcal{L}^{-1}\left[\frac{s+3}{s^2+2s+2}+\frac{1}{s-4}\right]=\mathcal{L}^{-1}\left[\frac{s+1}{(s+1)^2+1}+\frac{2}{(s+1)^2+1}+\frac{1}{s-4}\right]=e^{-t}\left(\cos(t)+2\sin(t)\right)+e^{4t}$

4.3.2a: $LaTeX: 2e^t+e^{-4t}$ ,

4.3.2b: $LaTeX: \mathcal{L}[y]=\frac{s-1}{s^2-1}=\frac{1}{s+1}\implies y=e^{-t}$

4.3.2c: $LaTeX: \mathcal{L}[y]=\frac{1}{s+2}+\frac{2}{(s+2)^2}\implies y=(1+2t)e^{-2t}$

4.3.2d: $LaTeX: 2\cos\left(2t\right)+\frac{1}{2}\sin\left(2t\right)$

4.3.2e: $LaTeX: \mathcal{L}[y]=\frac{s+5}{s^2+4s+7}=\frac{s+2}{(s+2)^2+3}+\frac{3}{(s+2)^2+3}\implies y=e^{-2t}\cos(\sqrt{3}t)+\sqrt{3}e^{-2t}\sin(\sqrt{3}t)$

4.3.2f: $LaTeX: \frac{\sin t}{t}$ (ledning: Använd $LaTeX: L\left\lbrack t\cdot y\left(t\right)\right\rbrack=-\frac{d}{ds}L\left\lbrack y\left(t\right)\right\rbrack$ från (7) i tabellen för Laplace transformer).

4.4.1a: $LaTeX: \mathcal{L}[1-u_T(t)]=\frac{1}{s}(1-e^{-sT})$

4.4.1b: $LaTeX: \mathcal{L}[u_1(t)(t-1)^2+u_1(t)(t-1)]=e^{-s}\left(\frac{2}{s^3}+\frac{1}{s^2}\right)$

4.4.1c: $LaTeX: \mathcal{L}[t^2-(t-2)^2u_2(t)+4u_2(t)]=\frac{2}{s^3}-e^{-2s}\frac{2}{s^3}+4e^{-2s}$

4.4.1d: $LaTeX: \mathcal{L}[t+(3-2t)u_7(t)+(t-2)u_8(t)]=\frac{1}{s^2}-e^{-7s}\left(\frac{2}{s^2}+\frac{11}{s}\right)+e^{-8s}\left(\frac{1}{s^2}+\frac{6}{s}\right)$

4.4.2a: $LaTeX: F(s) = \frac{e^{-s}(1-e^{-s})}{s(s^2 + 1)} = e^{-s} \left( \frac{1}{s} - \frac{s}{s^2+1} \right) - e^{-2s} \left( \frac{1}{s} - \frac{s}{s^2+1} \right)$ ger $LaTeX: f(t) = u(t-1) \left( 1 - \cos(t-1) \right) - u(t-2) \left( 1 - \cos(t-2) \right)$

4.4.2b: $LaTeX: \frac{1}{4}u_{16}(t)\left(1-\frac{1}{\sqrt{3}}e^{16-t}\sin(\sqrt{3}(t-16))-e^{16-t}\cos(\sqrt{3}(t-16))\right)$

4.4.4a: $LaTeX: y(t) = e^{-2t} \cos\sqrt{3} t + 2 e^{-2t} \frac{1}{\sqrt{3}}\sin \sqrt{3}t + u(t-1) \cdot \frac{1}{7} \left( 1 - e^{-2(t-1)} \cos\sqrt{3} (t-1) - 2 e^{-2(t-1)} \frac{1}{\sqrt{3}}\sin \sqrt{3}(t-1) \right)$

4.4.4b: $LaTeX: y(t) = u(t-\pi) e^{-(t-\pi)} \frac{1}{\sqrt{2}} \sin\sqrt{2}(t-\pi) + e^{-t} \cos\sqrt{2}t +e^{-t}\frac{1}{\sqrt{2}} \sin\sqrt{2}t$

4.4.4c: Write $LaTeX: \left(-1\right)^{\lfloor t\rfloor}=1-2u_1\left(t\right)+2u_2\left(t\right)-2u_3\left(t\right)+\cdots=1+2\sum_{n=1}^{\infty}(-1)^nu_n(t)$ . By transforming both sides of the equation and rearranging: $LaTeX: \mathcal{L}[y]=\frac{1}{s(s-1)^2}(1+2\sum_{n=1}^\infty (-1)^ne^{-ns})=:g(s)(1+2\sum_{n=1}^\infty (-1)^ne^{-ns})$ . Then $LaTeX: y=\mathcal{L}^{-1}[g(s)]+2\sum_{n=1}^{\infty}(-1)^n\mathcal{L}^{-1}[e^{-ns}g(s)]=1+(t-1)e^t+2\sum_{n=1}^\infty (-1)^n u_n(t)(1+(t-1-n)e^{t-n})$

4.4.4d: Transformed problem: $LaTeX: \left(s^2+2s+2\right)\mathcal{L}[y]-s-2=e^{-\pi s}$ . Then $LaTeX: \mathcal{L}[y]=\frac{s+1}{(s+1)^2+1}+\frac{1}{(s+1)^2+1}+\frac{e^{-\pi s}}{(s+1)^2+1}$ , so $LaTeX: y=e^{-t}\left(\cos(t)+\sin(t)\right)+u_{\pi}\left(t\right)e^{\pi-t}\sin(t-\pi)$

4.4.4e: $LaTeX: y(t) = u(t-\pi) \frac{1}{2} \sin(2(t-\pi)) - u(t-2\pi) \frac{1}{2} \sin(2(t-2\pi)) = \frac{1}{2} \big( u(t-\pi) - u(t-2\pi) \big) \sin(2t)$

4.4.5: Note: $LaTeX: y\left(t-1\right)=u_1\left(t\right)y\left(t-1\right)$ . Transformed problem: $LaTeX: \left(s+e^{-s}\right)\mathcal{L}[y]=\frac{2}{s^3}\implies \mathcal{L}[y]=\frac{2}{s^4}\frac{1}{1+(e^{-s}/s)}=\frac{2}{s^4}\sum_{n=0}^\infty (-\frac{e^{-s}}{s})^n=2\sum_{n=0}^\infty (-1)^n\frac{e^{-ns}}{s^{n+4}}$ (Recall: $LaTeX: \frac{1}{1-x}=\sum_{n=0}^\infty x^n$ ). Then $LaTeX: y=2\sum_{n=0}^{\infty}(-1)^n\mathcal{L}^{-1}[\frac{e^{-ns}}{s^{n+4}}]=2\sum_{n=0}^{\infty}(-1)^n u_n(t)\frac{(t-n)^{n+3}}{(n+3)!}$

4.5.1: Eftersom LaTeX: f(0) = 0 så är $LaTeX: \mathcal{L} [f'](s) = s\mathcal{L} [f](s) - f(0) = s\mathcal{L} [f](s)$ . Då $LaTeX: f'(t) = t^2 - t + e^{-t}$ har vi $LaTeX: \mathcal{L}[f'](s) = \frac{2}{s^3} - \frac{1}{s^2} + \frac{1}{s+1}$ och $LaTeX: \mathcal{L}[f](s) = \frac{1}{s} \mathcal{L}[f'](s) = \frac{2}{s^4} - \frac{1}{s^3} + \frac{1}{s(s+1)}$ .

4.5.2a: $LaTeX: F(s) = \frac{9(s^2 + 1)}{(s-2)(s+1)^2} = \frac{5}{s-2} + \frac{4}{s+1} - \frac{6}{(s+1)^2}$ ger $LaTeX: f(t)=4e^{-t}-6e^{-t}t+5e^{2t}$ .

4.5.2b: Laplace transform of problem: $LaTeX: \mathcal{L}[f\ast 1-f']=\mathcal{L}[f](\frac{1}{s}-s)=\frac{e^{-as}}{s}=\mathcal{L}[u_a]$ . This implies $LaTeX: \mathcal{L}[f]=-\frac{e^{-as}}{s^2-1}=-\mathcal{L}[\delta_a]\mathcal{L}[\sinh]$ which has solution $LaTeX: f=-\delta_a\ast\sinh=u_a(t)\sinh(t-a)$

4.5.2c: $LaTeX: F(s) = \frac{2s^2}{(s^2 + 4) (s^2 + 1)} = \frac{8}{3} \frac{1}{s^2 + 4} - \frac{2}{3} \frac{1}{s^2 + 1} = \frac{4}{3} \frac{2}{s^2 + 4} - \frac{2}{3} \frac{1}{s^2 + 1}$ ger $LaTeX: f(t) = \frac{4}{3} \sin 2t - \frac{2}{3} \sin t$ .

4.5.2d: Laplace transform of the problem: $LaTeX: \mathcal{L}[f'']=s^2\mathcal{L}[f]+s-1=\frac{1}{s^2}\mathcal{L}[f]=\mathcal{L}[t\ast f]$ . Solution: $LaTeX: f=\frac{1}{2}(\sin(t)-\cos(t)-e^{-t})$

4.5.2e: $LaTeX: \mathcal{L}[f+f\ast e^{-t}]=\mathcal{L}[f](1+\frac{1}{s+1})=\frac{1}{s}=\mathcal{L}[1]$ is solved by $LaTeX: f=\frac{1}{2}(e^{-2t}+1)$

4.5.2f: $LaTeX: F(s)^2 = \frac{9}{s^2(s-3)^2}$ ger $LaTeX: F(s) = \pm\frac{3}{s(s-3)} = \pm \left(\frac{1}{s-3} -\frac{1}{s}\right)$ och $LaTeX: f(t) = \pm(e^{3t} - 1)$ . Eftersom LaTeX: f'(0)>0 har vi sedan $LaTeX: f(t) = e^{3t} -1$ .

4.5.2m: (misprinted j in book) Transforming the problem gives $LaTeX: \mathcal{L}[f'\ast f]=s(\mathcal{L}[f])^2=\frac{144}{s^4}=\mathcal{L}[24t^3]$ , such that $LaTeX: \mathcal{L}[f]=\pm \frac{12}{s^{5/2}}$ and $LaTeX: f=\pm\frac{16}{\sqrt{\pi}}t^{3/2}$

4.5.3a: $LaTeX: Y(s) = \frac{1}{s^2+1} G(s)$ ger $LaTeX: y(t) = \sin(t) * g(t) = \int_0^t g(y) \sin(t-y) dy$ . Om $LaTeX: 0\leq t <1$ ger detta $LaTeX: y(t) = \int_0^t y \sin(t-y) dy = - \sin t + t$ . För $LaTeX: 1 \leq t$ får vi $LaTeX: y(t) = \int_0^1 y \sin(t-y) dy + \int_1^t 1\cdot \sin(t-y) dy = -\sin(t) + \sin(t-1) + 1$ .

4.5.3d: Transformed problem: $LaTeX: \mathcal{L}[y''+y]=(s^2+1)\mathcal{L}[y]-1=\mathcal{L}[\delta_\pi (t)\cos(t)]$ . Can be rewritten $LaTeX: \mathcal{L}[y]=\mathcal{L}[\sin(t)+\delta_\pi (t)\cos(t)\ast \sin(t)]$ . The solution is $LaTeX: y=\left(1+u_{\pi}\left(t\right)\right)\sin(t)$

4.5.3f: Transformed problem: $LaTeX: \mathcal{L}[y''+9y]=(s^2+9)\mathcal{L}[y]-as-b=\mathcal{L}[(t-c)u_c(t)]$ , in particular $LaTeX: \mathcal{L}[y]=\frac{1}{s^2+9}\mathcal{L}[(t-c)u_c(t)]+a\frac{s}{s^2+9}+b\frac{1}{s^2+9}$ . Thus, $LaTeX: y=\frac{1}{3}\sin(3t)\ast(t-c)u_c(t)+a\cos(3t)+\frac{b}{3}\sin(3t)$ , and furthermore $LaTeX: \sin(3t)\ast (t-c)u_c(t)=(\frac{1}{3}(t-c)-\frac{1}{9}\sin(3(t-c)))u_c(t)$

4.6.1a: Ekvationen är $LaTeX: f(t) * \frac{1}{\sqrt{t}} = 1 + t + t^2$ , vilket med Formelblad (12) ger $LaTeX: F(s) \frac{\Gamma(1/2)}{s^{1/2}} = \frac{1}{s} + \frac{1}{s^2} + \frac{2}{s^3}$ eller $LaTeX: F(s) = \frac{1}{\Gamma(1/2)} \left(\frac{1}{s^{1/2}} + \frac{1}{s^{3/2}} + \frac{2}{s^{5/2}} \right)$ . Från detta ger Formelblad (12) att $LaTeX: f(t) = \frac{1}{\Gamma(1/2)} \left( \frac{t^{-1/2}}{\Gamma(1/2)} + \frac{t^{1/2}}{\Gamma(3/2)} + 2\frac{t^{3/2}}{\Gamma(5/2)} \right)$ . Formler för Gamma-funktionen i detta avsnitt ger oss att $LaTeX: \Gamma(1/2) = \sqrt{\pi}$ , $LaTeX: \Gamma(3/2) = (1/2) \Gamma(1/2) = \sqrt{\pi}/2$ , $LaTeX: \Gamma(5/2) = (3/2) \Gamma(3/2) = 3\sqrt{\pi}/4$ . Tillsammans får vi $LaTeX: f(t) = \frac{1}{\pi} \left( t^{-1/2} + 2 t^{1/2} + 2\frac{4t^{3/2}}{3} \right) = \frac{1}{\pi t^{1/2}} \left( 1 + 2 t + \frac{8}{3} t^2 \right)$ .

4.6.7: Use the second equation from page 169: $LaTeX: \Gamma(p)\zeta(p)=\sum_{n=1}^\infty \frac{\Gamma(p)}{n^p}=\sum_{n=1}^\infty \mathcal{L}[t^{p-1}](n)=\sum_{n=1}^\infty\int_0^\infty t^{p-1}e^{-nt}\; dt=\int_0^\infty \frac{t^{p-1}}{e^t-1}\: dt$ , where we interchanged summation and integration, and used the sum-function of a geometric series: $LaTeX: \sum_{n=1}^\infty (e^{-t})^n=\frac{e^{-t}}{1-e^{-t}}=\frac{1}{e^{t}-1}$

4.8.11a: Let $LaTeX: F\left(x,s\right)$ be the Laplace transform of $LaTeX: u\left(x,t\right)$ w.r.t. the variable LaTeX: t . Then, transforming the differential equation:

$LaTeX: \frac{d}{dx}F(x,s)+2xsF(x,s)=2x(\frac{1}{s}+1)$ . Solving it as a linear first order equation: $LaTeX: F\left(x,s\right)=\frac{1}{s^2}+\frac{1}{s}+c(s)e^{-x^2s}$ , where LaTeX: c(s) is the constant of integration (which may be a function in the variable s). By the initial condition $LaTeX: F\left(0,s\right)=\mathcal{L}[1]=\frac{1}{s}$ , we get $LaTeX: c\left(s\right)=-\frac{1}{s^2}$ . Thus $LaTeX: u\left(x,t\right)=t+1-\left(t-x^2\right)\mathcal{U}_{x^2}(t)$ , where $LaTeX: \mathcal{U}$ is the Heaviside function.

4.8.11b: Let $LaTeX: F\left(x,s\right)$ be the Laplace transform of $LaTeX: u\left(x,t\right)$ w.r.t. the variable LaTeX: t . Then, transforming the differential equation:

$LaTeX: x\frac{d}{dx}F(x,s)+sF(x,s)=\frac{x}{s^2}$ . Solving as a linear first order equation: $LaTeX: F\left(x,s\right)=\frac{x}{s^2(s+1)}+c(s)x^{-s}$ . By the initial condition $LaTeX: F\left(0,s\right)=0$ , we get $LaTeX: c\left(s\right)=0$ . Thus $LaTeX: u\left(x,t\right)=x\mathcal{L}^{-1}\left[\frac{1}{s^2(s+1)}\right]=xt-x+xe^{-t}$ .