# Find The Laplace Transformation of Below Question

1. 7e^{2t}+9e^{-2t}+5cost+7t^{3}+5sint3t+2
2. sin2tcos3t
3. \left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3}
4. sinh^{2}2t
5. coshat-cosat

Q1. {{\color{Red} 7e^{2t}+9e^{-2t}+5cost+7t^{3}+5sint3t+2}}\\Applying Laplace Transformation\\\;\\L[{{\color{Black} 7e^{2t}+9e^{-2t}+5cost+7t^{3}+5sint3t+2}}]=\\\;\\L\left [ 7e^{2t} \right ]+L\left [ 9e^{-2t} \right ]+L\left [ 5cost \right ]+L\left [ 7t^{3} \right ]+L\left [ 5sint \right ]+L\left [ 2 \right ]\; \; \; \; \; \; \; \; \; \; \; (1)\\\;\;\;\\L\left [ 7e^{2t} \right ]=7\times L\left [ e^{2t} \right ]=7\times \frac{1}{s-2}=\frac{7}{s-2}\; \; \; \; \; \; \; \; \; \; \; \left ( \because L\left \lfloor e^{\pm at} \right \rfloor =\frac{1}{s\mp a}\right )\\\;\;\;\\L\left [ 9e^{-2t} \right ]=9\times L\left [ e^{-2t} \right ]=9\times\frac{1}{s+2}=\frac{9}{s+2}\\\;\;\;\\L[5cost]=5 \times L[cost]=5 \times \frac{s}{s^2+1^2}=\frac{5s}{s^2+1}\;\;\;\;\;\;\;\;\;(\because L[cosat]=\frac{s}{s^2+a^2})\\\;\;\;\\L[7t^3]= 7 \times L[t^3]=7 \times \frac{3!}{s^{3+1}}=\frac{42}{s^4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\because L[t^n]=\frac{n!}{s^{n+1}})\\\;\\L[5sin3t]=5 \times L[sin3t]=5 \times \frac{3}{s^2+3^2}= \frac{15}{s^2+9}\;\;\;\;\;\;\;(\because L[sinat]=\frac{a}{s^2+a^2})\\\;\;\;\\L[2]=L[2 \times 1]= 2 \times L[1]=2 \times \frac{1}{s}= \frac{2}{s}\;\;\;\;\;\;\;\;\;\;\;(\because L[1]= \frac{1}{s})\\\;\;\;\\Putting\;the\;Values\;of\\\;\\L[7e^{2t}],\;\;L[9e^{-2t}],\;\;L[5cost],\;\;L[7t^3],\;\;L[5sin3t]\;\;and\;L[2]\;in\;equation\;(1) \\\;\;\\L[{{\color{Black} 7e^{2t}+9e^{-2t}+5cost+7t^{3}+5sint3t+2}}]=\frac{7}{s-2}+\frac{9}{s+2}+\frac{5s}{s^2+1}+\frac{42}{s^4}+\frac{15}{s^2+9}+\frac{2}{s}

Q2. {\color{Red} sin2tcos3t}\\\;\\ First simplify the sin2tcos3t\\\;\\ sin2tcos3t =cos3tsin2t\\\;\\cos3tsin2t=\frac{1}{2}(2cos3tsin2t)=\frac{1}{2}(sin5t-sint)\\\;\;\;\;\;\;\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\; (\because sinC-sinD=2cos \frac{C+D}{2}sin \frac{C-D}{2})\\\;\\ Thus \\\;\\sin2tcos3t= \frac{1}{2}(sin5t-sint)\\\;\\Now Applying Laplace Transformation\\\;\\L[sin2tcos3t]=L[\frac{1}{2}(sin5t-sint)]\\\;\\ L[sin2tcos3t] =\frac{1}{2}(L[sin5t]-L[sint])\\\;\\ L[sin2tcos3t]= \frac{1}{2}(\frac{5}{s^2+5^2}-\frac{1}{s^2+1^2})\;\;\;\;\;\;\;\;\;\;\;\;(\because L[sinat]=\frac{a}{s^2+a^2})\\\;\\ L[sin2tcos3t]= \frac{1}{2}(\frac{5}{s^2+25}-\frac{1}{s^2+1})=\frac{2(s^2-5)}{(s^2+25)(s^2+1)}

Q3. {\color{Red}\left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3}}\\\;\\ First we will simplify \;\left(\sqrt{t}+\frac{1}{\sqrt{t}} \right )^3\\\;\\\left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3}=t^{\frac{3}{2}}+\frac{1}{t^{\frac{3}{2}}}+3 \times t^{\frac{1}{2}} \times \frac{1}{t^{\frac{1}{2}}}\left ( t^{\frac{1}{2}}+\frac{1}{t^{\frac{1}{2}}} \right )\\\;\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(\because \left(a+b\right)^3= a^3+b^3+3ab\left(a+b\right)\right)\\\;\\ \left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3} =t^{\frac{3}{2}}+t^{-\frac{3}{2}}+3t^{\frac{1}{2}}+3t^{-\frac{1}{2}}\\\;\\Now Applying Laplace Transformation\\\;\\L[ \left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3}]=L[ t^{\frac{3}{2}}+t^{-\frac{3}{2}}+3t^{\frac{1}{2}}+3t^{-\frac{1}{2}}]\\\;\\L\left[ \left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3} \right]=L\left[t^{\frac{3}{2}}\right]+L\left [ t^{-\frac{3}{2}} \right ]+3L\left [ t^{\frac{1}{2}} \right ]+3L\left [ t^{-\frac{1}{2}} \right ]\\\;\\ L\left[ \left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3} \right] =\frac{\Gamma \left ( \frac{3}{2}+1 \right )}{s^{\frac{3}{2}+1}}+\frac{\Gamma \left ( -\frac{3}{2}+1 \right )}{s^{-\frac{3}{2}+1}}+3\frac{\Gamma \left ( \frac{1}{2}+1 \right )}{s^{\frac{1}{2}+1}}+3\frac{\Gamma \left ( -\frac{1}{2}+1 \right )}{s^{-\frac{1}{2}+1}}\;\;\;\;\;\;\;\;\;\;(1)\\\;\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( \because L[t^n]=\frac{n!}{s^{n+1}}\;For\;n\;=0,1,2,3..\\\;\\\right)\\\;\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \left ( \because L[t^n]=\frac{\Gamma\left ( n+1 \right )}{s^{n+1}}\; if\;n\;is\;otherwise \right )\\\;\\First find the values of \;\Gamma(\frac{3}{2}+1),\;\;\;\Gamma(-\frac{3}{2}+1),\;\;\;\Gamma(\frac{1}{2}+1),\;\;\;\Gamma(-\frac{1}{2}+1)\\\;\\\Gamma (\frac{3}{2}+1)=\frac{3}{4}\sqrt{\pi}\;\;\;\;\;\;\;\;\left ( \because \Gamma(n+1)=n\Gamma(n)\;and\;\Gamma(\frac{1}{2})=\sqrt{\pi} \right )\\\;\\\Gamma (-\frac{3}{2}+1)=-2\sqrt{\pi}\\\;\\\Gamma (\frac{1}{2}+1)=\frac{1}{2}\sqrt{\pi}\\\;\\\Gamma (-\frac{1}{2}+1)=\sqrt{\pi}\\\;\\Putting the Values of\;\Gamma(\frac{3}{2}+1),\;\;\Gamma(-\frac{3}{2}+1),\;\;\Gamma(\frac{1}{2}+1),\;\;\Gamma(-\frac{1}{2}+1)\; in equation (1)\\\;\\ L\left[ \left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3} \right] =\frac{\frac{3}{4}\sqrt{\pi}}{s^{\frac{5}{2}}}+\frac{-2\sqrt{\pi}}{s^{-\frac{1}{2}}}+3\frac{\frac{1}{2}\sqrt{\pi}}{s^{\frac{3}{2}}}+3\frac{\sqrt{\pi}}{s^{\frac{1}{2}}}\\\;\\ L\left[ \left ( \sqrt{t}+\frac{1}{\sqrt{t}} \right )^{3} \right] =\frac{\frac{3}{4}\sqrt{\pi}}{s^{\frac{5}{2}}}-\frac{2\sqrt{\pi}}{s^{-\frac{1}{2}}}+\frac{\frac{3}{2}\sqrt{\pi}}{s^{\frac{3}{2}}}+\frac{3\sqrt{\pi}}{s^{\frac{1}{2}}}

Q4.\;\;{\color{Red} sinh^{2}2t}\\\;\\First Expand sinh2 2t\\\;\\sinh^{2}2t=\left ( \frac{e^{2t}-e^{-2t}}{2} \right )^{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( \because sinh^{2}\theta =\frac{e^\theta -e^{-\theta }}{2} \right )\\\;\\sinh^{2}2t=\frac{1}{4}\left ( e^{4t}+e^{-4t}-2e^{2t}e^{-2t} \right )\\\;\\sinh^{2}2t=\frac{1}{4}\left ( e^{4t}+e^{-4t}-2 \right )\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\because e^{2t}e^{-2t}=e^{2t-2t}=e^{0}=1)\\\;\\Now Applying Laplace Transformation\\\;\\L[sinh^{2}2t]=\frac{1}{4}\left ( L[e^{4t}]+L[e^{-4t}]-2L[1] \right )\\\;\\L[sinh^{2}2t]=\frac{1}{4}\left ( \frac{1}{s-4}+\frac{1}{s+4}-\frac{2}{s} \right )\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(\because L[e^{\pm at}]=\frac{1}{s \mp a}\right)

Q5\;\;{\color{Red} coshat-cosat}\\\;\\First Expand coshat\\\;\\coshat=\frac{e^{at}+e^{-at}}{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\because cosh\theta=\frac{e^{\theta}+ e^{\theta}}{2})\\\;\\Now Apply Laplace Transformation\\\;\\L[coshat-cosat]=L[coshat]-L[cosat]=\frac{1}{2}\left (L[e^{at}]+L[e^{-at}] \right )-L[cosat]\\\;\\L[coshat-cosat]=\frac{1}{2}\left (\frac{1}{s-a}+\frac{1}{s+a} \right )-\frac{s}{s^2+a^2}\\\;\\L[coshat-cosat]=\frac{s}{s^2-a^2}-\frac{s}{s^2+a^2}

#### Author: AJAZ UL HAQ

AJAZ UL HAQ has above 8 years of Experience in Electrical Power Transmission, Distribution and Substation. Presently He is working with KEI Industries Limited as Engineer-EPC/EHV.