Answer to Question #342620 in Differential Equations for ackey_1

Question #342620

Use the Laplace transform to solve the given initial-value problem.


y′ + 2y = sin 4t, y(0) = 1


1
Expert's answer
2022-05-19T14:52:40-0400

the Laplace transform for the first derivative:


L{y}=sY(s)y(0)L\{y'\}=sY(s)-y(0)

then for IVP:


L{y}+2L{y}=L{sin4t}L\{y'\}+2 L\{y\}=L\{sin⁡4t\}sY(s)y(0)+2Y(s)=4s2+16sY(s)-y(0)+2Y(s)=\frac 4 {s^2+16}Y(s)(s+2)=4s2+16+1Y(s)(s+2)=\frac 4 {s^2+16}+1Y(s)=4(s2+16)(s+2)+1s+2Y(s)=\frac 4 {(s^2+16)(s+2)}+\frac 1 {s+2}=s2+20(s2+16)(s+2)=As+Bs2+16+Cs+2=\frac {s^2+20} {(s^2+16)(s+2)}=\frac {As+B} {s^2+16}+\frac C {s+2}s2+20=As2+Bs+2As+2B+Cs2+16Cs^2+20=As^2+Bs+2As+2B+Cs^2+16Cs2+20=(A+C)s2+(2A+B)s+2B+16Cs^2+20=(A+C)s^2+(2A+B)s+2B+16C{A+C=12A+B=02B+16C=20A=0.2,B=0.4,C=1.2\begin{cases} A+C=1 \\ 2A+B=0 \\ 2B+16C=20 \end{cases} \rightarrow A=-0.2, B=0.4, C=1.2Y(s)=0.2s+0.4s2+16+1.2s+2Y(s)=\frac {-0.2s+0.4} {s^2+16}+\frac {1.2} {s+2}y(t)=L1{Y(s)}=L1{0.2s+0.4s2+16}+L1{1.2s+2}y(t)=L^{-1} \{Y(s)\}=L^{-1}\bigg\{\frac{-0.2s+0.4}{s^2+16}\bigg\}+L^{-1}\bigg\{\frac{1.2}{s+2}\bigg\}=0.2L1{ss2+16}+0.44L1{4s2+16}+1.2L1{1s+2}=-0.2L^{-1}\bigg\{\frac{s}{s^2+16}\bigg\}+\frac{0.4}{4}L^{-1}\bigg\{\frac{4}{s^2+16}\bigg\}+1.2L^{-1}\bigg\{\frac{1}{s+2}\bigg\}=0.2cos(4t)+0.1sin(4t)+1.2e2t=-0.2 cos(⁡4t)+0.1 sin⁡(4t)+1.2 e^{-2t}

Answer: y(t)=0.2cos(4t)+0.1sin(4t)+1.2e2ty(t)=-0.2 cos(⁡4t)+0.1 sin⁡(4t)+1.2 e^{-2t}




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