Answer to Question #342620 in Differential Equations for ackey_1

Question

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

y′ + 2y = sin 4t, y(0) = 1

Answer 1

the Laplace transform for the first derivative:

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

then for IVP:

$$L\{y'\}+2 L\{y\}=L\{sin⁡4t\}$$ $$sY(s)-y(0)+2Y(s)=\frac 4 {s^2+16}$$ $$Y(s)(s+2)=\frac 4 {s^2+16}+1$$ $$Y(s)=\frac 4 {(s^2+16)(s+2)}+\frac 1 {s+2}$$ $$=\frac {s^2+20} {(s^2+16)(s+2)}=\frac {As+B} {s^2+16}+\frac C {s+2}$$ $$s^2+20=As^2+Bs+2As+2B+Cs^2+16C$$ $$s^2+20=(A+C)s^2+(2A+B)s+2B+16C$$ $$\begin{cases} A+C=1 \\ 2A+B=0 \\ 2B+16C=20 \end{cases} \rightarrow A=-0.2, B=0.4, C=1.2$$ $$Y(s)=\frac {-0.2s+0.4} {s^2+16}+\frac {1.2} {s+2}$$ $$y(t)=L^{-1} \{Y(s)\}=L^{-1}\bigg\{\frac{-0.2s+0.4}{s^2+16}\bigg\}+L^{-1}\bigg\{\frac{1.2}{s+2}\bigg\}$$ $$=-0.2L^{-1}\bigg\{\frac{s}{s^2+16}\bigg\}+\frac{0.4}{4}L^{-1}\bigg\{\frac{4}{s^2+16}\bigg\}+1.2L^{-1}\bigg\{\frac{1}{s+2}\bigg\}$$ $$=-0.2 cos(⁡4t)+0.1 sin⁡(4t)+1.2 e^{-2t}$$

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