Answer to Question #235141 in Differential Equations for Lokesh bodele

Question

Solve the initial value problem

y" + 2y + 2y = 2, 7(0) = 0,1(0)=1

using the method of Laplace Transform.

Answer 1

$$y''+2y'+2y=2$$

$$L\{y''+2y'+2y\}=L\{2\}$$

$$(-y'(0)-sy(0)+s^2Y(s))+2(-y(0)+sY(s))+2Y(s)=\dfrac{2}{s}$$

Inserting the initial conditions and rearranging:

$$-1-0+s^2Y(s)+2(-0+sY(s))+2Y(s)=\dfrac{2}{s}$$

$$Y(s)(s^2+2s+2)=\dfrac{2+s}{s}$$

$$Y(s)=\dfrac{2+s}{s(s^2+2s+2)}$$

$$\dfrac{2+s}{s(s^2+2s+2)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+2s+2}$$

$$=\dfrac{As^2+2As+2A+Bs^2+Cs}{s(s^2+2s+2)}$$

$$s=0:2=2A=>A=1$$

$$s^2:A+B=0=>B=-1$$

$$s^1:2A+C=1=>C=-1$$

$$Y(s)=\dfrac{1}{s}+\dfrac{s-1}{s^2+2s+2}$$

$$y(t)=L^{-1}\{Y(s)\}=L^{-1}\{\dfrac{1}{s}-\dfrac{s+1}{(s+1)^2+1}\}$$

$$=1-e^{-t}\cos t$$