Answer to Question #102652 in Calculus for BIVEK SAH

Question

Using the method of Laplace transforms, solve the following initial value problem:

y"-3y'+2y=6e^-t; y(0)=3,y'(0)=3

Answer 1

Taking Laplace transform, we get;

$(s^2y^{-}(s)-sy(0)-y'(0))-3(sy^{-}(s)-y'(0))+2y^{-}(s)=L(6e^{-t})$

$(s^2y^{-}(s)-3s-3)-3(sy^{-}(s)-3)+2y^{-}(s)=6/(s+1)$

$(s^2-3s+2)y^{-}(s)=3(s-2)+6/(s+1)=(3s^2-3s)/(s+1)$

$\implies y^{-}(s)=3s/(s+1)(s-2)$

Taking inverse Laplace transform, we get;

$y(x)=L^{-1}(3s/(s+1)(s-2))$

$y(x)=L^{-1}(A/(s+1)+B/(s-2))$

$\implies 3s=A(s-2)+B(s+1)$

Equating the coefficients to solve for A,B we get;

$A+B=3;-2A+B=0$

Solving these equations, we get; $A=1;B=2$

$y(x)=L^{-1}((1/(s+1)+2/(s-2))$

$=L^{-1}(1/(s-(-1)))+2L^{-1}(1/(s-2))$

$=e^{-x}+2e^{2x}$ ---(Answer)