Solve the initial value problem $y''-6y'+13y=0,\;y(0)=y'(0)=1$ using the Laplace transform.

Question

Solve the initial value problem $$ \begin{cases} y''-6y'+13y=0 \\ y(0)=y'(0)=1 \end{cases} $$ using the Laplace transform.

I cannot figure out how to factor or get around factoring $L(Y)$.

Answer 1

Taking the LT gives you $$13Y+s^2Y-6(sY-y(0))-sy(0)-y'(0)=0.$$ Plugging in the IC's yields $$13Y+s^2Y-6(sY-1)-s-1=0.$$ Solving for $Y$ gives you $$ Y(13+s^2-6s)=s+1-6=s-5, $$ making $$Y=\frac{s-5}{s^2-6s+13}=\frac{s-5}{s^2-6s+9+4}=\frac{s-5}{(s-3)^2+4}.$$ Computing the inverse LT yields $$y(t)=[\cos(2t)-\sin(2t)][\cosh(3t)+\sinh(3t)].$$ The inverse LT you're going to have to complete the square on, as the denominator doesn't factor over the reals.