Laplace transform of ODE

Question

Solve for $y(x)$ given: $$y"-y'-6y=0$$ with $y(0)=6$ and $y'(0)=13$ by using Laplace transforms.

---

I tried solving it but I am getting the wrong answer. Please help me with this one.

Answer 1

$$y"-y'-6y=0$$ Apply Laplace Transform: $$s^2Y(s)-sy(0)-y'(0)-(sY(s)-y(0))-6Y(s)=0$$ $$Y(s)(s^2-s-6)=sy(0)+y'(0)-y(0)$$ Apply initial conditions: $$Y(s)(s^2-s-6)=6s+7$$ $$\implies Y(s)=\dfrac {6s+7}{s^2-s-6}$$ $$\implies Y(s)=\dfrac {6s+7}{(s+2)(s-3)}$$ Is it here that you are stuck, Steve ? $$ Y(s)=\dfrac {6s+7}{(s+2)(s-3)}$$ Decompose the fraction into simple fractions: $$ Y(s)=\dfrac {1}{(s+2)}+\dfrac {5}{(s-3)}$$ Apply inverse Laplace transform now. You have a table here Laplace Transform Table