Find the inverse Laplace transform of $\displaystyle F(s) = \frac{9s-24}{s^2-6s+13}$. I have tried factoring out a $3$ from the top and putting it into the form of $\displaystyle\frac{b}{(s-a)^2+b^2}$ but I can't seem to do that with this equation. Any help is appreciated, thank you in advance!
[Math] Inverse Laplace Transform for $F(s) = (9s-24)/(s^2-6s+13)$
inverselaplace transformordinary differential equations
Best Answer
Use $$L\left(e^{at}\cos bt\right)=\frac{s-a}{(s-a)^2+b^2}$$ and $$L\left(e^{at}\sin bt\right)=\frac{b}{(s-a)^2+b^2}$$
Using Partial fraction $$\frac{9s-24}{(s-3)^2+2^2}=\frac{A(s-3)}{(s-3)^2+2^2}+\frac{B\cdot 2}{(s-3)^2+2^2}$$ where $A,B$ are arbitrary constants
Can you take it from here?