I have the following second-order, linear differential equation: $y''-2y'-99y=0$ with the initial conditions $y(0)=12, y'(0)=-8$.
Converting the DE into a Laplace equation (I don't know what its actually called, professor didn't go over it), I get:
$$[s^2Y-y(0)s-y'(0)]-2[sY-y(0)]-99Y=0$$
Plugging in $y(0)$ and $y'(0)$, I get:
$$[s^2Y-12s+8]-2[sY-12]-99Y=0;$$ $$=[s^2Y-2sY-99Y]-12s+30.$$
Solving for $Y$, $$Y[s^2-2s-99]-12s+30=0,$$ $$Y[s^2-2s-99]=12s-30,$$ $$Y=\frac{12s-30}{s^2-2s-99}.$$
$y$ is the inverse Laplace of Y ($y=L^{-1}[Y]$ ).
$$y=L^{-1}[\frac{12s-30}{s^2-2s-99}]$$ $$y=L^{-1}[\frac{61}{10(s+9)}+\frac{51}{10(s-11)}]$$
Therefore,$y=\dfrac{69}{10}e^{-9t}+\dfrac{51}{10}e^{11t}$.
However, when I enter this into my math homework program I get flagged as incorrect. Where did I make my mistake?
Hint: You have a slight addition issue, what is $$24 + 8~?$$