I've been following an example from Slader, but I can't seem to find where I'm going wrong in my own problem.
Given $y''+4y'+3y=105e^{4t}, y(0)=-4, y'(0)=25$, solve using Laplace transforms.
I start: $$[s^2Y-sy(0)-y'(0)]+4[sY-y(0)]+3[Y]=\frac{105}{s-4}$$ $$Y[s^2+4s+3]+4s-25+4(4)=\frac{105}{s-4}$$ $$Y(s+3)(s+1)=\frac{105}{s-4}-4s+9$$ $$Y(s+3)(s+1)=\frac{105+(s-4)(-4s+9)}{s-4}$$
Isolating Y, $$Y=\frac{105-4s^2+9s-16s-36}{(s-4)(s+1)(s+3)}$$ $$Y=\frac{-4s^2-5s+69}{(s-4)(s+1)(s+3)}$$
Then to break it up into partial fractions: $$\frac{-4s^2-5s+69}{(s-4)(s+1)(s+3)}=\frac{A}{s-4}+\frac{B}{s+3}+\frac{C}{s+1}$$ $$-4s^2-5s+69=A(s+1)(s+3)+B(s-4)(s+1)+C(s-4)(s+3)$$
Solving for A, B, C: $$s=4, 35A=-15, A=\frac{-3}{7}$$ $$s=-3, 14B=18, B=\frac{7}{9}$$ $$s=-1, -10C=70, C=7$$
So then I get, $$Y=\frac{-3}{7(s-4)}+\frac{7}{9(s+3)}+\frac{7}{s+1}$$
Taking the inverse Laplace leaves me with: $$y=\frac{-3}{7}e^{4t}+\frac{7}{9}e^{-3t}-7e^{-t}$$
Please help me, I'm on the last try before I start to lose points!
$$Y=\frac{105-4s^2+9s-16s-36}{(s-4)(s+1)(s+3)}$$
Should have been
$$Y=\frac{105-4s^2+9s+16s-36}{(s-4)(s+1)(s+3)}$$