Problem:
Use the Laplace form to solve the following initial-value problem.
$$ y'' + 6y' + 8y = 16 $$ with $$ y(0) = 0 $$ $$ y'(0) = 10 $$ and the independent variable is $t$.
Answer:
\begin{align*} s^2Y(s) -sy(0) - y'(0) + 6 ( sY(s) - y(0)) + Y(s) &=\dfrac{16}{s} \\ s^2Y(s) - 10 + 6 ( sY(s) ) + Y(s) &=\dfrac{16}{s} \\ (s^2 + 6s+1)Y(s) &= \dfrac{ 16 }{s} + 10 \\ (s^2 + 6s+1)Y(s) &= \dfrac{ 10s + 16 }{s} \\ Y(s) &= \dfrac{ 10s + 16 }{s( s^2 + 6s+1 )} \end{align*} Using an online inverse Laplace calculator, the answer I get is a mess and wrong. The book's answer is: $$ y= e^{-2t} - 3e^{-4t} + 2 $$ Where did I go wrong?
I wrote the following line: $$ s^2Y(s) -sy(0) - y'(0) + 6 ( sY(s) - y(0)) + Y(s) =\dfrac{16}{s} $$ but it should have been: $$ s^2Y(s) -sy(0) - y'(0) + 6 ( sY(s) - y(0)) + 8Y(s) =\dfrac{16}{s} $$ with that change, I was able to solve the problem.