Question

Use Laplace transforms to solve the IVP: y''-2y'+5y=10 sin t y(0)=3, y'(0)=-4

Answer 1

Thats what i have so far.
\[(s^2-2s+5)Y-3s+10=\frac{ 10 }{ s^2+100 } \Rightarrow Y=\frac{ 3s^3-10s^2+200s-900 }{ (s^2+100)(s^2-2s+5) }\]

Answer 2

\[\begin{align*}
y''-2y'+5y&=10\sin t\\\\
\left(s^2Y-3s+4\right)-2\left(sY-3\right)+5Y&=\frac{10}{s^2+1}\\\\
\left(s^2-2s+5\right)Y&=\frac{10}{s^2+1}+3s-6\\\\
Y&=\frac{10+(3s-6)(s^2+1)}{(s^2+1)(s^2-2s+5)}\\\\
Y&=\frac{10+3s^3-6s^2+3s-6}{(s^2+1)(s^2-2s+5)}\\\\
Y&=\frac{3s^3-6s^2+3s+4}{(s^2+1)(s^2-2s+5)}
\end{align*}\]
Partial fractions:
\[\begin{align*}
\frac{3s^3-6s^2+3s+4}{(s^2+1)(s^2-2s+5)}&=\frac{As+B}{s^2+1}+\frac{Cs+D}{s^2-2s+5}\\\\
3s^3-6s^2+3s+4&=(As+B)(s^2-2s+5)+(Cs+D)(s^2+1)\\\\
3s^3-6s^2+3s+4&=(A+C)s^3+(-2A+B+D)s^2+(5A-2B+C)s\\&\quad+(5B+D)
\end{align*}\]
Find your constants:
\[\begin{cases}
\begin{align*}A+C&=3\\
-2A+B+D&=-6\\
5A-2B+C&=3\\
5B+D&=4\end{align*}
\end{cases}\]

Answer 3

@SithsAndGiggles
Why do you have \[\frac{ 10 }{ s^2+1 } \] on the \[10 \sin t\]

Answer 4

\[(s^2-2s+5)Y=\frac{ 10 }{ s^2+1 }+3s-10\]
Thats what i got after i transformed your solution. You forgot about the -4

Answer 5

Regarding your first reply: You're given \(10\sin t\), not \(\sin10t\).
\[\mathcal{L}\{\sin at\}=\frac{a}{s^2+a^2}\]

Answer 6

As for your second, you're right, sorry for the mix-up.

Answer 7

That means you have
\[Y=\frac{3s^3-10s^2+3s}{(s^2+1)(s^2-2s+5)}\]
PFs:\[\begin{align*}
\frac{3s^3-10s^2+3s}{(s^2+1)(s^2-2s+5)}&=\frac{As+B}{s^2+1}+\frac{Cs+D}{s^2-2s+5}\\\\
3s^3-10s^2+3s&=(As+B)(s^2-2s+5)+(Cs+D)(s^2+1)\\\\
3s^3-10s^2+3s&=(A+C)s^3+(-2A+B+D)s^2+(5A-2B+C)s\\&\quad+(5B+D)
\end{align*}\]
System:
\[\begin{cases}
\begin{align*}A+C&=3\\
-2A+B+D&=-10\\
5A-2B+C&=3\\
5B+D&=0\end{align*}
\end{cases}\]