Solving system of ODE with initial value problem $x′=−x+2y $
$y′=−2x−5y$
with initial values: $x(0)=0,y(0)=1$
I am not sure if my solution is correct:
from $x'=−x+2y$ we got:
$y=(x′+x)/2$
$=> y′=(x′′+x′)/2$
so $x′′(t)+6x′+9x=0$
result: $x=c_1e^{−3t}+c_2te^{−3t}$
and then from $y=(x′+x)/2$
$y=c_1∗−e^{−3t}+1/2∗c_2e−3t−c_2∗t∗e^{−3t}$
and finally by initial values $x(0)=0$ and$ y(0)=1$ i have found $c_1$ and $c_2$:
$c_1=0,1=c_1+c_2/2=>c_2=2$
Is this correct?
 A: Laplace transform is useful for system of differential equations.
Fundamental properties:
\begin{align}
\mathscr{L}\{f(t)\}&= F(s)\\
\mathscr{L}\{f'(t)\}&= sF(s)-f(0)\\
\mathscr{L}\{e^{at}\}&= F(s-a)\\
\mathscr{L}\{tf(t)\}&= -\frac{\mathrm{d}\,F(s)}{\mathrm{d}\,s}
\end{align}
Additional facts:
\begin{align}
\mathscr{L}\{1\}&=\frac{1}{s}\\
\mathscr{L}\{t\}&=\frac{1}{s^2}\\
\mathscr{L}\{e^{at}\}&=\frac{1}{s-a}\\
\mathscr{L}\{te^{at}\}&=\frac{1}{(s-a)^2}
\end{align}
Now proceed to your problems:
\begin{cases}
x'(t)=-x(t)+2y(t) \\
y'(t)=-2x(t)-5y(t)
\end{cases}
and
\begin{cases}
x(0)=0 \\
y(0)=1
\end{cases}
Transforming the given differential equations, we have:
\begin{cases}
sX(s)-x(0)=-X(s)+2Y(s) \\
sY(s)-y(0)=-2X(s)-5Y(s)
\end{cases}
\begin{cases}
sX(s)-0=-X(s)+2Y(s) \\
sY(s)-1=-2X(s)-5Y(s)
\end{cases}
\begin{cases}
(s+1)X(s)-2Y(s) =0 \\
2X(s)+(s+5)Y(s) =1
\end{cases}
You will get the following:
\begin{cases}
X(s)=\frac{2}{(s+3)^2} \\
Y(s)= \frac{1}{s+3}-\frac{2}{(s+3)^2}
\end{cases}
By inversing them, we have:
\begin{cases}
x(t)=2 t e^{-3t} \\
y(t)=e^{-3t}-2 t e^{-3t}
\end{cases}
