Coupled differential equations

[rain echo]

Das Boot

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Das Boot

Now use partial fraction decomposition

$\begin{align*}
\frac{s+a+b}{s\left(s+\left(2a+b\right)\right)}N_{0} & =\frac{A}{s}+\frac{B}{s+\left(2a+b\right)}\\
N_{0}\left(s+a+b\right) & =A\cdot\left(s+\left(2a+b\right)\right)+Bs
\end{align*}$

So 
$\begin{align*}
A+B & =N_{0}\\
N_{0}\left(a+b\right) & =A\left(2a+b\right)
\end{align*}$

thus $A=\frac{N_{0}\left(a+b\right)}{\left(2a+b\right)}$. For $B$
we have 
$\begin{align*}
B & =N_{0}-A\\
 & =N_{0}-\frac{N_{0}\left(a+b\right)}{\left(2a+b\right)}\\
 & =\frac{2N_{0}a+N_{0}b-N_{0}a-N_{0}b}{2a+b}\\
 & =\frac{N_{0}a}{2a+b}
\end{align*}$

Now we have 
$\begin{align*}
\hat{N}_{1} & =\left(\frac{N_{0}\left(a+b\right)}{\left(2a+b\right)}\right)\frac{1}{s}+\left(\frac{N_{0}a}{2a+b}\right)\frac{1}{s+2a+b}
\end{align*}$

then taking the inverse Laplace transform we have
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[crystal roost]

My solution attempt:

So my solution doesn't really match the given solution with regard to + and - signs but I can't see why

[crystal roost]

solved

[tough hound]

.close