Solving a differential equation with a linear solution and initial conditions The problem reads: 
The solution of a certain differential equation is of the form
$$y(t)= a\exp(5t) + b\exp(8t)$$
where $a$ and $b$ are constants.
The solution has initial conditions $y(0)=5$ and $y'(0)=5$
Find the solution by using the initial conditions to get linear equations for $a$ and $b$.
....................
What I did was solve using the initial conditions and I found that
$a + b = 5 $
and 
$
5a + 8b =5. $
Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step. 
 A: You're absolutely right.
If $y(t) = a\mathrm e^{5t} + b\mathrm e^{8t}$ then $y'(t) = 5a\mathrm e^{5t} + 8b\mathrm e^{8t}$. 
Hence $y(0) = a\mathrm e^0 + b\mathrm e^0 = a+b$ and $y'(0) = 5a\mathrm e^0 + 8b\mathrm e^0 = 5a+8b$.
To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.
Personally I would use matrix algebra, but it's up to you.
$$\left(\begin{array}{cc} 1 & 1 \\ 5 & 8 \end{array}\right)
\left(\begin{array}{c} a \\ b \end{array}\right)=
\left(\begin{array}{c} 5 \\ 5 \end{array}\right) $$
The two-by-two matrix on the right has determinant $1\times 8 - 5\times 1 = 3 \neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives
$$
\left(\begin{array}{c} a \\ b \end{array}\right)=
\frac{1}{3}\left(\begin{array}{cc} 8 & -1 \\ -5 & 1 \end{array}\right)
\left(\begin{array}{c} 5 \\ 5 \end{array}\right) $$
Expanding the right hand side gives
\begin{eqnarray*}
a &=& \frac{1}{3}(8\times 5 - 1\times 5) &=& \frac{35}{3} \\ \\
b &=& \frac{1}{3}(-5\times 5 + 1 \times 5) &=&-\frac{20}{3}
\end{eqnarray*}
Your final solution is then
$$\boxed{y(t) = \tfrac{35}{3}\mathrm e^{5t} - \tfrac{20}{3}\mathrm e^{8t}} $$
