Calculus 2 Equations: Expert Solutions and More

Solving 2nd order homogenous linear ODE with a squared coefficient
I am attempting to solve a differential equation of the form:
${\displaystyle \frac{d^{2}f}{d{x}^2}-{\gamma }^{2}f=0}$
I have set up and solved the characteristic equation as:
$\begin{array}{r}{z}^2-{\gamma }^{2}z+0z=0\\ z(z-{\gamma }^2)=0\end{array}$
which is satisfied when ${\displaystyle z=0}$ and when ${\displaystyle z={\gamma }^2}$
There are two distinct roots (${\displaystyle r_1\text{and}{r}_2}$) hence the general solution should be of the form
${\displaystyle f\left(x\right)=A{e}^{r_{1}x}+B{e}^{r_{2}x}}$
In this case:
$\begin{array}{rl}f\left(x\right)& =A{e}^{0\times x}+B{e}^{{\gamma }^{2}x}\\ & =A+B{e}^{{\gamma }^{2}x}\end{array}$
Apparently, this solution is incorrect as in the answers it is given as:
${\displaystyle f\left(x\right)=A{e}^{\gamma x}+B{e}^{-\gamma x}}$
I would like to know where I went wrong to give me the incorrect solution

Solving a system of differential equations with the variable x
How to find a basis of solutions of the system for
$y\text{'}=\left(\begin{array}{cc}3x-1& x-1\\ -x-2& x-2\end{array}\right)y$
where one solution is $y=\left(\begin{array}{c}{y}_1\\ -y_1\end{array}\right)$?

Solving a differential equation in matrix form but adding a constant
$$\begin{array}{rl}{x}_1^{\prime }& =2x_1+3x_2,\\ x_2^{\prime }& =x_1-x_2.\end{array}$$

Solve a differential equation involving matrices.
Let $A=\left[\begin{array}{cc}-1& 1\\ 0& -1\end{array}\right],b=\left[\begin{array}{c}2\\ 3\end{array}\right]$. 

Solve the differential equation
$\dot{x}=Ax+b; x\left(0\right)=\left[\begin{array}{c}1\\ 2\end{array}\right]$.
Use formula 

$\begin{array}{rl}x\left(t\right)& =e^{tA}x\left(0\right)+{\int }_0^t{e}^{(t-s)A}b\text{d}s\\ & =e^{-t}\left[\begin{array}{c}1+2t\\ 2\end{array}\right]+{\int }_0^t{e}^{-(t-s)}\left[\begin{array}{cc}1& t-s\\ 0& 1\end{array}\right]\left[\begin{array}{c}2\\ 3\end{array}\right]\text{d}s\\ & ⋮\end{array}$
My question is how do we get from the second summand on the RHS on line 1 to the second summand on the RHS of the equation on line 2? That is how does the equation come out?
${\int }_0^t{e}^{(t-s)A}b \text{d}s={\int }_0^t{e}^{-(t-s)}\left[\begin{array}{cc}1& t-s\\ 0& 1\end{array}\right]\left[\begin{array}{c}2\\ 3\end{array}\right]\text{d}s$

Solution to this differential equation that does not diverge at ${\displaystyle x=0}$.
$\tan x \frac{\text{d}y}{\text{d}x}+y={\mathrm{e}}^x\tan x$
By using the integrating factor ${\displaystyle \mu \left(x\right)=\cos x}$, I solved it as an equation in full differentials and got the solution
$y\left(x\right)=A\csc x+\frac{1}{2}{\mathrm{e}}^x(1-\cot x)$
However, the question I am solving asks for a solution that does not diverge at ${\displaystyle x=0}$, which this solution clearly does because of the cotx. How can I get a solution that converges?

X1 - x2 + x4 =5

Verify that the function is a solution to the differential equation
${\displaystyle u\left(t\right)=e^{t^2}{\int }_0^t{e}^{-s^2}ds+e^{t^2}}$
is a particular solution to the differential equation:
${\displaystyle \frac{du}{ dt }-2tu=1}$
My Attempt
I will verify this by differentiating the function with respect to t.
$\begin{array}{rl}\frac{du}{dt}& =\frac{d}{dt}(e^{t^2}{\int }_0^t{e}^{-s^2}ds+e^{t^2})\\ & =2t{e}^{t^2}\left({\int }_0^t{e}^{-s^2}ds\right)+e^{t^2}\frac{d}{dt}\left({\int }_0^t{e}^{-s^2}ds\right)+2t{e}^{t^2}\end{array}$
I'm having trouble solving the integral because it involves an error function. Could I get some pointers on how to evaluate this? Or is there a different way to verify that the function is a solution?