Solve $\frac{dx}{dt}=3x-5y$ $\frac{dy}{dt}=5x-3y$ Given that $x(0)=1$ and $y(0)=1$ 
Solve 
  $$\frac{dx}{dt}=3x-5y$$ $$\frac{dy}{dt}=5x-3y$$ Given that $x(0)=1$ and $y(0)=1$

So I've got to the stage where i have two general solutions and I now need the particular solutions, but I don't know how to find one of the variables and both of the general solutions for $x$ and $y$ are the same which makes me think I've made a mistake. My workings are below any help would be great.
$$\frac{dx}{dt}=3x-5y$$
$$\therefore \frac{d^2x}{dt^2} = 3\frac{dx}{dt} - 5\frac{dy}{dt}$$
$$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} -5 \left(5x-3y \right) = 3\frac{dx}{dt} -25x+15y$$
$$\frac{d^2x}{dt^2} = 3\frac{dx}{dt} -25x+15 \left(\frac 3 5 x - \frac 1 5 \frac{dx}{dt}\right)$$
$$\frac{d^2x}{dt^2} +16x=0$$
$$\lambda ^2 +16=0 $$
$$\therefore \lambda = 4i$$
$$x=A\sin (4t) + B\cos(4t)$$
I then did the same for $y$ and obtained the same differential equation of $$\frac{d^2y}{dt^2} +16y=0$$ and thus $y=A\sin (4t) + B\cos(4t)$
Now using the initial conditions of $x(0)=1$ and $y(0)=1$ i found that $B=1$ but how do I find $A$ and won't it just be the same for both equations? 
 A: There are several ways to solve this system. Here is yet another approach using eigenvalue/vector methods from linear algebra.
Let's re-write the system as:
$$ \frac{d\vec{r}}{dt} = \begin{pmatrix}3 &-5\\5 &-3 \end{pmatrix}\vec{r}(t) = \pmb{A}\,\vec{r}(t)$$
where $\vec{r}(t)=\begin{pmatrix}x(t)\\y(t)\end{pmatrix}$. Then, solve the characteristic equation,
$$ \det(\pmb{A}-\lambda\pmb{I}) = 0 \Rightarrow \begin{vmatrix}3-\lambda &-5\\5 &-3-\lambda \end{vmatrix} = 0 \Rightarrow \lambda^2 +16 = 0 \Rightarrow \lambda = \pm \,4i$$
We find eigenvalues of $\pm \,4i$. Choosing $\lambda = 4i $ we find an associated eigenvector solving 
$$(\pmb{A}-\lambda\pmb{I})\vec{v} = (\pmb{A}-4i\pmb{I})\vec{v}=\vec{0} \Rightarrow \begin{pmatrix}3-4i &-5\\5 &-3-4i \end{pmatrix}\vec{v} = \begin{pmatrix}0\\0\end{pmatrix}$$
where after some algebra $\vec{v} = \begin{pmatrix}5 \\ 3-4i\end{pmatrix}$. Looking at the real and imaginary parts of $\vec{v}\, e^{4it}$ will give use two linearly independent solutions to our system. With some algebra we find,
$$\vec{v}\, e^{4it} = \begin{pmatrix}5 \\ 3-4i\end{pmatrix}(\cos 4t +i\sin 4t) = \begin{pmatrix}5\cos 4t \\ 3\cos 4t + 4\sin 4t\end{pmatrix} +i \begin{pmatrix}5\sin 4t \\ 3\sin 4t -4\cos 4t\end{pmatrix} $$and thus our linear independent solutions are $\vec{r_1}(t) = \begin{pmatrix}5\cos 4t \\ 3\cos 4t + 4\sin 4t\end{pmatrix}$ and $\vec{r_2}(t) =\begin{pmatrix}5\sin 4t \\ 3\sin 4t -4\cos 4t\end{pmatrix} $. 
With these we now seek a solution to the system of the form $\vec{r}(t) = C_1\vec{r_1}(t)+C_2\vec{r_2}(t) $ where $C_1$ and $C_2$ are constants that solve the initial value problem $x(0)=1$ and $y(0)=1$, or matching our matrix/vector form $\vec{r}(0)=\begin{pmatrix}1\\1\end{pmatrix}$. This gives,
$$\vec{r}(0) = C_1\begin{pmatrix}5\\3\end{pmatrix} +C_2\begin{pmatrix}0\\-4\end{pmatrix} =\begin{pmatrix}1\\1\end{pmatrix} \Rightarrow \begin{pmatrix}5 &0\\3 &-4 \end{pmatrix}\begin{pmatrix}C_1\\C_2\end{pmatrix} = \begin{pmatrix}1\\1\end{pmatrix}$$ and solving these linear equations gives $C_1 = \frac{1}{5}, C_2 = \frac{-1}{10}$. Finally substituting these into the expression for $\vec{r}(t)$ gives
$$ \vec{r}(t) = \begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\frac{1}{5}\begin{pmatrix}5\cos 4t \\ 3\cos 4t + 4\sin 4t\end{pmatrix} - \frac{1}{10}\begin{pmatrix}5\sin 4t \\ 3\sin 4t -4\cos 4t\end{pmatrix} = \begin{pmatrix}\cos 4t - \frac{1}{2}\sin 4t\\\cos 4t + \frac{1}{2}\sin 4t\end{pmatrix}.$$
A: Here is another approach using Laplace transform for $t > 0$.
Let $\mathcal{L}(f(t)) = F(s)$  be the Laplace transform of $f(t)$, and $\mathcal{L}(f'(t)) = sF(s)-f(0^{+})$ be its derivative property. Applying this in your system we have:
$$sX(s)-x(0^{+}) = 3X(s)-5Y(s)$$
$$sY(s)-y(0^{+}) = 5X(s)-3Y(s),$$ which leads to the linear system (after using the initial conditions):
$$(s-3)X(s)+5Y(s) = 1$$
$$-5X(s)+(s+3)Y(s) =1.$$
The solutions are
$$X(s) = \dfrac{s-2}{s^2+16} = \dfrac{s}{s^2+16} - 0.5\dfrac{4}{s^2+16}$$ and $$ Y(s) = \dfrac{s+2}{s^2+16} = \dfrac{s}{s^2+16} + 0.5\dfrac{4}{s^2+16},$$
which are the Laplace transforms of cosine and sinus waves.
Thus,
$$x(t) = \cos(4t) -0.5\sin(4t)$$ and $$ y(t) = \cos(4t) +0.5\sin(4t),$$
for $t > 0$.
A: Be careful. The coefficients $A$ and $B$ in your general solution for $x(t)$ aren’t necessarily the same as the coefficients for $y(t)$. You should use different variables to denote them: let’s say $y(t)=C\sin{4t}+D\cos{4t}$ 
From the initial conditions, then, you have $x(0)=B=1$ and $y(0)=D=1$. To find the values of the other two coefficients, substitute back into the original system of equations: $$\dot x = 4A\cos{4t}-4\sin{4t} = (3A-5C)\sin{4t}-2\cos{4t} \\ \dot y = 4C\cos{4t}-4\sin{4t} = (5A-3C)\sin{4t}+2\cos{4t}.$$ This must hold for all values of $t$. Setting $t=0$ yields $4A=-2$ and $4C=2$.
