Is this DE solvable? So this might be a very simple question and I've got a knot in my brain but I've been at this for days and don't see how I could solve this problem.
I'm trying to find the solution for both x(t) and y(t) in this equation:
$$a\frac{d}{dt} x(t)+ b \frac{d}{dt}y(t)=c x(t) $$
a,b,c and d are all greater than zero and real values.
The initial conditions x(0) and y(0) are known.
(The tags probably aren't correct)
For context:
I've stumbled upon this problem trying to find the solution for an electrical network consisting of two parallel capacitors with an ESR and another parallel resistor.
The original set of equations for this problem are derived from this
[circuit][1].
The current trough the single resistor is defined as:
$$i_{sc}=i_A+i_B$$
The current $i_A$ of capacitor A and B respectively are defined as:
$$\frac{d}{dt}u_{cA}(t)=\frac{1}{C_A}i_{A}(t)$$
$$\frac{d}{dt}u_{cB}(t)=\frac{1}{C_B}i_{B}(t)$$
with
$$i_A=\frac{u_{cA}-i_{sc}\cdot R_{sc}}{R_A}$$
$$i_B=\frac{u_{cB}-i_{sc}\cdot R_{sc}}{R_B}$$
Using $i_A=C_A\cdot \frac{d}{dt}u_{cA}$ and $i_B=C_B\cdot \frac{d}{dt}u_{cB}$ in the equation for $i_{sc}$ I get:
$$ C_A\cdot \frac{d}{dt}u_{cA}=\frac{u_{cA}-(C_A\cdot \frac{d}{dt}u_{cA}+C_B\cdot \frac{d}{dt}u_{cB})\cdot R_{sc}}{R_A}$$
$$ \frac{d}{dt}u_{cA} \cdot C_A \cdot(R_A +R_{sc}) =u_{cA}-C_B\cdot \frac{d}{dt}u_{cB}\cdot R_{sc}$$
Based on this equation I get the simplified version with $x=u_{cA}$ and $y=u_cB$:
$$a\frac{d}{dt} x(t)+ b \frac{d}{dt}y(t)=c x(t) $$
[1]: https://i.stack.imgur.com/9F7eB.png
I hope this helps!
 A: At begining your system of equations is consistent : Five unknowns $i_{sc}$ , $i_A$ , $i_B$ , $u_{cA}$ , $u_{cB}$ , $ $ and five equations :
$$\begin{cases}
i_{sc}=i_A+i_B \\
\frac{d}{dt}u_{cA}(t)=\frac{1}{C_A}i_{A}(t)\\
\frac{d}{dt}u_{cB}(t)=\frac{1}{C_B}i_{B}(t)\\
i_A=\frac{u_{cA}-i_{sc}\cdot R_{sc}}{R_A}\\
i_B=\frac{u_{cB}-i_{sc}\cdot R_{sc}}{R_B}
\end{cases}$$
Note : I didn't check if the equations are correct even if it seems at first sight that they are some sign mistakes with regard to the direction of currents drawn on the shematic of the circuit. Since this isn't matematical but modelisation I will not discuss this in my answer. The above system of equations is taken as granted despite the doubt.
During your manipulations 3 unknows where eliminated as well as 4 equations instead of only 3 unfortunately.
It is easy to elimnate $i_{sc}$ and the first equation :
$$\begin{cases}
\frac{d}{dt}u_{cA}(t)=\frac{1}{C_A}i_{A}(t)\\
\frac{d}{dt}u_{cB}(t)=\frac{1}{C_B}i_{B}(t)\\
i_A=\frac{u_{cA}-(i_A+i_B)\cdot R_{sc}}{R_A}\\
i_B=\frac{u_{cB}-(i_A+i_B)\cdot R_{sc}}{R_B}
\end{cases}$$
Then $i_A$ and $i_B$ can be isolated from the two last equations.
$i_A=\frac{u_{cB}\left( 1+\frac{R_A}{R_{sc}} \right)-u_{cA}}{R_{A}\left( 1+\frac{R_A}{R_{sc}} \right)+R_{B}}\quad$ and $\quad i_B=\frac{u_{cA}\left( 1+\frac{R_B}{R_{sc}} \right)-u_{cB}}{R_{A}\left( 1+\frac{R_A}{R_{sc}} \right)+R_{B}}\quad$ that we eliminate in puting them in the two remaining equations :
$$\begin{cases}
\frac{d}{dt}u_{cA}(t)=\frac{1}{C_A}\frac{u_{cB}\left( 1+\frac{R_A}{R_{sc}} \right)-u_{cA}}{R_{A}\left( 1+\frac{R_A}{R_{sc}} \right)+R_{B}}\\
\frac{d}{dt}u_{cB}(t)=\frac{1}{C_B}\frac{u_{cA}\left( 1+\frac{R_B}{R_{sc}} \right)-u_{cB}}{R_{A}\left( 1+\frac{R_A}{R_{sc}} \right)+R_{B}}
\end{cases}$$
With simplified  notations :
$$\begin{cases}
\frac{d}{dt}x(t)=c_1(c_3\:y(t)-x(t))\\
\frac{d}{dt}y(t)=c_2(c_3\:x(t)-y(t))
\end{cases}$$
Now we have got a consistent system of two equations with two unknons $u_{cA}$ and $u_{cB}$.
Of course all this would be simpler with matrix calculus from the very start.
To continue I suppose that you have learned how to solve a system of several linear ODEs.
A: With two unknowns and a single equation you get a relation between $x(t),y(t)$ that can be solved/plotted. One more equation is needed for solving them both.
