# Solving a system of linear ODEs using variables

I am trying to solve the following system of linear ODEs.

$\dfrac{dx}{dt} = x-4y$

$\dfrac{dy}{dt} = 4y$.

The initial conditions are $x(0) = -1$ and $y(0) = -3$.

I have tried letting:

$u_1 = x \implies u_1' = u_2$

$u_2 = x' \implies u_2 = u_1 - u_3$

$u_3 = y \implies u_3' = u_4$

$u_4 = y' \implies u_4 = 4u_3$

I also know that $y' = x - x'$ however I am unsure how to tie all of this information together. Any help is appreciated. Thanks in advance.

$$\frac{dy}{dt} = 4y \rightarrow y = C_1e^{4t}$$

So then we can substitute into earlier where we have

$$\frac{dx}{dt} = x - 4y \rightarrow \frac{dx}{dt} = x - 4C_1e^{4t}$$

Now the equation

$$\frac{dx}{dt} = x - 4C_1e^{4t}$$

Is a standard first order linear ODE which can be written as

$$\frac{dx}{dt} - x = - 4C_1e^{4t}$$

By choosing the integration factor $e^{-t}$ It can be seen that

$$xe^{-t} = -4C_1 \int e^{-t}e^{4t} dt$$

Yielding

$$x = C_2e^{t} - 4C_1 e^{t} \int e^{3t} dt$$

Thus the final solution is:

$$x = C_2e^{t} - \frac{4C_1}{3} e^{4t}$$

$$y = C_1e^{4t}$$

• Thank you. That is a very clear answer and i see how it works. – letsmakemuffinstogether Jun 15 '15 at 20:02

We have $$y' = 4y \implies y(t) = C e^{4t} \quad \text{where} \quad C \in \mathbb{R}$$ This implies $$x' = x -4y = x - 4 C e^{4t}$$ which is a first order linear equation. Now you just have to solve $$x' -x = 4 C e^{4t}$$ which you've most likely seen.

Here is another way, let $A=\begin{bmatrix} 1 & -4 \\ 0 & 4\end{bmatrix}$ and note that $A (1,0)^T = (1,0)^T$ and $A(4,-3)^T = 4 (4,-3)^T$.

Hence starting from an initial condition $(1,0)^T$, the solution will be $t \mapsto e^t (1,0)^T$ and starting from $(4,-3)^T$ the solution will be $t \mapsto e^{4t} (4,-3)^T$.

Now write $(-1,-3)^T$ as a linear combination of $(1,0)^T$, $(4,-3)^T$.