MHB Understanding Systems of DEs with Non-Zero $b(t)$

[madflame991]

Hi!

Let's say we have a system of DEs
$$
\begin{cases}
\frac{dx}{dt} = y + e^t\\
\frac{dy}{dt} = x - t^2\\
\end{cases}
$$

One would write it in matrix form and compute the eigenvectors and stuff like in this tutorial (can't post links due to low post count - it's from "Paul's online math notes")

The only problem is that in that tutorial $b(t)$ is always $0$. In my case $b(t)$ is $(e^t ; -t^2)$
Where does $b(t)$ fit in when solving this type of system?

 

[Mathwebster]

There's a "variation of parameters" method for systems.

 

[Prove It]

Hi!

Let's say we have a system of DEs
$$
\begin{cases}
\frac{dx}{dt} = y + e^t\\
\frac{dy}{dt} = x - t^2\\
\end{cases}
$$

One would write it in matrix form and compute the eigenvectors and stuff like in this tutorial (can't post links due to low post count - it's from "Paul's online math notes")

The only problem is that in that tutorial $b(t)$ is always $0$. In my case $b(t)$ is $(e^t ; -t^2)$
Where does $b(t)$ fit in when solving this type of system?

In this case, note that $ \displaystyle \frac{dx}{dt} = y + e^t \implies \frac{d^2x}{dt^2} = \frac{dy}{dt} + e^t $, so that means

\[ \displaystyle \begin{align*} \frac{d^2x}{dt^2} &= x - t^2 + e^t \\ \frac{d^2x}{dt^2} - x &= e^t - t^2 \end{align*} \]

which is a second order linear constant coefficient nonhomogeneous ODE.

 

[Mathwebster]

In this case, note that $ \displaystyle \frac{dx}{dt} = y + e^t \implies \frac{d^2x}{dt^2} = \frac{dy}{dt} + e^t $, so that means

\[ \displaystyle \begin{align*} \frac{d^2x}{dt^2} &= x - t^2 + e^t \\ \frac{d^2x}{dt^2} - x &= e^t - t^2 \end{align*} \]

which is a second order linear constant coefficient nonhomogeneous ODE.

I think he wants a systems approach to this problem.

 

[HallsofIvy]

Either "variation of parameters" as Danny suggested or try "undetermined coefficients" trying a solution of the form $$\begin{bmatrix}Ae^t+ Bt^2+ Ct+ D \\ Ee^t+ Ft^2+ Gt+ Y\end{bmatrix}$$.