MHB Solutions of the ODEs - 2 first order linear equations

[Julio1]

Find the general solution of the ODE:

$\check{X_1}=X_1$

$\check{X_2}=aX_2$

where $a$ is a constant.

 

[HallsofIvy]

I find this hard to read. Is that symbol above the "X"s a double dot, the second derivative symbol? I will assume that it is.

You actually have two differential equations not one. And they are completely separate so you can solve them separately.

My question is "where did you get these?" Are you taking a Differential Equations class? If so you should have learned how to solve "second order linear equations with constant coefficients". Do you see that the "characteristic equations" are r^2= 1 for the first and s^2= a for the second? Do you know what to do with those?

 

[Julio1]

Find the general solution of the ODE:

$\dot{X_1}=X_1$

$\dot{X_2}=aX_2$

where $a$ is a constant.

Now fix it. Can you help me now?

 

[topsquark]

Hint: What is the derivative of [math]A e^{Bt}[/math]?

-Dan

 

[HallsofIvy]

So it's a single dot- a first derivative. That's even easier. Note that $X_1'= \frac{dX_1}{dt}= X_1$ can be written as $\frac{dX_1}{X_1}= dt$ and $X_2'= aX_2$ can be written as $\frac{dX_2}{X_2}= adt$.

To "solve" such a differential equation, to go from the derivative of a function to the function itself, integrate both sides! That's why topsquark asked "what is the derivative of $Ae^{Bt}$?".

 

[Julio1]

So it's a single dot- a first derivative. That's even easier. Note that $X_1'= \frac{dX_1}{dt}= X_1$ can be written as $\frac{dX_1}{X_1}= dt$ and $X_2'= aX_2$ can be written as $\frac{dX_2}{X_2}= adt$.

To "solve" such a differential equation, to go from the derivative of a function to the function itself, integrate both sides! That's why topsquark asked "what is the derivative of $Ae^{Bt}$?".

Thanks Hallsoflvy :)

My answer is

$X(t)=\begin{equation}
\begin{pmatrix}
X_1\exp(t)\\
X_2\exp(at)
\end{pmatrix}
\end{equation}$

is correct?

 

[HOI]

It's easy to check isn't it? If $X_1(t)= X_1e^t$ then $X_1(x)'= X_1e^r= X_1(x)$.
If $X_2(t)= X_2e^{at}$ then $X_2(x)'= X_2(ae^{at})= aX_2$.

Yes, those satisfy both equations. Well done!

(Personally, I would not use "X1" as the constant in a funtion I had already called X1(t)!)