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Solutions of the ODEs - 2 first order linear equations

Julio

Member
Feb 14, 2014
71
Find the general solution of the ODE:

$\check{X_1}=X_1$

$\check{X_2}=aX_2$

where $a$ is a constant.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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?
 
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Julio

Member
Feb 14, 2014
71
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,104
The Astral plane
Hint: What is the derivative of \(\displaystyle A e^{Bt}\)?

-Dan
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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 $Ae^{Bt}$?".
 
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Julio

Member
Feb 14, 2014
71
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?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
344
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.