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#### 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.

$\check{X_1}=X_1$

$\check{X_2}=aX_2$

where $a$ is a constant.

- Thread starter Julio
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- Thread starter
- #1

- 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.

$\check{X_1}=X_1$

$\check{X_2}=aX_2$

where $a$ is a constant.

- 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?

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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- #3

- Feb 14, 2014

- 71

Now fix it. Can you help me now?Find the general solution of the ODE:

$\dot{X_1}=X_1$

$\dot{X_2}=aX_2$

where $a$ is a constant.

Hint: What is the derivative of \(\displaystyle A e^{Bt}\)?

-Dan

-Dan

- 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 $Ae^{Bt}$?".

To "solve" such a differential equation, to go from the derivative of a function to the function itself,

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- #6

- Feb 14, 2014

- 71

Thanks HallsoflvySo 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,integrateboth sides! That's why topsquark asked "what is the derivative of $Ae^{Bt}$?".

My answer is

$X(t)=\begin{equation}

\begin{pmatrix}

X_1\exp(t)\\

X_2\exp(at)

\end{pmatrix}

\end{equation}$

is correct?

- Jan 30, 2018

- 344

If $X_2(t)= X_2e^{at}$ then $X_2(x)'= X_2(ae^{at})= aX_2$.

Yes, those satisfy both equations.