# Solutions of the ODEs - 2 first order linear equations

#### Julio

##### Member
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
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
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
Hint: What is the derivative of $$\displaystyle A e^{Bt}$$?

-Dan

#### HallsofIvy

##### Well-known member
MHB Math Helper
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
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

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

is correct?

#### Country Boy

##### Well-known member
MHB Math Helper
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.