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I am in Calc 4 and I am trying to solve the followings:

$\displaystyle \frac {dx} {dt}=-4x-y$

$\displaystyle \frac {dy} {dt}=x$

My handout from professor starts with

"We experiment with $y(t)=e^{at}$..."

My question is that

Is every system of linear differential equations with two variables can be solved by substituting $y=e^{at}$?

Or is it like the general solution is so difficult that some professors do not want to teach because it is taught in more higher classes

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3 Answers 3

up vote 1 down vote accepted

This is a linear system of differential equations with constant coefficients. The substitution suggested works here for much the same reasons it works for a single such equation. And just as in that case it can happen that you have to resort to $t^k e^{a t}$.

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If the system has a repeated eigenvalue corresponding to an eigenspace parallel with the $x$-axis, would there still be solutions even though $y = t^k e^{at}$ wouldn't find any? –  muzzlator Apr 13 '13 at 2:25

We are given:

$\tag 1 \displaystyle \frac {dx} {dt}=-4x-y$

$\tag 2 \displaystyle \frac {dy} {dt}=x$

We experiment with:

$\tag 3 y(t)=e^{at}.$

Well, lets use this and see where it leads us. Taking the derivative of $(3)$ yields:

$\tag 4 \displaystyle \frac{dy}{dt} = a e^{at}.$

We know that $\displaystyle x(t) = \frac{dy}{dt}$, so $\displaystyle \frac{dx}{dt} = a^2 e^{at}$, and so lets substitute this and $(4)$ into $(1)$, yielding:

$\displaystyle \frac {dx} {dt}= a^2 e^{at} = -4x-y = -4(a e^{at}) - e^{at}.$

After simplification, we have:

$$a^2 = -4a -1 \rightarrow a^2 + 4a + 1 = 0.$$

The roots are: $a_{1, 2} = \pm (\sqrt{3}) - 2.$

Do you know how to write the solutions from this?

You will learn many approaches to solve a system like this from eigenvalues, eigenvectors, characteristics and many others.

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Nice details! +1 –  amWhy Apr 13 '13 at 3:37
    
Thank you, I wanted to stay away from the matrix methods as this looks like a calculus class, so thought using the basic approach from the hint was more appropriate. –  Amzoti Apr 13 '13 at 3:41
    
Exactly...sometimes users answer at levels beyond the OP current level of exposure! So + 1 for OP sensitivity! –  amWhy Apr 13 '13 at 3:44
    
+1 nice and leading way, @amWhy. –  B. S. Apr 13 '13 at 5:09

Hint:$\frac{dx}{dy}=\begin{pmatrix} -4 & -1\\ 1 & 0\\ \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$

answer is $\begin{pmatrix}x\\y\end{pmatrix}=$$e^{t\delta_1}v_1+e^{t\delta_2}V_2$ such that $\delta_1,\delta_2$ are eigen value and $V_1,V_2$ are eigen vector correspond to $\delta_1,\delta_2$

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nice Maisam. +1.. –  B. S. Apr 13 '13 at 5:10

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