# Solving the system of differential equations

I need to solve the system of equations

$$tx'=\begin{pmatrix} 2 & -1 \\ 3 & -2 \\ \end{pmatrix}x$$ where $t>0$. I can solve this system if there was not any $t$ there. How do I treat that $t$?

Any help would be appreciated!

• Does $x'$ mean $dx/dt$ (so is the $t$ in front the same as the variable wrt which you're differentiating?) Commented Mar 10, 2015 at 21:08
• I guess so. The book gives no such an indication tho. I am using elementary differential equations by boyce and prima. page 406 @SmileySam Commented Mar 10, 2015 at 21:10
• Sorry, don't know that book. Has the book been doing linear DEs? It'd be worth looking through to see what notation the $'$ means. It will have a major effect on the answer! Commented Mar 10, 2015 at 21:13
• yes. Linear systems with constant coefficients Commented Mar 10, 2015 at 21:14
• I know. I couldn't decide that either Commented Mar 10, 2015 at 21:14

Change variable $\tau=\log(t)$ Then $$t dx/dt= dx/d(\log(t))=dx/d\tau$$ Solve the system with respect to $\tau$ and replace finally $\tau$ by $\log(t)$.

Ok, I'm going to suppose that the prime $'$ is not meaning $d/dt$, so the $t$ in front is just a constant.

Now, there are three ways of doing this; (1)(i), (1)(ii) and (2).

(1) The first is just brute force - bash through some algebra with a hammer. Write it as

$x'=\begin{pmatrix} 2/t & -1/t \\ 3/t & -2/t \\ \end{pmatrix}x.$

Now this is in your usual form. Just do all the algebra in whatever way you like to. This gives (1)(i) and (1)(ii).

(i) Find the eigenvalues and eigenvectors, then diagonalise the matrix. Define your new variable $u$ so that $u' = Du$, where $D$ is diagonal. This is immediate to solve as uncoupled. Then just undo the transform to get $x$ back. (Note that you will have to carry around the factor $t$.)

(ii) Expand out the matrix to give two equations. Differentiate one of them, then use the other to substitute in and eliminate one variable. This gives a second order ODE with constant coefficients - straight forward to solve in principle, just takes a bit of algebra. You'll get two solutions for it (as second order). These are your $x_1$ and $x_2$.

(2) Now this is the clever one. Let's suppose that $'$ is $\frac{d}{du}$. So $x' = \frac{dx}{du}$. Now, define $v = tu$. Then $\frac{d}{dv} = \frac{1}{t}\frac{d}{du}$ ($\frac{du}{dt} = \frac{1}{t}$). So now $$\frac{d}{dv}\left[t\frac{dx}{du}\right] = \frac{dx}{dv}.$$ This gets rid of the $t$, and you can solve as usual, then just change variables back.

Finally, if $t$ is the differentiation variable, then you get the answer as Alexander Vigodner has given: $\tau = \log(t)$ and you do what I've done but for this change of variables. The reason for choosing this change of variable is the following. You desire $\tau$ such that $\frac{d}{d\tau} = t\frac{d}{dt}$, and by the chain rule this is equivalent to requiring $\frac{dt}{d\tau} = t$, ie $t = e^\tau \iff \log(t) = \tau$.

Hope this helps! :)