# Thread: solving a system by substitution 3b

1. ## solving a system by substitution 3b

solve this system by sustituting$\displaystyle t=e^u$ and building a function $\displaystyle y(u)=x(e^u)$
$\displaystyle t\frac{dx}{dt}=(\begin{array}{cc} 2 & 2\\ 1 & 3\end{array})x$
$\displaystyle x(1)=(\begin{array}{c} 2\\ 3\end{array})$

how to do it?

2. Apply the chain rule to get: $\displaystyle \displaystyle \frac{dy}{du} = e^u \cdot \frac{dx}{dt} = t \cdot \frac{dx}{dt}$ ( $\displaystyle \frac{dx}{dt}$ evaluated at $\displaystyle e^u = t$ )

Then you have $\displaystyle \dot{y} = \begin{pmatrix} 2 & 2\\ 1 & 3 \end{pmatrix} \cdot y$ now you could solve it, say, by computing the matrix exponential.

For the initial values think what knowing $\displaystyle x(1)$ would mean in terms of $\displaystyle y$, at the end, when you have $\displaystyle y$ you can undo the change of variables to get the solution for $\displaystyle t>0$.

3. you sat that y=x(u)t(u)=x(u)e^y
so the derivative is must have sum of two member
why you have one?