solving a system by substitution 3b

**transgalactic** · January 5th 2011, 08:50 AM

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

how to do it?

**PaulRS** · January 6th 2011, 10:11 AM

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

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

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

**transgalactic** · January 18th 2011, 11:39 PM

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?

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