solving a system by substitution 3b

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January 5th 2011, 08:50 AM
transgalactic

solving a system by substitution 3b

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

how to do it?
January 6th 2011, 10:11 AM
PaulRS

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 $\dot{y} = \begin{pmatrix}<br /> 2 & 2\\ <br /> 1 & 3<br /> \end{pmatrix} \cdot y$ 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 $t>0$ .
January 18th 2011, 11:39 PM
transgalactic

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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