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Prove that

$\alpha e^{\lambda t} \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} + \beta e^{\lambda t} \begin{pmatrix} t \\ 1 \\ \end{pmatrix}$

is the general solution of

$X' = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \\ \end{pmatrix}X$

It was suggested I assume $Z(t)$ is a solution $Z = \begin{bmatrix} z_1 \\ z_2 \\ \end{bmatrix}$. Now I'm supposed to construct the function $f(t)=e^{-\lambda t}(z_1 - tz_2)$ and show that $f'(t)=0$.

First, how do I construct the function? I think you would start out $X'= \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \\ \end{pmatrix}Z$, which leads to $X'= \begin{pmatrix} \lambda z_1 + z_2 \\ \lambda z_2 \\ \end{pmatrix}$, right? And I think $\lambda z_1 = \alpha e^{\lambda t}z_1$, correct? But I'm not really sure if any of this is correct or where to go from there...

Bryyo
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    If you know that the space of solutions has dimension $2$, it suffices to show that the solutions are linearly independent. – Git Gud Feb 09 '15 at 17:00

1 Answers1

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Here is what I think your professor is getting at.

If $Z$ satisfies the ODE, then $\dot{z_2} = \lambda z_2$ and so $z_2$ has the form $z_2(t) = z_2(0) e^{\lambda t}$. To get $z_1$ you can solve the equation $\dot{z_1} = \lambda z_1+z_2$, which is why (I think) your professor gave you the function $f$.

Show that $\dot{f}(t) = 0$. This follows because $Z$ satisfies the ODE, and means that $f$ is constant. So $z_1(t)-t z_2(t) = c_0 e^{\lambda t}$ for some constant. That is, $z_1(t) = c_0 e^{\lambda t} + t z_2(0) e^{\lambda t}$.

To see why $\dot{f}(t) = 0$, differentiate to get $\dot{f}(t) = e^{-\lambda t} (\dot{z_1}(t)-t \dot{z_2}(t) - z_2(t)) - \lambda e^{-\lambda t} ( z_1(t) - t z_2(t))$. Now use the fact that $\dot{z_1}(t) = \lambda z_1(t)+ z_2(t)$ and $\dot{z_2}(t) = \lambda z_2(t)$.

Now choose $\alpha=c_0, \beta = z_2(0)$.

Note: This solution glossed over one significant point, which is to show that $t \mapsto z_2(0) e^{\lambda t}$ is the only (unique) solution to $\dot{z_2} = \lambda z_2$ (with the specified initial condition). And if one is willing to accept that the solution of the differential equation (passing through a specified initial point) is unique, then there is a simpler solution as Git Gud's comment points out.

To show that the single dimensional equation has a unique solution (this can be extended to the general $\dot{x} = Ax$ case, but needs to use the matrix exponential), suppose $x,y$ are two solutions to $\dot{z_2} = \lambda z_2$ with $x(0) = y(0)$. Then let $\delta = x-y$ and note that $\delta$ satisfies the same differential equation. Then consider $\phi(t) = e^{-\lambda t} \delta(t)$. We see that $\dot{\phi}(t) = 0$, so $\phi(t) = \phi(0) = 0$ and so $\delta(t) = 0$ for all $t$, hence $x=y$.

Git Gud's comment below gives a more general version of this approach.

copper.hat
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    Tho whoever may care, the point copper.hat glossed over follows from this elementary approach. – Git Gud Feb 09 '15 at 17:08
  • I'm assuming $\dot z = z'$? And where does $z_2(t)$ come from? After that, how do you construct $f(t)$? And finally, why would $Z$ satisfying the ODE mean that $f(t)$ is constant? – Bryyo Feb 09 '15 at 17:15
  • Yes, I should have written $z'$, old habits. You are given a solution $Z$, I am writing $Z(t) = (z_1(t), z_2(t))^T$. If $Z$ satisfies the ODE, then $z_1'(t) = \lambda z_1(t) + z_2(t)$ and $z_2'(t) = \lambda z_2(t)$. The $f$ is given by your professor, and assumes the $Z$ is a solution to the ODE. – copper.hat Feb 09 '15 at 17:20
  • Ah. So when my professor said to "construct" $f(t)$ she merely meant define it that way to try using that? – Bryyo Feb 09 '15 at 19:23
  • She did. ${}{}{}$ – copper.hat Feb 09 '15 at 19:31
  • In that case I think I understand how to do this now. Thank you! – Bryyo Feb 09 '15 at 19:53