Let $x = x(t), y = y(t)$ be the solution to the initial-value problem

$$\frac{dx}{dt} = -x - y, \hspace{1em} \frac{dy}{dt} = 2x - y, \hspace{1em} x(0)=y(0)=1.$$

Suppose that we make an error of magnitude $10^{-4}$ in measuring $x(0)$ and $y(0)$. What is the largest error we can make when evaluating $x(t), y(t)$ for $0 \leq t \leq \infty$?

This is a question in my textbook that I am solving for practice for an upcoming quiz. It is in the section on systems of differential equations, and equilibrium values.

To find an equilibrium value, you can set $\dot{x} = \dot{y} = 0,$ and solve for $x, y$. But I don't see how that helps here.

I don't really know how to approach this problem, and error wasn't mentioned in the chapter. I would really appreciate some help.


You solve the system first. Substituting $y=-\dot x-x$ into another equation gives $$ \ddot x+2\dot x+3x=0。 $$ The general solution is $$ x=e^{-t}(A \cos(\sqrt 2 t)+B\sin (\sqrt 2 t)). $$ Now we can put in the initial conditions. $$ x=e^{-t}(x(0) \cos(\sqrt 2 t)-\frac{y(0)}{\sqrt 2}\sin (\sqrt 2 t)). $$ Let $\epsilon_x,\epsilon_y$ be the error in $x(0)$ and $y(0)$ respectively.

Error in $x=$ $$ |e^{-t}(\epsilon_x \cos(\sqrt 2 t)-\frac{\epsilon_y}{\sqrt 2}\sin (\sqrt 2 t))|\\ \leq e^{-t}(\epsilon_x |\cos(\sqrt 2 t)|+\frac{\epsilon_y}{\sqrt 2}|\sin (\sqrt 2 t)|)\\ \leq e^{-t}\sqrt{\epsilon_x^2+\frac{\epsilon_y^2}{2}}\leq \sqrt{\epsilon_x^2+\frac{\epsilon_y^2}{2}}=\sqrt{1.5\times 10^{-8}}=1.225\times 10^{-4}. $$ You can obtain more accuatate estimate by performing different operations with the inequalities. Your question should specify the degree of accuracy it wants.

Repeat this process for $y(t)$, and you are done.

  • $\begingroup$ Thanks so much! I am still confused about how to get the error $y$? Wouldn't it just be the exact same? $\endgroup$ – user662628 Apr 22 at 12:51
  • $\begingroup$ @effunna9 Yes of course. You can repeat the argument above. $\endgroup$ – Holding Arthur Apr 22 at 14:13
  • $\begingroup$ So both of the errors are the same? $1.225 \times 10^{-4}$? $\endgroup$ – user662628 Apr 22 at 15:10

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