# Differential equations error of magnitude question

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.

## 1 Answer

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.

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