Science:Math Exam Resources/Courses/MATH102/December 2013/Question A 03

MATH102 December 2013

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Question A 03
Consider the differential equation and initial condition ${\frac {dy}{dt}}=2-y^{2},\quad y(0)=1$ Use Euler's method with one step of size $\Delta t=0.1$ to approximate the value of the solution at time $t=0.1$ (a) $y(0.1)=2$ (b) $y(0.1)=2.1$ (c) $y(0.1)=2.2$ (d) $y(0.1)=1.2$ (e) $y(0.1)=1.1$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Euler's method is an iterative method whereby we start with the initial value, plug it into our differential equation to find the slope, then use the slope of this tangent line to find the next point on the curve that is $\Delta t$ units away.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We start at the point $(0,1)$, where the slope of the tangent line at this point is ${\frac {dy}{dx}}=2-y^{2}=2-[y(0)]^{2}=2-1=1$ By applying Euler's method with a step size of $\Delta t=0.1$, our next point would be at the value $y(0.1)=y(0)+\left.{\frac {dy}{dx}}\right|_{x=0}(0.1-0)=1+0.1=1.1$, which is answer ${\color {blue}{\text{(E)}}}$.

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