Course:MATH102/Question Challenge/2002 December Q1e

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Question

If Euler's Method is used to solve the differential equation

$\displaystyle {\frac {dy}{dt}}=y(4-y),\quad y(0)=2$

with step size $\Delta t=h=0.1$, then the value of $y_{1}$ (i.e. the approximate solution at the first time) step is

${\displaystyle }$

Hints

[show]Hint
Euler's Method is as follows: $y1=y0+h*(y0(4-y0))$ Where $y0$ is the previous $y$ value, $y1$ is the next value, and $h$ is the step value you increment by.

Solutions

[show]Solution
$dy/dt=y(4-y),y(0)=y_{0}=2$ Let $f(y)$ represent the function on the right-hand side of the differential equation. Applying Euler's Method: $y(t_{1})=y_{0}+h\cdot f(y_{0})$ $y(t_{1})=y_{0}+h\cdot (y_{0}(4-y_{0}))$ $y(t_{1})=2+0.1\cdot (2(4-2))$ $y(t_{1})=2+0.1\cdot 4$ $y(t_{1})=2+0.4$ $y(t_{1})=2.4$