Author Topic: Euler's method of approximation (Read 1839 times) Tweet Share

NE2000 · « **on:** April 03, 2009, 05:54:10 pm »

Sketch the graph of y(x) over domain [0,4] if y(x) is solution of

y' = y/(y-x), y(0) = 3

I don't know where to start, just a hint would be good for starters, thanks

kamil9876 · « **Reply #1 on:** April 03, 2009, 09:12:44 pm »

I'm pretty sure this graph is an increasing one. Since y'(0)>0 it's increasing at x=0 but in order for it change into a decreasing function, it must first reach a turning point but the only way it can reach a turning point is if y=0 but it can't reach y=0 before it reaches a turning point since the y value is intially greater than y=0 and increasing.

Mao · « **Reply #2 on:** April 03, 2009, 11:53:44 pm »

This is effectively a differential equation: $\frac{dy}{dx} = \frac{y}{y-x}$

Remembering euler's formula for linear approximation: $y_{n+1} \approx \frac{dy}{dx}\cdot \Delta x + y_{n}$ , given that $(x_0,y_0) = (0,3)$

Hence taking a $\Delta x$ of 1: $y_1 \approx \frac{y_0}{y_0 - x_0}\cdot \Delta x + y_0 = 6$

$y_2 \approx \frac{y_1}{y_1 - x_1} \cdot \Delta x + y_1 = \frac{6}{5}+6 = 7.2$

etc, plot the graph till x=4

Note that the accuracy of this graph depends on the value of $\Delta x$ , the smaller this increment is, the more accurate the approximation is, as $\Delta x \to dx$ , the approximation becomes closer and closer to the solution to the differential equation.

[The original equation, essentially, allow you to plot a slope field on the cartesian plane. Euler's method can be graphically thought of as joining the gradients together to form a 'curve']

/0 · « **Reply #3 on:** April 04, 2009, 12:02:51 am »

Is it basically linear approximation?

shinny · « **Reply #4 on:** April 04, 2009, 12:14:25 am »

Pretty much. It's the same formula, but I guess it's rather an application of linear approximation. Just as a tip, it definitely helps to make the mental links between how it's even possible to use linear approximation to solve integration and such, rather than just walking blindly into using the formula.

NE2000 · « **Reply #5 on:** April 05, 2009, 06:29:44 pm »

Quote from: Mao on April 03, 2009, 11:53:44 pm

This is effectively a differential equation: $\frac{dy}{dx} = \frac{y}{y-x}$

Remembering euler's formula for linear approximation: $y_{n+1} \approx \frac{dy}{dx}\cdot \Delta x + y_{n}$ , given that $(x_0,y_0) = (0,3)$

Hence taking a $\Delta x$ of 1: $y_1 \approx \frac{y_0}{y_0 - x_0}\cdot \Delta x + y_0 = 6$

$y_2 \approx \frac{y_1}{y_1 - x_1} \cdot \Delta x + y_1 = \frac{6}{5}+6 = 7.2$

etc, plot the graph till x=4

Note that the accuracy of this graph depends on the value of $\Delta x$ , the smaller this increment is, the more accurate the approximation is, as $\Delta x \to dx$ , the approximation becomes closer and closer to the solution to the differential equation.

[The original equation, essentially, allow you to plot a slope field on the cartesian plane. Euler's method can be graphically thought of as joining the gradients together to form a 'curve']

OK, thanks Mao, I initially had thought it would be a more hardcore problem as compared to just finding points and connecting the dots.

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Author Topic: Euler's method of approximation (Read 1839 times) Tweet Share

NE2000

Euler's method of approximation

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