dy/dx = y' = f(x,y(x)) = 2e^(-x^2)

Solve this differential equation using Heun's method and plot solution over -5 < x < 5. You'll need to derive the appropriate initial condition y(x0) = y0

:confused:

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- May 8th 2007, 07:25 PM12_bladeznumerical solution to differencial equation
dy/dx = y' = f(x,y(x)) = 2e^(-x^2)

Solve this differential equation using Heun's method and plot solution over -5 < x < 5. You'll need to derive the appropriate initial condition y(x0) = y0

:confused: - May 9th 2007, 12:06 AMCaptainBlack
Heun's method is the basic Predictor Corrector method. This uses the Euler

step to predict the value of y(x+h) using y(x) and y'(x). Then using the

prediction of y(x+h) calculate the derivative y'(x+h), then a new estimate

of the derivative over the interval is taken as the average of the two

derivatives found and y(x+h) is then found using an Euler step with this

new estimate of the derivative.

Here f(x,y(x)) is independent of y so that makes things a bit simpler, since

now we can calculate y'(x+h) without having to estimate y(x+h). So in

this case the Heun step is:

y(x+h) = y(x) + h [y'(x) + y'(x+h)]/2

RonL - May 9th 2007, 12:20 AMCaptainBlack