I have this differential equation y′=y(sint)/t, y(0)=2, and h=1/4.

ndevunidt 2022-10-23 Answered
I have this differential equation y = y ( sin t ) t , y ( 0 ) = 2, and h = 1 4 . The first set of values, the inital, are (0,2). For the next iteration would it be y 1 = 2 + ( 1 4 ) ( 2 sin ( 0 ) 0 ) ? I know you can't divide by zero, so what would should I do in this case?
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Answers (1)

Audrey Russell
Answered 2022-10-24 Author has 16 answers
Use the fact that
lim x 0 sin ( x ) x = 1.
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asked 2022-08-19
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b) Use part a) and Euler's method with h = 0.1 to find x(0.2).
x ( 0.1 ) = 1 + ( 0.1 ) ( 0 1 + 0 ) = 0.9
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asked 2022-08-22
I have a question from my book and it says essentially, consider the IVP x = x with x ( 0 ) = 1, what is the exact value of x ( 1 ), then using Eulers method with step size1 , estimate x ( 1 ) call this x ( 1 ) , then repeat for step sizes of 10 n for n = 1 , 2 , 3 , 4 then finally plot E = | x ( 1 ) x ( 1 ) | as a function of step size and then as l n E vs l n t.
Now I am having some issues and ill explain,
for the first part I get that x ( 1 ) = e 1
We have f ( x ) = x and x 0 = 1
so I have that by Euler method
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which would imply that for t = 1 , x 1 = 0
and then for the second part it would imply x 1 = 0.9, then 0.99, then 0.999 and finally 0.9999.
But this doesn't seem to make any sense to me. Seeing as none of these are close to e 1
So I am confused in regard to where I am making mistakes, or where everything is going wrong. For the plotting part, I am also stuck because of this. I am looking for any help and advice.
asked 2022-11-11
I'm trying to solve this 1st order ODE numerically by bringing it into an explicit form, but I don't think it is valid because of the dependency on x_n in the final expression.
d y d x + x = 0 d y d x = x
Alpha is the angle from point with index (n) to point with index (n+1).
t a n ( α ) = d y d x t a n ( α n ) = x n
I call h the step size.
t a n ( α n ) = y n y n 1 x n x n 1 t a n ( α n ) = y n y n 1 h
When I rearrange this I obtain the following form.
y n = y n 1 + h t a n ( α n ) y n = y n 1 + h ( x n ) y n = y n 1 h x n
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y n = y n 1 h x n
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y 1 = y 0 h x 1 y 2 = y 1 h x 2 = y 0 h x 1 h x 2 y 3 = y 2 h x 3 = y 0 h x 1 h x 2 h x 3
So that the final solution is the following. y n = y 0 h ( x 1 + . . . + x n )
asked 2022-05-09
I just wanted to see if I went about solving this correctly. The problem didn't provide a step so I'm using .1:
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I did the following:
y 1 = 0 + .1 [ 1 + ( .1 ) sin ( 1 ( 0 ) ) ] = .1
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Continuing the process, F ( 10 h ) = h f ( 0 ) + h f ( h ) + h f ( 2 h ) + . . . . . h f ( 9 h )
This resembles the Riemann sum: Σ i = 1 n f ( x i ) ( x i x i 1 )
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Example: 3 3.09 f ( x ) d x = 0.81. Approximate f ( 3 ) . F ( x ) = h f ( x )
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