Consider this differential equation, d y / d x = x + sin ⁡ ( y )with initial condition y = 0.5 when x = 1.2:1. Write down the recurrence relation for Euler's numerical method applied above.2. With step size h = 0.1, calculate the approximations to y(1.3) and y(1.4).1. This is my answer: Y i + 1 = Y i + 0.1 ( X i + sin ⁡ ( Y i ) )2. I have problem solving this one...From my txtbk reference, I sub i = 1 into the eqn thus giving me Y 1 = Y 0 + 0.1 ( 1.2 + sin ⁡ ( 0.5 ) ) = 0.1679Then, i = 2 Y 2 = Y 1 + 0.1 ( 1.3 + sin ⁡ ( 0.6 )Im sure my method in part 2 is wrong and I don't really understand how to solve it via my txtbk.

Question

Consider this differential equation,  d  y      /    d  x  =  x  +  sin  ⁡  (  y  )with initial condition   y  =  0.5 when   x  =  1.2:1. Write down the recurrence relation for Euler&#039;s numerical method applied above.2. With step size h = 0.1, calculate the approximations to y(1.3) and y(1.4).1. This is my answer:      Y    i    +  1  =      Y    i    +  0.1  (      X    i    +  sin  ⁡  (      Y    i    )  )2. I have problem solving this one...From my txtbk reference, I sub   i  =  1 into the eqn thus giving me      Y    1    =      Y    0    +  0.1  (  1.2  +  sin  ⁡  (  0.5  )  )  =  0.1679Then,   i  =  2      Y    2    =      Y    1    +  0.1  (  1.3  +  sin  ⁡  (  0.6  )Im sure my method in part 2 is wrong and I don&#039;t really understand how to solve it via my txtbk.

Sanai Douglas · Accepted Answer

We have the differential equation       y    ′    (  x  )  =  x  +  sin  ⁡  (  y  ) with initial condition y  y  (  1.2  )  =  0.5.Euler&#039;s method states that we may approximate the value of the function at   y  (  x  +  δ  x  ) for some &#039;small step&#039;   δ  x by assuming the function is approximately linear between   x and   x  +  δ  x. I.e:  y  (  x  +  δ  x  )  ≈  y  (  x  )  +      y    ′    (  x  )  ⋅  δ  xThus, for a step size of   δ  x  =  0.1, starting at   x  =  1.2 we would have:  y  (  1.2  +  0.1  )  ≈  y  (  1.2  )  +      y    ′    (  1.2  )  ⋅  0.1  y  (  1.3  )  =  0.5  +  (  1.2  +  sin  ⁡  (  0.5  )  )  ⋅  0.1  =  0.5  +  0.167942554So you dropped the 0.5 term in your calculation, i.e - what you call       Y    0  . Hopefully this makes things clearer for you. Can you figure out   y  (  1.4  ) now?

Consider this differential equation, dy/dx=x+sin(y) with initial condition y=0.5 when x=1.2: Write down the recurrence relation for Euler's numerical method applied above. With step size h = 0.1, calculate the approximations to y(1.3) and y(1.4).

Open question

Answer & Explanation

New Questions in Integral Calculus