I'm trying to approximate the solution to this equation using the Euler's method: y ′ ( x ) = 3 − tan ⁡ ( x ) ⋅ y ( x ) , y ( 2 ) = 4When solving for the step of 0.2, I don't know what to calculate when plugging y ( 2.2 ).What is y ( x )? Is it the derivative of 3 − tan ⁡ ( x ) ⋅ y ( x )?

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I&#039;m trying to approximate the solution to this equation using the Euler&#039;s method:      y    ′    (  x  )  =  3  −  tan  ⁡  (  x  )  ⋅  y  (  x  )  ,  y  (  2  )  =  4When solving for the step of   0.2, I don&#039;t know what to calculate when plugging   y  (  2.2  ).What is   y  (  x  )? Is it the derivative of   3  −  tan  ⁡  (  x  )  ⋅  y  (  x  )?

percolarse2rzd · Accepted Answer

You are solving a special case of the initial value problem      u    ′    (  x  )  =  f  (  x  ,  u  (  x  )  )  ,    u  (      x    0    )  =      u    0    .In your case   f is defined by   f  (  x  ,  y  )  =  3  −  tan  ⁡  (  x  )  y, you have       x    0    =  2 and       u    0    =  4.When applying Euler&#039;s (explicit) method you are attempting to approximate   u  (      x    0    +  n  h  ) with       y    n   where   h  &amp;gt;  0 is your time step, and      y    0    =      u    0    ,        y          n      +      1        =      y    n    +  h  f  (      x    n    ,      y    n    )  .In your case, you have       y    0    =  4,   h  =  0.2, and so      y    1    =      y    0    +  h  f  (      x    0    ,      y    0    )  =  4  +  0.2  ⋅  f  (  2  ,  4  )  .

I'm trying to approximate the solution to this equation using the Euler's method: y

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