Use Euler's method to approximate the solution for the following initial value problem. y ′ = t e 3 t − 2 y , 0 ≤ t ≤ 1 , y ( 0 ) = 0 with h = 0.5h is the step size The initial condition is y ( 0 ) = 0 ⇒ w 0 = 0 t 0 = 0 thus w 1 = w 0 + h ( t 0 e 3 t 0 − 2 w 0 ) 0 + 0.5 [ ( 0 ) e 3 ( 0 ) − 2 ( 0 ) = 0 t = a + i h common distance t 1 = .5From this we can create this Table: i t i w i y ( t i ) 0 0 0 0 1 .5 0 2 1 1.2 To find the the actual solutions requires one to solve the ordinary differential equation IVP problem.A 1st order linear ODE has the form y ′ ( x ) + p ( x ) y = q ( x ) y ( x ) = ∫ e ∫ p ( x ) d x q ( x ) d x + C e ∫ p ( x ) d t ∫ e ∫ p ( x ) d x q ( x ) d x + C e ∫ p ( x ) d x ∫ e ∫ 2 d t e 3 t t d t + C e ∫ 2 d t ∫ e 5 t t d t e 2 t = t e 3 t 5 − e 3 t 25 + e − 2 t C 1 I was unable to reach the correct answer to the ODE which was 1 5 t e 3 t − 1 25 e 3 t + 1 25 e − 2 t How do I correct my work and arrive to the correct conclussion?

Question

Use Euler&#039;s method to approximate the solution for the following initial value problem.       y    ′    =  t      e          3      t        −  2  y  ,  0  ≤  t  ≤  1  ,  y  (  0  )  =  0   with   h  =  0.5h is the step size The initial condition is  y  (  0  )  =  0  ⇒      w    0    =  0        t    0    =  0    thus        w    1    =      w    0    +  h  (      t          0            e          3              t        0              −  2      w    0    )    0  +  0.5  [  (  0  )      e          3      (      0      )        −  2  (  0  )  =  0    t  =  a  +  i  h   common distance        t    1    =  .5From this we can create this Table:                    i                              t          i                                      w          i                            y        (                  t          i                )                            0                    0                    0                    0                            1                    .5                    0                                  2                    1                    1.2                  To find the the actual solutions requires one to solve the ordinary differential equation IVP problem.A 1st order linear ODE has the form      y    ′    (  x  )  +  p  (  x  )  y  =  q  (  x  )    y  (  x  )  =            ∫              e                  ∫          p          (          x          )          d          x                    q      (      x      )      d      x      +      C              e              ∫        p        (        x        )        d        t                          ∫              e                  ∫          p          (          x          )          d          x                    q      (      x      )      d      x      +      C              e              ∫        p        (        x        )        d        x                        ∫              e                  ∫          2          d          t                            e                  3          t                    t      d      t      +      C              e              ∫        2        d        t                          ∫              e                  5          t                    t      d      t              e              2        t              =            t              e                  3          t                      5    −            e              3        t              25    +                    e                  −          2          t                    C        1  I was unable to reach the correct answer to the ODE which was      1    5    t      e          3      t        −      1    25        e          3      t        +      1    25        e          −      2      t      How do I correct my work and arrive to the correct conclussion?

Kelbelol · Accepted Answer

Just insert the initial condition  0  =  0  −      1    25    +  C    ⟹    C  =      1    25    .The table is then                              i                                      t            i                                                w            i                                    y          (                      t            i                    )                                      0                          0.0                          0.000000                          0.000000                                      1                          0.5                          0.000000                          0.283617                                      2                          1.0                          1.120422                          3.219099                                      3                          1.5                          10.042768                          23.406446                                      4                          2.0                          67.512848                          145.235098

Ivan Buckley · Accepted Answer

Thanks for answer

Use Euler's method to approximate the solution for the following initial value problem. y′=te^(3t)−2y,0<=t<=1,y(0)=0 with h=0.5

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