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Find approximate values of the solution of the given initial value problem at $t = 0.1, 0.2, 0.3, 0.4$. a) Use the Euler method with h = 0.05

$$y' = 5t-3\sqrt{y}$$ $$y(0)=2$$

$$y_{i+1}=y_i+h(5t_{i}-3\sqrt{y_{i}} )$$

by using equation,

$y_1 = 1.78787$, $t= 0$

$y_2 = 1.599801196, t = 0.05$

$y_3 = 1.433076324, t = 0.1$

$y_4 = 1.292884317, t = 0.15$

$y_5 = 1.172326711, t = 0.20$

$y_6 = 1.072415655 ,t = 0.25$

$y_7= 0.9920793998, t = 0.3$

$y_8 = 0.9301746258, t = 0.35$

$y_9 = 0.8855062841, t = 0.40$

My answers are $1.433, 1.1723, 0.9921, 0.8855$ for $0.1,0.2,0.3,0.4$.

However, the answers from the book are $1.59980, 1.29288, 1.07242, 0.930175$.

I don't know what I did wrong.

I used $y_0 = 2$

$t_0 = 0$

$t_1 = 0 + 0.05$

$t_2 = 0+ 2(0.05)$

$t_3 = 0 + 3(0.05)$ ....

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    I changed the first line to make it easier to understand. Perhaps you can change the rest? – 2017-02-14

1 Answers 1

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Your time steps are off (at least in the upper half of your question, which is very confusing as written because the upper and lower half do not match).

Regardless, you should have

  • $t_0 = 0.00, y_0 = 2.00000$
  • $t_1 = 0.05, y_1 = 1.78787$
  • $t_2 = 0.10, y_2 = 1.59980$
  • $\ldots$
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    oh....so t1 follows y1.. – 2017-02-14
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    Even though I used t1 = 0.05 to get y2, y2 is not at t1. y2 is at t2 right..? – 2017-02-15
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    The calculation is correct - it is just offset, but that is wrong - technically speaking. – 2017-02-15