Given the initial value problem: y'= 3√(t+y), y(0) = 4 Use...

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00:01 So in this question, they say let y of x be the solution to the initial value problem.

00:06 Y prime is equal to x over y plus 1.

00:12 And they tell us that y of 0 is equal to 1.

00:16 I want to use oilers method with a step size of 1 half to estimate y of 1.

00:22 So what is my oilers method formula? i know that y of x plus h is approximately equal.

00:32 To y of x plus y prime of x times h.

00:39 So my x in my initial step is zero and they said my step size is one -half.

00:45 So y of zero plus one -half is approximately equal to y of zero plus y -prime of zero times one -half.

00:59 Now my y of zero we said was one.

01:02 I need my y prime of zero.

01:06 So i need my y -prime of zero.

01:07 We need y prime when x is 0, and if x is 0, i know that y is 1.

01:14 According to my formula, that means we have x over y, 0 over 1, plus 1, which is just 1.

01:24 So 1 times 1 1⁄2, i am getting 3 halves.

01:33 Now in my second step of oilers method, my x is 1 half, and my h is still 1 half.

01:40 So y of a half plus a half is approximately y of a half plus y prime of a half times a half.

01:54 My y of one half, we just said, is about three halves...

Given the initial value problem: y'= 3\sqrt{t+y}, y(0) = 4 Use improved Euler's method with step size h = 0.01 to estimate: y(0.01) = y(0.02) =