Use Euler's method with step size ( 0.5 ) to compute the...

Use Euler's method with step size ( 0.5 ) to compute the approximate ( y )-values ( y_{1}, y_{2}, y_{3} ), and ( y_{4} ) of the solution of the initial-value problem [ y^{prime}=-1+4 x+4 y, quad y(0)=3 ]

Transcript

00:01 For this problem, you're working with finding oilers or using euler's method to find four y values.

00:09 So i have the formula for euler's method, and it says y sub n is equal to the initial y value, or excuse me, the previous y value, plus the derivative times the value of dx.

00:29 Alright, so we're going to start by finding y sub 1 using the initial y times the derivative, times the change in x.

00:41 All right, so we'll substitute in the values, so we'll use this point to start, and then we will never use that point again.

00:51 All right, so we'll put in into the derivative.

00:55 We're using those values, negative 1 plus 4 times 0.

01:01 Plus 4 times 3.

01:08 All right, so that is going to give a value of 11 for the d -y -d -x.

01:19 So then to get the y value, we're going to use the initial y plus the derivative times the d -x, which is 0 .5.

01:38 So that y value is going to be 3 plus 11 times 0 .5.

01:47 Oops three times 11 3 plus 11 times 0 .5 and you're going to get 8 .5 for your y value.

02:04 All right, that is y sub 1.

02:09 All right, so that is the first y value.

02:12 Now we're going to use this point to answer the next line.

02:21 All right, we will never use that point again.

02:24 We'll always use the previous point to get the new values.

02:31 So now we're going to substitute in 0 .5 for the x and 8 .5 for the y.

02:39 So we'll go back to the derivative, negative 1 plus 4 times 0 .5 plus 4 times 8 .5.

03:01 So negative 1 plus 2 plus 4 times 8 .5.

03:08 8 .5 and we get a derivative value of 35.

03:17 All right so now we're not using this one anymore because that was the initial so we're going to go back up we're going to use this form that says so now we're on y sub 2 we're going to use y sub 1 to answer this question so y sub 1 is 8 .5 plus the derivative that we found is 35 we're using the dx does not change .5 so we have 8 .5 plus 35 times 0 .5 and that gives you a value of 26.

04:07 That is your y sub 2 value.

04:11 All right so now as we go through the problem we will never use the point again that we just used so for that next line we're going to use this point.

04:24 All right, so we'll go ahead and substitute in the 1 and the 26...

Question

Use Euler's method with step size ( 0.5 ) to compute the approximate ( y )-values ( y_{1}, y_{2}, y_{3} ), and ( y_{4} ) of the solution of the initial-value problem [ y^{prime}=-1+4 x+4 y, quad y(0)=3 ]

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