00:01
In this problem, we are given the differential equation y ' equal to minus 3x plus y squared with the initial condition y0 equal to minus 1.
00:14
And we are going to approximate y at x equal to 0 .5 using the earlier method with a step size of 0 .1.
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The earlier method says yn plus 1 is equal to yn plus fn times the step size.
00:37
Here fn is just the right -hand side of this differential equation evaluated at xn and yn.
00:52
Now, n goes like 0, 1, 2, and so on.
00:58
So the initial condition corresponds to y0.
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And of course, this initial value of x is x0.
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So now let's make this table.
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Here the n values 0, 1, 2, 3, 4, 5.
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We will have the xn values 0, 0 .1, 0 .2, 0 .3, 0 .4, 0 .5.
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We will have the y values.
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The starting value is 0 .1.
01:33
And we will prepare a column for fn.
01:36
And we will also prepare a column for yn plus 1.
01:39
So we will find some value here and copy it over here and so on.
01:45
Okay, now the problem tells us to round our answer to 3 decimal places.
01:52
So working with a spreadsheet and setting the number of decimal places to 3, i have obtained the following values...