00:01
Hello everyone, in this problem we are given with the initial value problem which is of y ' to be is equal to minus 2 plus 2x plus 4y with the y initial value y of 1 to be equal to 3.
00:15
So, here in this problem we need to use euler's method with the step size h is given as 0 .5 to compute the approximate values of y1, y2, y3 and y4.
00:30
So, the given differential equation can also be written as f of x y which will be equals to y ' to be equal to minus 2 plus 2x plus 4y and with the initial values we can say x0 to be equal to 1.
00:46
When x0 to be equal to 1 we have the value of y0 to be equal to and by euler's method we have the formula yn plus 1 to be equal to yn plus h of f of xn yn.
01:03
So, now taking n to be equal to 0 in this euler's formula we can have y1 to be equal to y0 plus h multiplied by f of x0 y0.
01:21
Now, substituting the corresponding values y0 is 3 plus h is 0 .5 and f of x0 y0 is substituting the values of x0 and y0 in the given function.
01:33
So, it is of minus 2 plus 2 multiplied by 1 plus 4 multiplied by.
01:41
So, simplifying this we have this value to be 3 plus of 0 .5 multiplied by 12 and simplifying this further we have the value of y1 to be 9.
01:55
Now, similarly when n to be equal to 1 in euler's formula we get y2 to be equal to y1 plus h of f of x1 y1.
02:08
Since we are given with the h value to be 0 .5 and with the x1 can be find with the value of x0 as 1 plus 0 .5 which is the step size is 1 .5 and y1 is from the above value.
02:25
So, now substituting its values here similarly by the above step.
02:29
So, it would be 9 plus of 0 .5.
02:34
So, f of x1 y1 value here is of 37...