Suppose that we use Euler's method to approximate the...

Suppose that we use Euler's method to approximate the solution to the differential equation dy/dx = x^5/y; y(0) = 3. Let f(x, y) = x^5/y. We let x0 = 0 and y0 = 3 and pick a step size h = 0.2. Euler's method is the the following algorithm. From xn and yn, our approximations to the solution of the differential equation at the nth stage, we find the next stage by computing xn+1 = xn + h, yn+1 = yn + h * f(xn, yn). Complete the following table: n xn yn 0 0 3 1 2 3 4 5 The exact solution can also be found using separation of variables. It is y(x) = Thus the actual value of the function at the point x = 1 y(1) =

Transcript

00:01 Hi in this problem we are given with the differential equation d .y by d x which is equal to x power 5 by by by with y of 0 is equal to 3.

00:11 Now we need to find the solution for this differential equation by using oil as method so let us take f of x comma y as x per 5 by y and x not as 0 0 is equal to 3 and given the step size has 0 .2 also we are given with the values x and plus 1 as equal to x n plus h and y n plus 1 is equal to y n plus h into f of x n comma y n so for the values of n from 0 to 5 we are to find the values of x n and y n also x not is given as 0 and y0 is given as 3 now we need to find the value of x 1 so x1 will be x0 plus 1 which is equal to x0 plus h so it will be x1 which is equal to x not is 0 plus h is 0 .2 0 plus 0 .2 is 0 .2 is 0 .2 is 0 .2 so x1 x0 plus 0 .2 so that will be x1 which is equal to 0 .2.

01:26 Similarly for y n y 1 so y0 plus 1 is equal to y0 plus h into f of x n comma y n so here n is 0 so it will be x0 power 5 divided by y0 so substituting the values and simplifying we get y1 as 3 so it is 3 so similarly for all the remaining values we obtain so these are the corresponding values now we need to find the exact solution of the difficult given differential equation using separation of variables.

02:22 The differential equation, d .y by d x, which is equal to x power 5 by y.

02:36 Also with the initial condition, y of 0 is equal to 3.

02:43 So with the differential equation dy by t x, which is equal to x par 5 by y.

02:51 Now using separation of variables, so we can rewrite this.

02:55 As y into dy which is equal to x power 5 into d x now taking integration on both sides so integrating this we obtained y square by 2 which is equal to x power 6 by 6 plus c so simplifying it further we obtain y square which is equal to x power 6 plus c by 3.

03:29 So this can be again written as y is equal to square root of x power 6 plus c by 3...

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