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find the exact solution of the differential equation analytically

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(a) Use Euler's Method with a step size of h = 0 .1 to approximate the particular solution of the initial value problem at the given x - value, (b) find  the exact solution of the differential equation analytically, and (c) compare the solutions at the given x - value.

Differential Equation               Initial condition                      x - value

asked Feb 16, 2015 in CALCULUS by anonymous
reshown Feb 16, 2015 by goushi

4 Answers

0 votes

(a)

Step 1:

The differential equation is .

The initial condition is .

Step size is .

Euler's method is a numerical approach to approximate the  particular solution of the differential equation.

Let that passes through the point .

From this starting point, one can proceed in the direction indicated by the  slope.

Use a small step , move along the tangent line.

and .

Step 2:

Use step size , , and .

So we have , , , ,.....and,

answered Feb 17, 2015 by yamin_math Mentor
0 votes

Contd....

Step 3:

Proceeding with similar calculations, we get the values in the table:

 From the table particular solution at x = 2 is 3.031.

 Solution:

The particular solution at x = 2 is 3.031.

answered Feb 17, 2015 by yamin_math Mentor
0 votes

(b)

Step 1:

The differential equation is .

The initial condition is .

Solution to the differential equation :

Integrate on each side.

Substitute initial conditions , .

The exact solution is .

 Solution:

The exact solution is .

answered Feb 17, 2015 by yamin_math Mentor
0 votes

(c)

Step 1:

The differential equation is .

From Euler's method, particular solution at x = 2 is 3.031.

From the exact solution :

The exact solution is .

Substitute x = 2 in exact solution.

Solutions to the 3rd degree equation are .

Imaginary values are neglected.

So the particular solution at x = 2 is 3.

Therefore the particular solution at x = 2 is almost same in both the methods.

 Solution:

The particular solution at x = 2 is almost same in both the methods.

answered Feb 17, 2015 by yamin_math Mentor

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