The solution to the initial value problemdydx=(x+y+2)2,y(0)=-2crosses the x-axis at a point in the interval [0,14].By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

Short Answer

Expert verified

The solution on the given conditions and on the interval cross x-axis at x = 1.27.

Step by step solution

01

Find the equation of approximation value

Heredydx=(x+y+2)2,y(0)=-2 ,

For,xo=0,yo=-2,M = 280, h = 0.005, interval =0,14

F=f(x,y)=(x+y+2)2G=f(x+h,y+hF)=(x+y+2+0.005(1+(x+y+2)2))2

02

Solve for x and y

Apply initial pointsx=0,y=-2,h=0.005

F(0,-2)=0G(0,-2)=0.000025

x=(x+h)y=x+h2(F+G)x=0.005y=-2

03

Determine the all other values

Apply the same procedure for all other values and the values are

(x = 1.270, y = -0.047)

(x = 1.275, y = 0.006)

Hence, the solution on the given conditions and on the interval cross x-axis at x=1.27.

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