Question

Consider the initial value problem given below.Apply Taylor'sseries method, Euler's method,and RK-4th order method. Make comparision by plotting the solution in a single graph.dy/dx = e^-1.1x + xy, y(0) = 1, step size h = 0.1,evaluate y(0..1) to y (0.6)​

Answer 1

Answer:

y′=f(x,y),y(x0)=y0

Step-by-step explanation:

y′+2y=x3e−2x,y(0)=1

at x=0.1,0.2,0.3.

Solution

We rewrite Equation 3.1.5 as

y′=−2y+x3e−2x,y(0)=1,

which is of the form Equation 3.1.1, with

f(x,y)=−2y+x3e−2x,x0=0,andy0=1.

Euler’s method yields

y1 =y0+hf(x0,y0)

=1+(0.1)f(0,1)=1+(0.1)(−2)=0.8,

y2 =y1+hf(x1,y1)

=0.8+(0.1)f(0.1,0.8)=0.8+(0.1)(−2(0.8)+(0.1)3e−0.2)=0.640081873,

y3 =y2+hf(x2,y2)

=0.640081873+(0.1)(−2(0.640081873)+(0.2)3e−0.4)=0.512601754.