Consider the following. (A computer algebra system is recommended. Round your an
ID: 3665928 • Letter: C
Question
Consider the following. (A computer algebra system is recommended. Round your answers to four decimal places.)
y' = 3cos(t) - 7y, y(0) = 1
(a) Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.1.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
(b) Find approximate values of the solution of the given initial value problem at
t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.05.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
(c) Find approximate values of the solution of the given initial value problem at
t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.025.
y(0.1) =
y(0.2)
y(0.3) =
y(0.4) =
(c) Find approximate values of the solution of the given initial value problem at
t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.025.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
(d) Find the solution y = (t) of the given problem.
y(t) =
Evaluate (t) at t = 0.1, 0.2, 0.3, and 0.4. (Round your answers to five decimal places.)
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
Explanation / Answer
y' = 3cos(t) - 7y, y(0) = 1
(a) Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.1.
SOL : given data
y'=3cos(t)-7y, y(0)=1,
t=0.1,0.2,0.3,0.4, h=0.1.
let f(t,y)=3 cos(t)-7y
y(0)=1
t0=0, y0=1. equ..............1
sub equ........1 in the following
f0(0,1)=3cos0-7(1)
=3-7 = - 4
y(0.1) = ?
y1(0.1)=y0+hf0
y1(0.1)= 1+(0.1)(-4)
= 1-0.4= 0.6.
y(0.2) =?
y2(0.2)= y1 +hf1
f1 = f(0.1,0.6) =3Cos(0.1) - 7(0.6)
=2.9999 - 4.2
= -1.200000.
y2(0.2) = 0.6 +(0.1)(-1.20000)
= 0.48.
y(0.3) = ?
y3(0.3) =y2 +hf2
f2(0.2,0.48) = 3Cos(0.20) -7(0.48)
= 2.99998 - 3.36
= - 0.36001.
y3(0.3)= y2 +hf2
= 0.48 + (0.1) (- 0.36001)
=0.44399
y(0.4) = ?
y4(0.4) = y3+ hf3
f3(0.3,0.44399)= 3Cos(0.30)-7(0.44399)
= -0.10797.
y4 (0.4) = 0.44399 + (0.1) (-0.10797)
= 0.433193.
y(0.1) = 0.6
y(0.2) = 0.48
y(0.3) = 0.44399
y(0.4) = 0.433193.
(b) Find approximate values of the solution of the given initial value problem at
t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.05.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
SOL : given data
y'=3cos(t)-7y, y(0)=1,
t=0.1,0.2,0.3,0.4, h=0.05
let f(t,y)=3 cos(t)-7y
y(0)=1
t0=0, y0=1. equ..............1
sub equ........1 in the following
f0(0,1)=3cos0-7(1)
=3-7 = - 4
y(0.1) = ?
y1(0.1)=y0+hf0
y1(0.1)= 1+(0.05)(-4)
= 1-0.2= 0.8
y(0.2) =?
y2(0.2)= y1 +hf1
f1 = f(0.1,0.8) =3Cos(0.1) - 7(0.8)
= - 2.600004
y2(0.2) = 0.8+(0.05)(-2.600004)
= 0.66699
y(0.3) = ?
y3(0.3) =y2 +hf2
f2(0.2,0.66699) = 3Cos(0.20) -7(0.66699)
= - 1.68994
y3(0.3)= y2 +hf2
= 0.66999 + (0.05) (- 1.68994)
=0.585493
y(0.4) = ?
y4(0.4) = y3+ hf3
f3(0.3,0.585493)= 3Cos(0.30)-7(0.585493)
= - 1.098492
y4 (0.4) = 0.585493 + (0.05) (-1.098492)
= 0.53056
y(0.1) = 0.8
y(0.2) = 0.66999
y(0.3) = 0.585493
y(0.4) = 0.53056.
(c) Find approximate values of the solution of the given initial value problem at
t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.025.
y(0.1) =
y(0.2)
y(0.3) =
y(0.4) =
SOL : given data
y'=3cos(t)-7y, y(0)=1,
t=0.1,0.2,0.3,0.4, h=0.025
let f(t,y)=3 cos(t)-7y
y(0)=1
t0=0, y0=1. equ..............1
sub equ........1 in the following
f0(0,1)=3cos0-7(1)
=3-7 = - 4
y(0.1) = ?
y1(0.1)=y0+hf0
y1(0.1)= 1+(0.025)(-4)
= 1-0.1= 0.9
y(0.2) =?
y2(0.2)= y1 +hf1
f1 = f(0.1,0.9) =3Cos(0.1) - 7(0.9)
= - 3.30000
y2(0.2) = 0.9+(0.025)(-3.30000)
= 0.8175
y(0.3) = ?
y3(0.3) =y2 +hf2
f2(0.2,0.66699) = 3Cos(0.20) -7(0.8175)
= - 2.72251
y3(0.3)= y2 +hf2
= 0.8175 + (0.025) (- 2.72251)
=0.74943
y(0.4) = ?
y4(0.4) = y3+ hf3
f3(0.3,0.74943)= 3Cos(0.30)-7(0.74943)
= - 2.24605
y4 (0.4) = 0.74943 + (0.025) (-2.24605)
= 0.69327
y(0.1) = 0.9
y(0.2) = 0.8175
y(0.3) = 0.74943
y(0.4) = 0.69327.
(c) Find approximate values of the solution of the given initial value problem at
t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.025.
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
SOL : given data
y'=3cos(t)-7y, y(0)=1,
t=0.1,0.2,0.3,0.4, h=0.025
let f(t,y)=3 cos(t)-7y
y(0)=1
t0=0, y0=1. equ..............1
sub equ........1 in the following
f0(0,1)=3cos0-7(1)
=3-7 = - 4
y(0.1) = ?
y1(0.1)=y0+hf0
y1(0.1)= 1+(0.025)(-4)
= 1-0.1= 0.9
y(0.2) =?
y2(0.2)= y1 +hf1
f1 = f(0.1,0.9) =3Cos(0.1) - 7(0.9)
= - 3.30000
y2(0.2) = 0.9+(0.025)(-3.30000)
= 0.8175
y(0.3) = ?
y3(0.3) =y2 +hf2
f2(0.2,0.66699) = 3Cos(0.20) -7(0.8175)
= - 2.72251
y3(0.3)= y2 +hf2
= 0.8175 + (0.025) (- 2.72251)
=0.74943
y(0.4) = ?
y4(0.4) = y3+ hf3
f3(0.3,0.74943)= 3Cos(0.30)-7(0.74943)
= - 2.24605
y4 (0.4) = 0.74943 + (0.025) (-2.24605)
= 0.69327
y(0.1) = 0.9
y(0.2) = 0.8175
y(0.3) = 0.74943
y(0.4) = 0.69327.
(d) Find the solution y = (t) of the given problem.
y(t) =
Evaluate (t) at t = 0.1, 0.2, 0.3, and 0.4. (Round your answers to five decimal places.)
y(0.1) =
y(0.2) =
y(0.3) =
y(0.4) =
SOL: IN given problem sufficient data is not availiable.