O »(x) = Given the differential equation y' = 8x + 4 y' and y (0) = 1 and its next few implicit derivatives y" = 8 + 4 y' y'' = 4 y" %3D Use the initial value "y(0) = 1" to find the value of "y'(0)", "y"(0)", and "y"(0)", then plug them into the Taylor Series Polynomial formula. 160 (A y(x) 0! 8 x' + 1! 40 x2 + ... 2! 3! B y(x) 0! 8. x! 8. x2 + x3 + 1! 2! 3! y(x) O! 12 x + 1! 12 x2 + 4 r3 + ... 3! - 2! 12 D y(x) x0 0! 54 x2 + ? x + ... | 1! 2! 3! E E y(x) O! 4 x! 12 x2 + 4 x3 + 3! - - 1! 2! y(x) = 24 x2 + 2! 96 -r3 +... F x° + 0! x'+ 1! 3! +

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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q4

Given the differential equation
y' = 8x + 4 y' and y(0) = 1
and its next few implicit derivatives
y" = 8 + 4 y'
y"' = 4 y"
Use the initial value "y(0) = 1" to find the value of "y'(0)", "y"(0)", and "y"(0)", then plug them into the Taylor Series Polynomial
formula.
y(x)
x° +
1!
40
x! +
2!
160
x +
A)
=
-
...
O!
3!
8.
8
y(x) :
x0
0!
B
x! +
x2
+
3!
r3 + ...
1!
2!
12
x! +
I!
12
x2 +
2!
4
y(x)
0!
+3
3!
+...
|
y(x)-
x° +
O!
12
x! +
1!
54
x +
x3
+
-
...
2!
3!
y(x)
4
x!
12
x2 +
2!
4
r +...
+
|
|
O!
1!
3!
96
y(x)
4
x! +
1!
24
x2 +
(F)
=
O!
x° +
+
|
...
2!
3!
%3D
Transcribed Image Text:Given the differential equation y' = 8x + 4 y' and y(0) = 1 and its next few implicit derivatives y" = 8 + 4 y' y"' = 4 y" Use the initial value "y(0) = 1" to find the value of "y'(0)", "y"(0)", and "y"(0)", then plug them into the Taylor Series Polynomial formula. y(x) x° + 1! 40 x! + 2! 160 x + A) = - ... O! 3! 8. 8 y(x) : x0 0! B x! + x2 + 3! r3 + ... 1! 2! 12 x! + I! 12 x2 + 2! 4 y(x) 0! +3 3! +... | y(x)- x° + O! 12 x! + 1! 54 x + x3 + - ... 2! 3! y(x) 4 x! 12 x2 + 2! 4 r +... + | | O! 1! 3! 96 y(x) 4 x! + 1! 24 x2 + (F) = O! x° + + | ... 2! 3! %3D
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