yo(x) = −2+ 4e²x² 5. yo(x) = 2 + 4e²x² (iii) For the particular solution yo in (ii), find the value of yo(√In 2). 1. yo(√In 2) = 12 2. yo(√In 2) = 14 3. yo(√In 2) = 11 4. yo(√In 2) = 13 (5.) yo(√In 2) = 10 f(x + y) 018 A differentiable function f has the property that = lim f(x) x 0 2+4 e = 0, In 2 is f(x) + f(y) 1- f(x) f(y) holds for all x, y in the domain of f. It is known also that f(x) lim x →0 X = 9. (i) Use the definition of the derivative of f to determine f'(x) in terms of f(x). 019 (ii) By solving the differential equation in Part ㅠ (i) for f, find the value of f( 1. f( 2. $(7) ƒ (27) = √3 3. ƒ (17) = 4. f ㅠ 27 5. ƒ (77) 1 √2 న ప/ /వ 2xy √√3 (²7) = √₂ If y₁ satisfies the equations dy dx determine the value of y₁(e). 1. y₁(e) = (8 + 4e²)¹/2 2. y₁(e) = (5 + 4e²)¹/2 3. y₁(e) = 54e² 4. y₁(e) = (8 — 4e²)¹/2 2 + 12x² lnx, y(1) = e,

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Question

What is the answer for Q19 

Q18 is = 9(1+f(x)^2)

Yo (x)
= −2+4e²x²
5. yo(x) = 2 + 4e²r²
(iii) For the particular solution yo in (ii), find
the value of yo(√In 2).
In 2
1. yo(√In 2)
2. yo(√In 2) = 14
= 12
3. yo(√In 2) = 11
4. yo(√In 2)
V
(5.) yo(vln 2) = 10
= 13
f(x + y)
lim f(x)
x → 0
018
A differentiable function f has the property
that
-
=
=
2+4e
holds for all x, y in the domain of f. It is
known also that
0,
f(x) + f(y)
1 - f(x) f(y)
y f'(x) = 1 + f(x)²
2₂. f'(x)
is
lim
x → 0
f(x)
X
(i) Use the definition of the derivative of ƒ to
determine f'(x) in terms of f(x).
9(1 - f(x)²)
= 9.
019
(ii) By solving the differential equation in Part
π
(i) for f, find the value of f(7)
1
27. √2
2. ƒ (77) = √3
1. f(
77
ㅠ
3. f 27.
4. f
-
ㅠ
27
-
1
√√3
5. ƒ (277) = √₂
√3
2
If Y1 satisfies the equations
dy
2xy-
dx
determine the value of y₁(e).
1. y₁(e) = (8 + 4e²)¹/2
2. y₁(e) = (5 + 4e²)¹/2
3. y₁(e) = 54e²
4. y₁(e) = (8 — 4e²)¹/2
5. y₁(e) = 5 + 4e²
=
2 + 12x² Inx, y(1) = e,
Transcribed Image Text:Yo (x) = −2+4e²x² 5. yo(x) = 2 + 4e²r² (iii) For the particular solution yo in (ii), find the value of yo(√In 2). In 2 1. yo(√In 2) 2. yo(√In 2) = 14 = 12 3. yo(√In 2) = 11 4. yo(√In 2) V (5.) yo(vln 2) = 10 = 13 f(x + y) lim f(x) x → 0 018 A differentiable function f has the property that - = = 2+4e holds for all x, y in the domain of f. It is known also that 0, f(x) + f(y) 1 - f(x) f(y) y f'(x) = 1 + f(x)² 2₂. f'(x) is lim x → 0 f(x) X (i) Use the definition of the derivative of ƒ to determine f'(x) in terms of f(x). 9(1 - f(x)²) = 9. 019 (ii) By solving the differential equation in Part π (i) for f, find the value of f(7) 1 27. √2 2. ƒ (77) = √3 1. f( 77 ㅠ 3. f 27. 4. f - ㅠ 27 - 1 √√3 5. ƒ (277) = √₂ √3 2 If Y1 satisfies the equations dy 2xy- dx determine the value of y₁(e). 1. y₁(e) = (8 + 4e²)¹/2 2. y₁(e) = (5 + 4e²)¹/2 3. y₁(e) = 54e² 4. y₁(e) = (8 — 4e²)¹/2 5. y₁(e) = 5 + 4e² = 2 + 12x² Inx, y(1) = e,
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