U₁₂₁ = X u 3.12 Plug √²₂ = x² -3/2 3 √²₁ x ² + x² • X²³² = 0 3/2 2 √ ₁ x ² + x Su ₁₂ = Sx -3/2 Yp 3/2 U ²₁ x ² = -X²³² 2 Sv₁ = -√x 1/2+1 Y2+1 -3/2+1 into (1) = U, Y₁ + U₂ 9₂ -1/4 -3/27/16 U₂ = x 3/2 Yp=-2/3 x ¹² ( x ²) - 2x ¹/2 (x²) 712 512 2/₁ = -2/32 ²/2 = 2 × ²/2 ? -1/₂ -1/₂ Choose 6₁26₂ = 0 -1/2 3 -2х +C2 3/2-2 √√₁=-x → 0₁/₁ = = x 3/2 → U₁² 0₁ = - ( X²-²2), 0₁ = -3/3x² + + 5₁ 3/2 1/2
U₁₂₁ = X u 3.12 Plug √²₂ = x² -3/2 3 √²₁ x ² + x² • X²³² = 0 3/2 2 √ ₁ x ² + x Su ₁₂ = Sx -3/2 Yp 3/2 U ²₁ x ² = -X²³² 2 Sv₁ = -√x 1/2+1 Y2+1 -3/2+1 into (1) = U, Y₁ + U₂ 9₂ -1/4 -3/27/16 U₂ = x 3/2 Yp=-2/3 x ¹² ( x ²) - 2x ¹/2 (x²) 712 512 2/₁ = -2/32 ²/2 = 2 × ²/2 ? -1/₂ -1/₂ Choose 6₁26₂ = 0 -1/2 3 -2х +C2 3/2-2 √√₁=-x → 0₁/₁ = = x 3/2 → U₁² 0₁ = - ( X²-²2), 0₁ = -3/3x² + + 5₁ 3/2 1/2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Can you check to see where I made a mistake
as I was trying to solve this question
which is using variation of parameters to find a particular solution, given the solutions y1, y2 of
the complementary equation.
Transcribed Image Text:1
_312
U ₁₂₁ = x
X
Plug №²₂
2
12
U =
3/2
√ ₁ x ² + x²
- 3/2 3
√²₁x² + x² ·x² = 0
X
Su ₂₁ = √ x
-3/2
x
X
20
2
U ₁ X ² = = √3/2
2
Sv ₁ = -√√x ²1/2+1
Y2+1
واد
Yp = = 2/3x²
ур
Ур 9,
Yo zUY + 0,92
36 10
-3/2+1
into (1)
2x
-3/2+/-10
U₁
= -X
Yp= 2/3x 220
-1/2
3
2
Yp=-² / 3 x 1 2 ( x ²) - 2x ¹/2 (x²)
7/2
5/2
1
3/2
³-1
?
X
U₂ = x
2
1/1/2
-1/₂
12
Choose
C₁=C₂ = 0
U₂
-1/2
prost-1₂
= -2x +²₂
3/2-2
√√=-x → √√²₁ =
1
Hot
X
U
3/²2²2²) + 0₁ = -2/38/³9/2 + (₁
1/2
√ ₁ = = x
(
Transcribed Image Text:2 11
x²y² - 4xy² +6y= x
Y₂ = V₁ Y₁ + 0₂ 9₂
q
Ур
5/2
3
1 x ³0 = y₁ = x ²₁ y₂ = x ³ (Question 11)
Уг
1
Y₁ = U₁₂ x ² +U₂₂x²
P
2
2
Y p = 0 ₁ x ² + 2√₁x +0₂² x³ +30 ₂ x ²
ур
tu
STOLC;,
√ ₁ x² + √²₂ x ³ = 0
tu
2
2
2
Then y'p= 20₂x +30₂x²
y" p =
2
20²₁ x + 20₁ + 30 ²₂ x ² + 60 ₂ x
LS = x²y³p = 4Xy'p + 6yp
гур
ур
= x²(Qu²₁ x +20₂ + 3√ ₂ x ² + 60₂ x) = 4x (20₁ x + 30₂x²) + 6(U₁x² + √₂x ²³ )
2 u²₁ +3
+20₁ à ² / 30²₂ x ¹¹460₂x²³/1-80₁ x ²-120₂x³ +, 60, x² +60₂83
2
L
2
1
3
3
2
3
20 ²₁ +³ +30²1₂2 x
20₁²
3
20
22
(1) U₁x² + 0 ₂ x ³ = 0
4= X
5/2
3
20₁ x ³ + 30²₁ x ² zd
+3
X X
2 x (U₁x² + U₂₁ x ³² = 01
x ³ + 2 0 ₂2 x ² = 0 5/2
3, 2
O
2
5/2
# 2 0 ₁ x ²³² + 3√²₂2 x ²4 = x²
+302
- 20₁4 ³ + 20 % x ¹ = 0
3
0₂284285/2 $.4-2
43
89
U/₁² = x $12-4
X
-3/2
9/2
0₂= = X
NIG
gold min
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