du Each integration requires a simple substitution. Let u = 4y + 3 and v = 8x + 9. Then dy = 4 1 √ (4y + 3)² dy = ²1/12/20 du = - 1²/27/1/20 dv -1-¹+₁=(-1/2) (4y + 3)-¹ + C₁ J = 1 (8x + 9)² 8 1 8 dx )x= 1 + C₂ )(8x + 9)-¹ + C₂ and dx = dv To finish, replace the constant C₂ C₁ by C to write the general solution to the given differential equation.
du Each integration requires a simple substitution. Let u = 4y + 3 and v = 8x + 9. Then dy = 4 1 √ (4y + 3)² dy = ²1/12/20 du = - 1²/27/1/20 dv -1-¹+₁=(-1/2) (4y + 3)-¹ + C₁ J = 1 (8x + 9)² 8 1 8 dx )x= 1 + C₂ )(8x + 9)-¹ + C₂ and dx = dv To finish, replace the constant C₂ C₁ by C to write the general solution to the given differential equation.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![du
Each integration requires a simple substitution. Let u = 4y + 3 and v= 8x + 9. Then dy =
4
1
√ (4y + 3)²
dy =
1/12/20
·]
du =
-u-¹ + C₁
-(4y + 3)-¹ + C₁
=
1
(8x + 9)²
1/1/20
1
8
1
8
dv
dx
) v- 1 + C₂
)(8x + 9)-1 + C₂
and dx =
dv
To finish, replace the constant C₂ C₁ by C to write the general solution to the given differential equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F335131ff-484f-4768-821d-726b05458836%2F2cff40aa-3c2e-46d9-9e69-791b076aaae0%2Fknq3d59_processed.jpeg&w=3840&q=75)
Transcribed Image Text:du
Each integration requires a simple substitution. Let u = 4y + 3 and v= 8x + 9. Then dy =
4
1
√ (4y + 3)²
dy =
1/12/20
·]
du =
-u-¹ + C₁
-(4y + 3)-¹ + C₁
=
1
(8x + 9)²
1/1/20
1
8
1
8
dv
dx
) v- 1 + C₂
)(8x + 9)-1 + C₂
and dx =
dv
To finish, replace the constant C₂ C₁ by C to write the general solution to the given differential equation.
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