Solve the initial value problem 2yy' + 6 = y² + 6x with y(0) = 7. a. To solve this, we should use the substitution u = y^2 help (formulas) With this substitution, y = u^(1/2) help (formulas) y' = 1/2 u^(-1/2) u' help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y' for dx -). b. After the substitution from the previous part, we obtain the following linear differential equation in x, u, u'. u'-u = 6x -6 help (equations) c. The solution to the original initial value problem is described by the following equation in x, y. help (equations) y = sqrt (69 e^x -6x)
Solve the initial value problem 2yy' + 6 = y² + 6x with y(0) = 7. a. To solve this, we should use the substitution u = y^2 help (formulas) With this substitution, y = u^(1/2) help (formulas) y' = 1/2 u^(-1/2) u' help (formulas) dy Enter derivatives using prime notation (e.g., you would enter y' for dx -). b. After the substitution from the previous part, we obtain the following linear differential equation in x, u, u'. u'-u = 6x -6 help (equations) c. The solution to the original initial value problem is described by the following equation in x, y. help (equations) y = sqrt (69 e^x -6x)
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
only part c please
![Entered
y^2
u^(1/2)
(1/2)*[u^(-1/2)]*u'
u'-u = 6*x-6
y = sqrt(69*(e^x)-6*x)
At least one of the answers above is NOT correct.
Answer Preview
1
2
3²
U 2
-1
u žu
u' – u = 6x − 6
y = √69eº - 6x
Solve the initial value problem 2yy' + 6 = y² + 6x with y(0) = 7.
a. To solve this, we should use the substitution
u= y^2
help (formulas)
With this substitution,
y = u^(1/2)
help (formulas)
y' = 1/2 u ^(-1/2) u'
help (formulas)
Enter derivatives using prime notation (e.g., you would enter y' for dy).
dx
Result
c. The solution to the original initial value problem is described by the following equation in x, y.
help (equations)
y = sqrt ( 69 e^x -6x)
correct
correct
correct
correct
incorrect
x,
b. After the substitution from the previous part, we obtain the following linear differential equation in
u'-u = 6x -6
help (equations)
u, u'.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe70859f-d1ee-4d19-93f5-d3c21b44393e%2F2f6b955c-c96d-4285-b9b2-2560969e3d38%2Fahyg3z8_processed.png&w=3840&q=75)
Transcribed Image Text:Entered
y^2
u^(1/2)
(1/2)*[u^(-1/2)]*u'
u'-u = 6*x-6
y = sqrt(69*(e^x)-6*x)
At least one of the answers above is NOT correct.
Answer Preview
1
2
3²
U 2
-1
u žu
u' – u = 6x − 6
y = √69eº - 6x
Solve the initial value problem 2yy' + 6 = y² + 6x with y(0) = 7.
a. To solve this, we should use the substitution
u= y^2
help (formulas)
With this substitution,
y = u^(1/2)
help (formulas)
y' = 1/2 u ^(-1/2) u'
help (formulas)
Enter derivatives using prime notation (e.g., you would enter y' for dy).
dx
Result
c. The solution to the original initial value problem is described by the following equation in x, y.
help (equations)
y = sqrt ( 69 e^x -6x)
correct
correct
correct
correct
incorrect
x,
b. After the substitution from the previous part, we obtain the following linear differential equation in
u'-u = 6x -6
help (equations)
u, u'.
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