Finding the Expression for v Using Partial Integration

Your solution is correct.In summary, the expression for v of an analytic function with u=(3x^2) - (3y^2), considering K to be a constant, is v=6xy+K. There is an error in the book's solution where they integrated with respect to x instead of y.
  • #1
LeitHunt

Homework Statement


For an analytic function, f(x+iy)=u(x,y)+iv(x,y),u is given by u=(3x^2) -(3y^2). The expression for v, considering K to be a constant is?

Homework Equations


δu/δx=δv/δy
δu/δy=-δv/δx

[/B]

The Attempt at a Solution



My attempt :-
u=(3x^2) -(3y^2)

δu/δx=6x & δu/δy=-6y...(1)

From Relevant equations,

δu/δx=δv/δy

From (1)
6x=δv/δy

δv=∫6x δy

v=6xy+f(x)

From Relevant equations,

δu/δy=-δv/δx

From (1)
-6y=-δv/δx

-6y=-[δ/δx(6xy+f(x)]

6y=6y+f'(x)

f'(x)=0

Integrating,

f(x)=0+K

∴v=6xy+0+K

v=6xy+K[/B]

https://m.imgur.com/a/CDQlL
I know Partial derivative but never heard about Partial integration so first time came across this type of problem.
I checked the procedure on Internet and try to solve according to that.
In book they may have solved it wrong.
In picture on right side of red line I solved the problem according to Internet procedures and I got the same answer as book. But in book the (6x) should be partially integrate with respect to y by mistake they may have integrated with respect to x.
Which one is correct in picture solved by book or by me?
 
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  • #2
LeitHunt said:

Homework Statement


For an analytic function, f(x+iy)=u(x,y)+iv(x,y),u is given by u=(3x^2) -(3y^2). The expression for v, considering K to be aconstant is?

Homework Equations


δu/δx=δv/δy
δu/δy=-δv/δx

[/B]

The Attempt at a Solution


https://imgur.com/a/S82cs[/B]
I know Partial derivative but never heard about Partial integration so first time came across this type of problem.
I checked the procedure on Internet and try to solve according to that.
In book they may have solved it wrong.
In picture on right side of red line I solved the problem according to Internet procedures and I got the same answer as book. But in book the (6x) should be partially integrate with respect to y by mistake they may have integrated with respect to x.
Which one is correct in picture solved by book or by me?

Do not post images---especially sideways ones. Just type out the solution (which is the PF standard!)

You should not expect people to lie down sideways on their desks in order to read your pictures.
 
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  • #3
Ray Vickson said:
Do not post images---especially sideways ones. Just type out the solution (which is the PF standard!)

You should not expect people to lie down sideways on their desks in order to read your pictures.
Sorry I was on mobile. I'll edit it soon :)
 
  • #4
Edited :)
 
  • #5
LeitHunt said:
Edited :)

You are correct. Clearly, the book integrated part i) wrong.
 
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1. What is partial integration?

Partial integration is a mathematical technique used to solve integrals by breaking them into smaller, simpler integrals. It is also known as integration by parts, and is the reverse process of differentiation.

2. When is partial integration used?

Partial integration is typically used when the integral of a function cannot be easily solved by other integration techniques, such as substitution or trigonometric identities.

3. What is the formula for partial integration?

The formula for partial integration is ∫u dv = uv - ∫v du, where u and v are functions of x and du and dv are their respective differentials.

4. What are the steps for partial integration?

The steps for partial integration are: 1. Identify u and dv in the given integral. 2. Calculate du and v using differentiation and integration respectively. 3. Substitute the values of u, du, v, and dv into the formula ∫u dv = uv - ∫v du. 4. Solve for the remaining integral. 5. If necessary, repeat the process until the integral is fully solved.

5. What are some common mistakes to avoid when using partial integration?

Some common mistakes to avoid when using partial integration include: - Forgetting to add the constant of integration when solving the remaining integral. - Incorrectly identifying u and dv, or not using the correct formula for the given integral. - Making a mistake in the calculation of du or v. - Not simplifying the final answer.

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