'guess' solution, differential question.

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In summary, the conversation involves solving the equation dy/dx = x/y for the given condition y = 1 and x = 2. The approach of finding a complementary function does not make sense and the equation is actually separable. The particular integral is found by rearranging the equation and integrating, resulting in the general function y = C + x + F and the answer y = sqrt(x^2 - 3), which checks with the initial conditions y = 0 and x = 2.
  • #1
keith river
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dy/dx = x/y
Solve the equation (get general form of y) for the given condition y=1 and x=2

I've tried finding the complementary function, dy/dx = 0.
So I assume y = C (a constant)

Now I'm trying the find the particular Integral.
dy/dx = x/y

rearrange for LHS containing only y and RHS containing only x

dy y = dx x
I integrate I get (y^2) / 2 = (x^2)/2 + D(constant due to integration)

y^2 = 2(x^2)/2 + 2D
y^2 = (x^2) + E (2D= E)
y = Sqrt (x^2) + Sqrt (E)
y = x + F

General function
y = C + x + F
y = G + x

The answer (given onnsheet) is y = Sqrt ((x^2) - 4)
 
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  • #2
keith river said:
Now I'm trying the find the particular Integral.
dy/dx = x/y

rearrange for LHS containing only y and RHS containing only x

dy y = dx x
I integrate I get (y^2) / 2 = (x^2)/2 + D(constant due to integration)

y^2 = 2(x^2)/2 + 2D
y^2 = (x^2) + E (2D= E)
y = Sqrt (x^2) + Sqrt (E)
y = x + F

If you were going to get the particular integral by solving the equation, it sort of made no sense to get a complementary solution to y' = 0.

But from this last line: if a2= b + c then a ≠ √b + √ c,

a = √(b+c)
 
  • #3
keith river said:
dy/dx = x/y
Solve the equation (get general form of y) for the given condition y=1 and x=2

I've tried finding the complementary function, dy/dx = 0.
So I assume y = C (a constant)
I can't see how this approach would work.

Your equation is separable:
y dy = x dx


keith river said:
Now I'm trying the find the particular Integral.
dy/dx = x/y

rearrange for LHS containing only y and RHS containing only x

dy y = dx x
I integrate I get (y^2) / 2 = (x^2)/2 + D(constant due to integration)

y^2 = 2(x^2)/2 + 2D
y^2 = (x^2) + E (2D= E)
y = Sqrt (x^2) + Sqrt (E)
y = x + F

General function
y = C + x + F
y = G + x

The answer (given onnsheet) is y = Sqrt ((x^2) - 4)

I get y = sqrt(x^2 - 3), and my answer checks. Are you sure you have the right initial conditions?
 
  • #4
thanks, I can't believe I forgot something as simple as that.
and Initial conditions were y=0, x=2
Sorry about the typo, there are a lot on the worksheet.
But seeing the sqrt all under one bracket made me realize what to do.
I've got it now.
 
Last edited:

1. What is a 'guess' solution in differential equations?

A 'guess' solution in differential equations refers to an initial estimate or tentative function that is used to solve a differential equation. It is usually an approximation that can be refined using other techniques such as numerical methods.

2. How is a 'guess' solution different from an exact solution?

A 'guess' solution is an approximation of the exact solution and may not satisfy all the conditions of the differential equation. An exact solution, on the other hand, is a function that satisfies all the conditions of the differential equation and is considered to be the most accurate representation of the problem.

3. Can a 'guess' solution be used to solve any type of differential equation?

Yes, a 'guess' solution can be used to solve any type of differential equation. However, the accuracy of the solution may vary depending on the complexity of the equation and the chosen method of approximation.

4. What are some common methods for obtaining a 'guess' solution?

Some common methods for obtaining a 'guess' solution include the method of undetermined coefficients, variation of parameters, and the method of separation of variables. These methods involve making an educated guess based on the form of the differential equation and then refining the solution using other techniques.

5. Is it necessary to have a 'guess' solution before attempting to solve a differential equation?

No, it is not necessary to have a 'guess' solution before attempting to solve a differential equation. There are many other methods and techniques that can be used to solve differential equations without the need for a 'guess' solution. However, having an initial estimate can often make the process easier and more efficient.

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