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'guess' solution, differential question.

 
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Feb24-11, 01:19 PM   #1
 

'guess' solution, differential question.


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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Feb24-11, 01:30 PM   #2
 
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Quote by keith river View Post
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)
Feb24-11, 01:35 PM   #3
 
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Quote by keith river View Post
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


Quote by keith river View Post

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?
Feb24-11, 01:50 PM   #4
 

'guess' solution, differential question.


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 realise what to do.
I've got it now.
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