'guess' solution, differential question.

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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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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)
 
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?
 
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:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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