Still stuck on diffrential equations

sara_87
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Homework Statement



let x + y = u and y = uv
Expand dx and dy in terms of du and dv

Homework Equations





The Attempt at a Solution



i got this answer:

dy = udv + vdu

and

dx = du - udv - vdu


is this correct?
 
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Looks correct to me. Using the product rule on "y = uv", you get dy, and then a simple substitution in the equation "x + y = u" gives you dx. You got it.
 
ok thanx
now that makes things harder

we have w=(1 - y e^(y/x+y))dx + (1 + xe^(y/x+y)dy

find an integrating factor 'mu' in terms of u and v such that 'mu'w is exact

after subing that lot in for x and y and dx and dy, and rearranging a little i got this:


'mu' = [ (1+u(1-v)d'mu' ... something long and horrible!

how do i do this?
 
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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