What Is the Correct Integrating Factor for the Equation ay dx + bx dy = 0?

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EXACT equation !

hi! to all
I want little help about exact equation..
that the given equation is
>> ay dx + bx dy= 0
I had solved and found integrating factor of all of my equations by formula
P(x) = My-Nx/N
but for this equation I have got IF for this equation is x((a-b)/b) and after multiplying this factor by equation I can't get exact equation once again :mad:
and in book answers IF for this equation is x^a-1.y^b-1
So., can anyone tell me how can I solve this eq by method of inspection >>
pleasezz help me!
 
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Why not just divide by xy?
 


In order to determine what the integrating factor is you must first normalize the equation.
\frac{dy}{dx}+P(x)y=f(x)
is an example of a normalized differential equation, the IF being
e^{\int P(x)dx}.

Does this help?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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