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I'm having trouble with this problem, and was wondering if someone could lead me in the right direction.

Here is the question. Show that $x^ay^b$ is an integrating factor of the equation $$(b+1)x {dy\over dx} + (a+1)y = 0$$, and find its solution

My attempt: $${dy\over dx}={-(a+1)y\over (b+1)x}$$ $${1\over (a+1)y}dy+{1\over (b+1)x}dx = 0$$ $A = {1\over (a+1)y}$, $B = {1\over (b+1)x}$ $${{dA\over dy}-{dB\over dx}\over B}={x(b+1)\over y^2(a+1)}+{1 \over x}$$ However, I'm kind of lost right now, where do I go from here, or am I even doing this right? Thanks

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  • $\begingroup$ Multiply $x^ay^b$ to your equation then by using $\frac{dA}{dx}=\frac{dB}{dy}$ find $a,b$ $\endgroup$ – R.N Nov 16 '15 at 1:12
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$$(b+1)x {dy\over dx} + (a+1)y = 0\Rightarrow (b+1)x {dy } + (a+1)y dx = 0$$ let $N=(b+1)x, M=(a+1)y$ then $\frac{\delta N}{\delta x}=b+1, \frac{\delta M}{\delta y}=a+1$ so $$\frac{\frac{\delta N}{\delta x}-\frac{\delta M}{\delta y}}{M}=\frac{b-a}{(a+1)y}$$

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