# Finding an integrating factor and solving the differential equation

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

• Multiply $x^ay^b$ to your equation then by using $\frac{dA}{dx}=\frac{dB}{dy}$ find $a,b$ – R.N Nov 16 '15 at 1:12

$$(b+1)x {dy\over dx} + (a+1)y = 0\Rightarrow (b+1)x {dy } + (a+1)y dx = 0$$ 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 $$\frac{\frac{\delta N}{\delta x}-\frac{\delta M}{\delta y}}{M}=\frac{b-a}{(a+1)y}$$