Common integrating factor of two differential equations.

Question

The following differential equations have a common integrating factor.

$$(3y+4xy^2)dx+(4x+5x^2y)dy=0$$ $$(6y+x^2y^2)dx+(8x+x^3y)dy=0$$

I have to find this integrating factor.

I have arrived to the solution of this problem assuming that the integrating factor I was trying to find was of the form $x^ny^m$, and in this way, I have found that the common integrating factor is $x^2y^3$.

However, I was wondering if there might be a more direct way to find this integrating factor, without having to suppose that is has the form $x^ny^m$. Any ideas?

Thanks.

Answer 1

This may help

Let us re-write the first one as: $$\frac{xdy}{ydx}=-\frac{3+6xy}{4+5xy}\implies\frac{xdy}{ydx}+1=-\frac{3+6xy}{4+5xy}+1. $$ $$\implies \frac{xdy+ydx}{ydx}=\frac{1-xy}{4+5xy}$$ $$\implies \frac{dv}{ydx}=\frac{1-v}{4+5v}, xy=v \implies \frac{dv}{dx}=\frac{v-v^2}{4+5v} \frac{1}{x}.$$ $$\implies \int \frac{4+5v}{v-v^2}dv=\int \frac{dx}{x}$$ $$\implies \ln \left( \frac{v^4}{(1-v)^9} \right)=\ln Cx \implies x^4y^4=Cx(1-xy)^9.$$

The second one can be re-written as: $$\frac{xdy}{ydx}=-\frac{6+x^2y}{8+x^2y} \implies \frac{x^2dy}{2xydx}=-\frac{1}{2}~\frac{6+x^2y}{8+x^2y}. $$ $$\implies \frac{x^2dy}{2xydx}+1=1-\frac{1}{2}~\frac{6+x^2y}{8+x^2y}$$ $$\implies \frac{x^2dy+2xydy}{2xydx}=\frac{10+x^2y}{2(8+x^2y)}$$ Let $x^2y=u$, then $$\int \frac{2(8+u)du}{10u+u^2}=\int \frac{dx}{x}$$ $$\implies \ln u^8 +\ln (10+u)^2=\ln (Cx)^5$$ $$\implies u^8(10+u)^2=(Cx)^5, u=x^2y$$

Answer 2

$$(6y+x^2y^2)dx+(8x+x^3y)dy=0$$ Rearrange terms: $$(6ydx+8xdy)+(x^2y^2dx+x^3ydy)=0$$ Factorize: $$2(3ydx+4xdy)+\frac 12x(2xy^2dx+2yx^2dy)=0$$ $$4(3ydx+4xdy)+xd(xy)^2=0$$ $$4(3y^4x^2dx+4x^3y^3dy)+(xy)^3d(xy)^2=0$$ $$4d(y^4x^3)+(xy)^3d(xy)^2=0$$ $$2d(y^4x^3)+(xy)^4d(xy)=0$$ Finally, integrate: $$10y^4x^3+(xy)^5=K$$