# Solving differential equation without finding integrating factor

I have been trying to solve this equation

$$\left( 8y+10x \right) \mathrm{d}x+\left( 5y+7x \right) \mathrm{d}y=0\\$$

by finding integrating factor, though is there more rational way to do it?

• Hint: Let $$y = v x \implies y' = v + v' x$$ Now substitute and solve. – Moo May 28 '17 at 22:36
• Thank you, that should do it! – Jack Nickolson May 28 '17 at 22:39

Turning Moo's comment into an answer. In general if we have something of the form $$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{8y + 10x}{5y + 7x} = -\frac{8\frac{y}{x} + 10}{5\frac{y}{x} + 7}$$ i.e $y' = f(y/x)$ then it is a good idea to try the substitution $y=vx \implies y' = v + v'x$ to get $$v + x\frac{\mathrm{d}v}{\mathrm{d}x}=-\frac{8v+10}{5v +7}\implies x\frac{\mathrm{d}v}{\mathrm{d}x} = -\left(\frac{8v+10}{5v+7} +\frac{5v^2+7v}{5v+7}\right)=-\frac{5(v+2)(v+1)}{5v+7}$$ which is a separable equation $$\int\frac{5v+7}{5(v+2)(v+1)}\, \mathrm{d}v=-\log x + c \implies \log(v+1)^2(v+2)^3=-5 \log x + c$$ via partial fractions giving $$(v+1)^2(v+2)^3=\frac{A}{x^5} \implies (y+x)^2(y+2x)^3 = A.$$