Question

x3y3(2ydx+ xdy) = (5ydx+7xdy)

Answer 1

Solution.

$$x^3y^3(2ydx+xdy)=5ydx+7xdy.$$

Divide both parts of the equation by $dx:$

$x^4y^3\frac{dy}{dx}+2x^3y^4-5y-7x\frac{dy}{dx}=0,$

or $x^4y^3y'+2x^3y^4-5y-7xy'=0.$

Make a replacement

$u(x)=xy(x),$ from here $y(x)=\frac{u(x)}{x},$ and $y'=\frac{u'(x)}{x}-\frac{u(x)}{x^2}.$

Substitute the obtained expressions in our equation:

$x^4\frac{u^3}{x^3}(\frac{u'}{x}-\frac{u}{x^2})+2x^3\frac{u^4}{x^4}-5\frac{u}{x}-7x(\frac{u'}{x}-\frac{u}{x^2})=0,$

or $u^3u'-7u'+\frac{u^4}{x}+\frac{2u}{x}=0.$

Divide both parts of the equation by $u^4+2u$ and write the equation in the form:

$$\frac{(u^3-7)u'}{(u^3+2)u}=-\frac{1}{x}.$$

Thus we obtain a differential equation with dividing terms. Let's solve it:

$$\frac{dx(u^3-7)\frac{du}{dx}}{(u^3+2)u}=-\frac{dx}{x},$$ $$\frac{du(u^3-7)}{(u^3+2)u}=-\frac{dx}{x},$$ $$\int\frac{u^3-7}{(u^3+2)u}du=\int(-\frac{1}{x})dx, \newline \int\frac{9u^2}{2(u^3+2)}du-\int\frac{7}{2u}du=\int(-\frac{1}{x})dx,$$ $$\frac{3}{2}\ln{|u^3+2|}-\frac{7}{2}\ln(|u|)=-\ln{|x|}+C,$$

where C is some constant.

Returning to the replacement we obtain the solution of the given equation:

$$\frac{3}{2}\ln{|x^3y^3(x)+2|}-\frac{7}{2}\ln(|xy(x)|)=-\ln{|x|}+C.$$

Answer.

$$\frac{3}{2}\ln{|x^3y^3(x)+2|}-\frac{7}{2}\ln(|xy(x)|)=-\ln{|x|}+C.$$