Answer to Question #162393 in Differential Equations for Kanchan Roy

Question

(3x² +y²)dy = (-x²- 3y²)dx

Answer 1

Solution.

$$(3x^2+y^2)dy=(-x^2-3y^2)dx.$$

This is a homogeneous nonlinear differential equation of the first order. Divide both parts of the equation by $dx:$

$(3x^2+y^2)\frac{dy}{dx}=-x^2-3y^2, \text{or } (3x^2+y^2)y'=-x^2-3y^2.$

Make a replacement

$u(x)=\frac{y(x)}{x},$ from here $y(x)=xu(x),$ and $y'=u(x)+xu'(x).$

Substitute the obtained expressions in our equation:

$(3x^2+x^2u^2)(u+xu')+x^2+3x^2u^2=0,$

$x^3u^2u'+3x^3u'+x^2u^3+3x^2u^2+3x^2u+x^2=0,$

$x^3u'(u^2+3)+x^2(u^3+3u^2+3u+1)=0.$

Divide both parts of the equation by $u^3+3u^2+3u+1$ and write the equation in the form:

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

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

$$\frac{dx(u^2+3)\frac{du}{dx}}{u^3+3u^2+3u+1}=-\frac{dx}{x},$$

$$\frac{du(u^2+3)}{u^3+3u^2+3u+1}=-\frac{dx}{x},$$

$$\int\frac{u^2+3}{u^3+3u^2+3u+1}du=\int-\frac{1}{x}dx,$$

$$\frac{2u}{u^2+2u+1}+\ln(u+1)=-\ln{x}+C,$$

where C - some constant.

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

$$\frac{2y(x)}{x(1+\frac{2y(x)}{x}+\frac{y^2(x)}{x^2})}+\ln(1+\frac{y(x)}{x})=-\ln{x}+C,$$

or

$$\frac{2xy(x)}{(y(x)+x)^2}+\ln(1+\frac{y(x)}{x})=-\ln{x}+C.$$

Answer.

$$\frac{2xy(x)}{(y(x)+x)^2}+\ln(1+\frac{y(x)}{x})=-\ln{x}+C.$$