Answer to Question #197110 in Differential Equations for Taimoor Arshad

Question

find the differential equation:

2x + 6 y y' =(x^2+3y^2/y)y'

Answer 1

Solution.

$$2x+6yy'=(x^2+3y^2)/y•y'$$

Let be $y=x•v(x),$ then

$y'=xv'(x)+v.$

We will have

$$2x+6x(xv'+v)•v=\frac{(x^2+3x^2v^2)(xv'+v)}{xv}$$

$$2x+6x(xv'+v)•v=\frac{x(1+3v^2)(xv'+v)}{v}$$

$$2xv+6x^2v^2v'+6xv^3=3x^2v^2v'+x^2v'+3xv^3+xv$$

$$xv+3x^2v^2v'+3xv^3=x^2v'$$

$$x(3v^2-1)v'=-3v^3-v$$

$$\frac{3v^2-1}{-3v^3-v}v'=\frac{1}{x}$$

$$\int\frac{3v^2-1}{-3v^3-v}dv=\int\frac{1}{x}dx$$

$$-\int(\frac{6v}{3v^2+1}-\frac{1}{v})dv=\int\frac{1}{x}dx$$

$$-\ln (3v^2+1)+\ln v=\ln x+C,$$

where C is some constant.

$$v=x(3v^2+1)e^C$$

$$y=(3y^2+x^2)e^C$$

$$-3e^Cy^2+y-x^2e^C=0$$

$$y=\frac{1}{6}(e^C\pm\sqrt{e^{2C}-12x^2}).$$