Question

1.y(3x^3-x+y)dx+x^2(1-x^2)dy=0
2.y(2x+y^2)dx+x(y^2-x)dy=0

Answer 1

Solution.

1.

$$y(3x^3-x+y)dx+x^2(1-x^2)dy=0$$

This is the first order nonlinear ordinary differential equation.

Divide on $x^2(1-x^2)dx.$

$$y'=\frac{3x^3y-xy+y^2}{x^2(x^2-1)},$$ $$y'=\frac{y^2}{x^2(x-1)(x+1)}+\frac{3xy}{(x-1)(x+1)}-\frac{y}{x(x-1)(x+1)},$$

$$y'-\frac{3x^2-1}{x^3-x}y=\frac{y^2}{x^4-x^2}.$$

We have Bernuli's equation for n=2.

Divide by $y^2.$

$$\frac{y'}{y^2}-\frac{3x^2-1}{(x^3-x)y}=\frac{1}{x^4-x^2}.$$

Replacement $u=\frac{1}{y},$ then $u'=-\frac{y'}{y^2}, y'=-u'y^2.$

$$-\frac{u(3x^2-1)}{x^3-x}-u'=\frac{1}{x^4-x^2},$$

$$u'+(\frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)})u=-\frac{1}{x^4-x^2}$$

This is the first order linear equation.

Let be $P(x)=\frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)}.$

$\int P(x)dx=\int \frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)}dx=\ln (x^3-x)+C,$

where C is some constant.

Solve

$$u'+P(x)u=0,$$

$$\frac{du}{u}=-P(x)dx,$$

$$\int\frac{du}{u}=-\int P(x)dx,$$

$$\ln u =\ln (x^3-x)+C,$$

$$u=\frac{C}{x^3-x}.$$

The solution of $u'+(\frac{3x}{(x-1)(x+1)}-\frac{1}{x(x-1)(x+1)})u=-\frac{1}{x^4-x^2}$ find by the method of variation constant in the form $u=\frac{C(x)}{x^3-x}.$

$$C(x)=\int(-\frac{x^3-x}{x^4-x^2})dx=-\ln x+C.$$

Thus $u=\frac{-\ln x+C}{x^3-x}.$

From here $y=\frac{x-x^3}{\ln x+C}.$

Answer. $y=\frac{x-x^3}{\ln x+C}.$

2.

$$x(y^2-x)dy=y(y^2+2x)dx=0.$$

This is the first order nonlinear ordinary differential equation.

Replacement $y=\sqrt{z},$ then $dy=\frac{dz}{2\sqrt{z}}.$ We will have

$$(\frac{c\sqrt{z}}{2}-\frac{x^3}{2\sqrt z})dz=(-z^{\frac{3}{2}}-2x\sqrt z)dx.$$

Replacement $u=\frac{z}{x},$ then

$x=ux, dz=udx+xdu.$

We will have

$$(\frac{\sqrt {u}}{2}-\frac{1}{2\sqrt u}x^{3/2}(udx+xdu)=(-u^{3/2}-2\sqrt u)x^{3/2}dx,$$

$$(\frac{\sqrt {u}x^{5/2}}{2}-\frac{x^{5/2}}{2\sqrt u})du=(-\frac{3u^{3/2}x^{3/2}}{2}-\frac{3\sqrt u u^{3/2}}{2})dx.$$

Divide by $x^{5/2}$ and $-\frac{3u^{3/2}x^{3/2}}{2}-\frac{3\sqrt u u^{3/2}}{2}.$

We will have

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

$\int(\frac{1}{3u(u+1)}-\frac{1}{3(u+1)})du=\int\frac{dx}{x}.$

$\frac{\ln u}{3}-\frac{2\ln (u+1)}{3}=\ln x+C,$

where C is some constant.

From here

$$\frac{u}{(u+1)^2}=Cx^3.$$

Thus

$$\frac{z}{(z+x)^2}=Cx^2.$$

So we have answer

$$y^2=Cx^2(y^2+x)^2.$$

Answer. $y^2=Cx^2(y^2+x)^2.$