Answer to Question #269452 in Differential Equations for Aljon

Question

Homogeneous differential equation

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

2. (x^2+y^2)dx+xydy=0

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

4. (3x+2y)dx-2xdy=0

Answer 1

1.

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

$$(xy^2)y'-(x^3+y^3)=0$$

$$y'-\dfrac{1}{x}y=x^2y^{-2}$$

First order Bernoulli ODE

$$u=y^{1-(-2)}=y^3$$

$$u'=3y^2y'$$

$$\dfrac{1}{3}u'-\dfrac{1}{x}u=x^2$$

$$u'-\dfrac{3}{x}u=3x^2$$

Integration factor

$$\mu(x)=e^{\int(-3/x)dx}=x^{-3}$$

$$x^{-3}u'-\dfrac{3}{x^4}u=\dfrac{3}{x}$$ $$d(x^{-3}u)=\dfrac{3}{x}dx$$

Integrate

$$\int d(x^{-3}u)=\int\dfrac{3}{x}dx$$

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

$$y^3=3x^3\ln x+Cx^3$$

$$y=x\sqrt[3]{3\ln x+C}$$

2.

$$(x^2+y^2)dx+xydy=0$$

$$x^2+y^2+xyy'=0$$

$$y'+\dfrac{1}{x}y=-xy^{-1}$$

First order Bernoulli ODE

$$u=y^{1-(-1)}=y^2$$

$$u'=2yy'$$

$$\dfrac{1}{2}u'+\dfrac{1}{x}u=-x$$

$$u'+\dfrac{2}{x}u=-2x$$

Integration factor

$$\mu(x)=e^{\int(2/x)dx}=x^2$$

$$x^2u'+2xu=-2x^3$$

$$d(x^2u)=-2x^3dx$$

Integrate

$$\int d(x^2u)=-\int2x^3dx$$

$$x^2u=-\dfrac{1}{2}x^4+C$$

$$y^2=-\dfrac{1}{2}x^2+\dfrac{C}{x^2}$$

3.

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

$$y^2-x^2+2xyy'=0$$

$$y'+\dfrac{1}{2x}y=\dfrac{1}{2}xy^{-1}$$

First order Bernoulli ODE

$$u=y^{1-(-1)}=y^2$$

$$u'=2yy'$$

$$\dfrac{1}{2}u'+\dfrac{1}{2x}u=\dfrac{1}{2}x$$

$$u'+\dfrac{1}{x}u=xy^{-1}$$

Integration factor

$$\mu(x)=e^{\int(1/x)dx}=x$$

$$xu'+u=x^2$$

$$d(xu)=x^2dx$$

Integrate

$$\int d(xu)=\int x^2dx$$

$$xu=\dfrac{1}{3}x^3+C$$

$$y^2=\dfrac{1}{3}x^2+\dfrac{C}{x}$$

4.

$$(3x+2y)dx-2xdy=0$$

$$y'-\dfrac{1}{x}y=\dfrac{3}{2}$$

Integration factor

$$\mu(x)=e^{-\int(1/x)dx}=\dfrac{1}{x}$$

$$\dfrac{1}{x}y'-\dfrac{1}{x^2}y=\dfrac{3}{2x}$$

$$d(\dfrac{1}{x}y)=\dfrac{3}{2x}dx$$

Integrate

$$\int d(\dfrac{1}{x}y)=\int\dfrac{3}{2x}dx$$

$$\dfrac{1}{x}y=\dfrac{3}{2}\ln x+C$$

$$y=\dfrac{3}{2}x\ln x+Cx$$