Answer to Question #173965 in Differential Equations for Randal Rodriguez

Question

Integrating Factors found by Inspection.

1. y(2xy + 1)dx − xdy = 0

2. y(x^4 − y^2)dx + x (x^4 + y^2)dy = 0

3. (x^3y^3 + 1)dx + x^4y^2dy = 0

4. y(x^2y^2 − 1)dx + x(x^2y^2 + 1)dy = 0

5. y(2x + y^2)dx + x(y^2 − x)dy = 0

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

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

8. y^2(1 − x^2)dx + x(x^2y + 2x + y) = 0

Answer 1

Solution.

1. $y(2xy + 1)dx − xdy = 0$

$2xdx+\frac{ydx-xdy}{y^2}=0$

$2xdx+d(\frac{x}{y})=0$

$\int 2xdx+\int d(\frac{x}{y})=0$

$x^2y+x=Cy,$ where $C$ is some constant.

2. $y(x^4 − y^2)dx + x (x^4 + y^2)dy = 0$

$x^4(ydx+xdy)+y^2(xdy-ydx)=0$

$ydx+xdy+\frac{y^2}{x^2}\frac{xdy-ydx}{x^2}=0$

$d(xy)+(\frac{y}{x})^2d(\frac{y}{x})=0$

$\int d(xy)+\int(\frac{y}{x})^2d(\frac{y}{x})=0$

$3x^4y+y^3=Cx^3,$ where $C$ is some constant.

3. $(x^3y^3 + 1)dx + x^4y^2dy = 0$

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

$x^2y^2d(xy)+\frac{dx}{x}=0$

$x^3y^3+3\ln{x}=C,$ where $C$ is some constant.

4. $y(x^2y^2 − 1)dx + x(x^2y^2+1)dy = 0$

$x^2y^3dx-ydx+x^3y^2dy+xdy=0$

$\frac{xdy-ydx}{x^2}+y^2(ydx+xdy)=0$

$d(\frac{y}{x})+y^2d(xy)=0 \newline \frac{x}{y}d(\frac{y}{x})+xyd(xy)=0$

$2\ln{\frac{y}{x}}+x^2y^2=C,$ where $C$ is some constant.

5. $y(2x + y^2)dx + x(y^2 − x)dy = 0$

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

$\frac{2xydx-x^2dy}{y^2}+ydx+xdy=0$

$d(\frac{x^2}{y})+d(xy)=0$

$x^2+xy^2=Cy,$ where $C$ is some constant.

6. and 7.

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

$x^3-x=Cy-y\ln{x},$ where $C$ is some constant.

8. $y^2(1 − x^2)dx + x(x^2y + 2x + y) = 0$

$y^2dx-x^2y^2dx+x^3ydy+2x^2dy+xydy=0\newline ydx+xdy+x^3dy-x^2ydx+\frac{2x^2}{y}dy=0 \newline d(xy)+x^4d(\frac{y}{x})+2x^2\frac{dy}{y}=0 \newline (2Cxy+x^2+1)^2=-4Cx^2+x^4+2x^2+1, \text{ where C is some constant.}$