Definición: Sean P(x, y) y Q(x, y) funciones reales continuas en un dominio D. Se dice que la ecuación

P(x, y)dx+Q(x, y)dy=0

Es diferencial exacta si existe una función real F(x, y) tal que en el dominio D cumple:

${\displaystyle \begin{array}{l}\frac{\partial F}{\partial x}=P(x,y)\\ \frac{\partial F}{\partial y}=Q(x,y)\end{array}}$

La función F(x, y) es una primitiva de la ecuación y la integral general es:

F(x, y)=cte

Teorema.- La condición necesaria y suficiente para que la ecuación P·dx+Q·dy=0 sea diferencial exacta en un dominio D, siendo P(x, y), Q(x, y) y sus derivadas parciales continuas en D es que se cumpla:

${\displaystyle \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}}$

Ejemplos:

1. ${\displaystyle e^x(y^3+x{y}^3+1)dx+3y^2(x{e}^x-6)dy=0}$

${\displaystyle \begin{array}{l}\frac{\partial P}{\partial y}=e^x(3y^2+3y^{2}x)=e^{x}3y^2(1+x)\\ \frac{\partial Q}{\partial x}=3y^2(e^x+x{e}^x)=e^{x}3y^2(1+x)\end{array}}$

D=R²

Resolución.- Por lo anterior se sabe que:

${\displaystyle \begin{array}{l}\frac{\partial F}{\partial x}=P(x,y)=e^x(y^3+x{y}^3+1)\Rightarrow F(x,y)={\displaystyle \int e^x(y^3+x{y}^3+1)dx}=\\ =e^x(y^3+x{y}^3+1)-{\displaystyle \int e^x{y}^{3}dx=}{e}^x(y^3+x{y}^3+1)-e^x{y}^3=e^x(x{y}^3+1)+\alpha (y)\\ \frac{\partial F}{\partial y}=3x{y}^2{e}^x+\alpha \text{'}(y)\end{array}}$

deberá coincidir con

${\displaystyle \begin{array}{l}Q(x,y)=3y^2(x{e}^x-6)dy\Rightarrow \\ \Rightarrow \alpha \text{'}(y)=-18y^2\Rightarrow \alpha (y)=-6y^3+\mathrm{cte}\end{array}}$

Por tanto,

${\displaystyle F(x,y)=e^x(x{y}^3+1)-6y^3+\mathrm{cte}}$

La integral general

${\displaystyle e^x(x{y}^3+1)-6y^3=K}$

2. ${\displaystyle (2xy+y^2)dx+(x^2+2xy-y)dy=0}$

${\displaystyle \begin{array}{l}\frac{\partial P}{\partial y}=2x+2y\\ \frac{\partial Q}{\partial x}=2x+2y\end{array}}$

Diferencial exacta

${\displaystyle \begin{array}{l}F(x,y)={\displaystyle \int (2xy+y^2)dx=x^{2}y+y^{2}x+\alpha (y)}\\ \frac{\partial F}{\partial y}=x^2+2xy+\alpha \text{'}(y)=Q(x,y)=x^2+2xy-y\Rightarrow \alpha \text{'}(y)=-y\\ \Rightarrow \alpha (y)=-\frac{y^2}{2}+\mathrm{cte}\end{array}}$

Integral general

${\displaystyle x^{2}y+y^{2}x-\frac{y^2}{2}=K}$

3. ${\displaystyle (x+y\cos x)dx+\mathrm{sen}xdy=0}$

P_y=cosx, Q_x=cosx. Diferencial exacta

${\displaystyle \begin{array}{l}F(x,y)={\displaystyle \int \mathrm{sen}xdy=}y\mathrm{sen}x+\alpha (x)\\ \frac{\partial F}{\partial x}=y\cos x+\alpha \text{'}(x)=P(x,y)=x+y\cos x\\ \Rightarrow \alpha \text{'}(x)=x\Rightarrow \alpha (x)=\frac{x^2}{2}+\mathrm{cte}\end{array}}$

Integral general

${\displaystyle y\mathrm{sen}x+\frac{x^2}{2}=K}$

4. ${\displaystyle (2x{e}^y+e^x)dx+(x^2+1)e^{y}dy=0}$

P_y=2xe^y, Q_x==2xe^y. Diferencial exacta

${\displaystyle \begin{array}{l}F(x,y)={\displaystyle \int (2x{e}^y+e^x)dx=x^2{e}^y+}{e}^x+\alpha (y)\\ \frac{\partial F}{\partial y}=x^2{e}^y+\alpha \text{'}(y)=Q(x,y)=(x^2+1)e^y\Rightarrow \alpha \text{'}(y)=e^y\Rightarrow \alpha (y)=e^y+\mathrm{cte}\end{array}}$

Integral general

${\displaystyle x^2{e}^y+e^x+e^y=(x^2+1)e^y+e^x=K}$

5. ${\displaystyle \left(e^x+\ln y+\frac{y}{x}\right)dx+\left(\frac{x}{y}+\ln x+\mathrm{sen}y\right)dy=0}$

${\displaystyle P_y=\frac{1}{y}+\frac{1}{x},Q_x=\frac{1}{y}+\frac{1}{x}}$

Diferencial exacta

${\displaystyle \begin{array}{l}F(x,y)={\displaystyle \int \left(e^x+\ln y+\frac{y}{x}\right)dx=}{e}^x+x\ln y+y\ln x+\alpha (y)\\ \frac{\partial F}{\partial y}=\frac{x}{y}+\ln x+\alpha \text{'}(y)=Q(x,y)=\frac{x}{y}+\ln x+\mathrm{sen}y\\ \Rightarrow \alpha \text{'}(y)=\mathrm{sen}y\Rightarrow \alpha (y)=-\cos y+\mathrm{cte}\end{array}}$

Integral general

${\displaystyle e^x+x\ln y+y\ln x-\cos y=K}$

6. $\left(\frac{x}{\sqrt{x^2+y^2}}+\frac{1}{x}+\frac{1}{y}\right)dx+\left(\frac{y}{\sqrt{x^2+y^2}}+\frac{1}{y}-\frac{x}{y^2}\right)dy=0$

$\begin{array}{l}{P}_y=x\left(\frac{-1}{2}\right){\left(x^2+y^2\right)}^{-\frac{3}{2}}2y-\frac{1}{y^2}=-xy{\left(x^2+y^2\right)}^{-\frac{3}{2}}-\frac{1}{y^2}\\ Q_x=y\left(\frac{-1}{2}\right){\left(x^2+y^2\right)}^{-\frac{3}{2}}2x-\frac{1}{y^2}=-yx{\left(x^2+y^2\right)}^{-\frac{3}{2}}-\frac{1}{y^2}\end{array}$

Diferencial exacta

Factores integrantes

Definición.- Dada la ecuación P(x, y)dx+Q(x, y)dy=0, se dice que es un factor integrante en un dominio D, siendo P(x, y), Q(x, y) y sus derivadas parciales continuas en D, si multiplicado por la ecuación la convierte en diferencial exacta. Es decir si la ecuación

μ(x, y)P(x, y)dx+μ(x, y)Q(x, y)dy=0

es exacta. Por lo tanto deberá de cumplir:

${\displaystyle \frac{\partial \left(\mu P\right)}{\partial y}=\frac{\partial \left(\mu Q\right)}{\partial x}\Rightarrow {\mu }_{y}P+\mu P_y={\mu }_{x}Q+\mu Q_x}$, (*)

1. El factor integrante es solo función de x

${\displaystyle \mu =\mu (x)\Rightarrow \{\begin{array}{l}{\mu }_y=0\\ {\mu }_x=\frac{d\mu }{dx}=\mu \text{'}\end{array}}$

Sustituyendo en (*)

${\displaystyle \mu (P_y-Q_x)=\mu \text{'}Q\Rightarrow \frac{\mu \text{'}}{\mu }=\frac{P_y-Q_x}{Q}}$

Para poder integrar y obtener así el factor integrante:

${\displaystyle \frac{P_y-Q_x}{Q}}$

debe ser solo función de x

Ejemplo: (1-x²y)dx+x²(y-x)dy=0

${\displaystyle P_y=-x^2,Q_x=2xy-3x^2}$

No es exacta

${\displaystyle \frac{P_y-Q_x}{Q}=\frac{2x^2-2xy}{x^2(y-x)}=\frac{2x(x-y)}{x^2(y-x)}=-\frac{2}{x}}$

Admite factor integrante. Integrando

${\displaystyle \ln \mu =-2\ln x\Rightarrow \mu =\frac{1}{x^2}}$

La ecuación

${\displaystyle \frac{1}{x^2}(1-x^{2}y)dx+\frac{1}{x^2}{x}^2(y-x)dy=0=\left(\frac{1}{x^2}-y\right)dx+(y-x)dy}$

Exacta. Solución

${\displaystyle -\frac{1}{x}-yx+\frac{y^2}{2}=K}$

2. El factor integrante es solo función de y

${\displaystyle \mu =\mu (y)\Rightarrow \{\begin{array}{l}{\mu }_x=0\\ {\mu }_y=\frac{d\mu }{dy}=\mu \text{'}\end{array}}$

Sustituyendo en (*)

${\displaystyle \mu (Q_x-P_y)=\mu \text{'}P\Rightarrow \frac{\mu \text{'}}{\mu }=\frac{Q_x-P_y}{P}}$

Para poder integrar y obtener así el factor integrante $\frac{Q_x-P_y}{P}$, debe ser solo función de y

Ejemplo. x²y²dx+(x³y+y+3)dy=0

P_y=2yx², Q_x=3yx². No es exacta

${\displaystyle \frac{Q_x-P_y}{P}=\frac{3x^{2}y-2y{x}^2}{x^2{y}^2}=\frac{1}{y}}$

Integrando

${\displaystyle \ln \mu =\ln y\Rightarrow \mu =y}$

3. El factor integrante es solo función de x+y

μ=μ(x+y), se hace x+y=t

${\displaystyle \{\begin{array}{l}{\mu }_x=\frac{d\mu }{dt}\frac{\partial t}{\partial x}=\frac{d\mu }{dt}·1=\mu \text{'}\\ {\mu }_y=\frac{d\mu }{dt}\frac{\partial t}{\partial y}=\frac{d\mu }{dt}·1=\mu \text{'}\end{array}}$

Sustituyendo en (*)

${\displaystyle \begin{array}{l}\mu \text{'}P+\mu P_y=\mu \text{'}Q+\mu Q_x\Rightarrow \mu \text{'}(P-Q)=\mu (Q_x-P_y)\\ \frac{\mu \text{'}}{\mu }=\frac{(Q_x-P_y)}{P-Q}\Rightarrow \frac{(Q_x-P_y)}{P-Q}\end{array}}$

debe ser solo función de x+y

4. El factor integrante es solo función de x-y

μ=μ(x-y), se hace x-y=t

${\displaystyle \{\begin{array}{l}{\mu }_x=\frac{d\mu }{dt}\frac{\partial t}{\partial x}=\frac{d\mu }{dt}·1=\mu \text{'}\\ {\mu }_y=\frac{d\mu }{dt}\frac{\partial t}{\partial y}=\frac{d\mu }{dt}·(-1)=-\mu \text{'}\end{array}}$

Sustituyendo en (*)

${\displaystyle \begin{array}{l}-\mu \text{'}P+\mu P_y=\mu \text{'}Q+\mu Q_x\Rightarrow \mu (P_y-Q_x)=\mu \text{'}(P+Q)\\ \frac{\mu \text{'}}{\mu }=\frac{(P_y-Q_x)}{P+Q}\Rightarrow \frac{(P_y-Q_x)}{P+Q}\end{array}}$

debe ser solo función de x-y

5. El factor integrante es solo función de x·y

μ=μ(x·y), se hace x·y=t

${\displaystyle \{\begin{array}{l}{\mu }_x=\frac{d\mu }{dt}\frac{\partial t}{\partial x}=\frac{d\mu }{dt}·y=\mu \text{'}·y\\ {\mu }_y=\frac{d\mu }{dt}\frac{\partial t}{\partial y}=\frac{d\mu }{dt}·x=\mu \text{'}·x\end{array}}$

Sustituyendo en (*)

${\displaystyle \begin{array}{l}P+\mu P_y=\mu \text{'}yQ+\mu Q_x\Rightarrow \mu \text{'}(xP-yQ)=\mu (Q_x-P_y)\\ \frac{\mu \text{'}}{\mu }=\frac{(Q_x-P_y)}{xP-yQ}\Rightarrow \frac{(Q_x-P_y)}{xP-yQ}\end{array}}$

debe ser solo función de x·y

Ejemplo: (y+xy²)dx+(x-yx²)dy=0

${\displaystyle \begin{array}{l}{P}_y=1+2xy,Q_x=1-2xy\Rightarrow \frac{(Q_x-P_y)}{xP-yQ}=\frac{-4xy}{(xy+x^2{y}^2)-(xy-x^2{y}^2)}=\frac{-2}{xy}\\ \Rightarrow \ln \mu ={\displaystyle \int \frac{-2}{t}dt=-2\ln t\Rightarrow \mu =\frac{1}{{(xy)}^2}}\end{array}}$

6. El factor integrante es solo función de x²+y²

μ=μ(x²+y²), se hace x²+y²=t

${\displaystyle \{\begin{array}{l}{\mu }_x=\frac{d\mu }{dt}\frac{\partial t}{\partial x}=\frac{d\mu }{dt}·2x=2\mu \text{'}·x\\ {\mu }_y=\frac{d\mu }{dt}\frac{\partial t}{\partial y}=\frac{d\mu }{dt}·2y=2\mu \text{'}·y\end{array}}$

Sustituyendo en (*)

${\displaystyle \begin{array}{l}2\mu \text{'}yP+\mu P_y=2\mu \text{'}xQ+\mu Q_x\Rightarrow 2\mu \text{'}(yP-xQ)=\mu (Q_x-P_y)\\ \frac{\mu \text{'}}{\mu }=\frac{(Q_x-P_y)}{2(yP-xQ)}\Rightarrow \frac{(Q_x-P_y)}{2(yP-xQ)}\end{array}}$

debe ser solo función de x²+y²

Ejemplo. (x³+xy²-y)dx+(y³+yx²+x)dy=0

${\displaystyle \begin{array}{l}P=x^3+x{y}^2-y\Rightarrow P_y=2xy-1\\ Q=y^3+x^{2}y+x\Rightarrow Q_x=2xy+1\\ \frac{\mu \text{'}}{\mu }=\frac{(Q_x-P_y)}{(2yP-2xQ)}=\frac{2}{2(yP-xQ)}=\frac{1}{(x^{3}y+x{y}^3-y^2)-(x{y}^3-x^{3}y+x^2)}=\\ \frac{1}{-(x^2+y^2)}=-\frac{1}{t}\Rightarrow \frac{d\mu }{\mu }=-\frac{1}{t}dt\Rightarrow \ln \mu =-\ln t\Rightarrow \mu =\frac{1}{t}=\frac{1}{x^2+y^2}\end{array}}$

Integrar las ecuaciones:

1. ${\displaystyle (x^3+x{y}^2)dx+(x^{2}y+y^3)dy=0}$

2. ${\displaystyle (2x^3+4y)dx+(4x+y+2)dy=0}$

3. ${\displaystyle (x^2+y^2+x)dx+2xydy=0}$

4. ${\displaystyle \left(\frac{\mathrm{sen}2x}{y}+x\right)dx+\left(y-\frac{{\mathrm{sen}}^{2}x}{y^2}\right)dy=0}$

5. ${\displaystyle \left(\mathrm{sen}y+y\mathrm{sen}x+\frac{1}{x}\right)dx+\left(x\cos y-\cos x+\frac{1}{y}\right)dy=0}$

6. ${\displaystyle \left(3x^2\mathrm{tg}y-\frac{2y^3}{x^3}\right)dx+\left(x^3{\sec }^{2}y+4y^3+\frac{3y^2}{x^2}\right)dy=0}$

7. ${\displaystyle (3x^2-2x-y)dx+(2y-x+3y^2)dy=0}$

8. ${\displaystyle \left(\frac{xy}{\sqrt{1+x^2}}+2xy-\frac{y}{x}\right)dx+\left(\sqrt{1+x^2}+x^2-\ln x\right)dy=0}$

9. ${\displaystyle (x+y^2)dx-2xydy=0}$

10. ${\displaystyle 2xydx+(y+y^2-x^2)dy=0}$

11. ${\displaystyle \frac{y}{x}dx+(y^3-\ln x)dy=0}$

12. ${\displaystyle (2x{y}^2-3y^3)dx+(7-3x{y}^2)dy=0}$

13. ${\displaystyle (3+y^3\mathrm{sen}x)dx+3y^2\cos xdy=0}$

14. ${\displaystyle \left(x^3{y}^2-e^{\frac{1}{x}}\right)dx+x^{4}ydy=0}$

15. ${\displaystyle (x{y}^2+x^2{y}^2+3)dx+(x^{2}y)dy=0}$

16. ${\displaystyle (x+\mathrm{sen}x+\mathrm{sen}y)dx+\cos ydy=0}$

17. ${\displaystyle (x\cos y-y\mathrm{sen}y)dy+(x\mathrm{sen}y+y\cos y)dx=0}$

18. ${\displaystyle (2x{y}^4{e}^y+2x{y}^3+y)dx+(x^2{y}^4{e}^y-x^2{y}^2-3x)dy=0}$

19. ${\displaystyle y^2{x}^{2}dx+(x^{3}y+y+3)dy=0}$

20. ${\displaystyle (3+y^3\mathrm{sen}x)dx+3y^2\cos xdy=0}$

21. ${\displaystyle y\text{'}=-\frac{(1+x\mathrm{tg}x)y^2}{2(xy+\cos x)}}$

22. ${\displaystyle xdx+ydy+x(xdy-ydx)=0,\mu =\mu (x^2+y^2)}$

23. ${\displaystyle (3y^2-x)dx+(2y^3-6xy)dy=0,\mu =\mu (x+y^2)}$

24. ${\displaystyle (xy-2y^2)dx-(x^2-3xy)dy=0,\mu =\mu (x·y^2)}$

25. ${\displaystyle ydx+x(x^{2}y-1)dy=0,\mu =\mu \left(\frac{y}{x^3}\right)}$

26. ${\displaystyle y(x^2+y^2+1)dx+x(1-x^2-y^2)dy=0,\mu =\mu (x·y)}$

27. ${\displaystyle (2y-3x{y}^2)dx-xdy=0,\mu =\mu \left(\frac{x}{y^2}\right)}$