Definición: Una función f(x, y) se dice que es homogénea de grado n respecto a sus variables si cumple:

${\displaystyle \forall \lambda ,f(\lambda x,\lambda y)={\lambda }^{n}f(x,y)}$

Ejemplos:

1. ${\displaystyle f(x,y)=3x^{2}y-y^2\sqrt{x^2+2y^2}}$, es una función homogénea de grado tres

${\displaystyle f(\lambda x,\lambda y)=3{\left(\lambda x\right)}^2(\lambda y)-{\left(\lambda y\right)}^2\sqrt{{\left(\lambda x\right)}^2+2{\left(\lambda y\right)}^2}={\lambda }^3\left(3x^{2}y-y^2\sqrt{x^2+2y^2}\right)}$

2. ${\displaystyle f(x,y)=3\text{\hspace{0.17em}sen}\left(\frac{y}{x}\right)}$, es homogénea de grado cero

${\displaystyle f(\lambda x,\lambda y)=3\text{\hspace{0.17em}sen}\left(\frac{\lambda x}{\lambda y}\right)={\lambda }^{0}3\text{\hspace{0.17em}sen}\left(\frac{y}{x}\right)}$

3. ${\displaystyle f(x,y)=x{e}^{\frac{2x}{x+y}}-5y}$, es homogénea de grado uno

Ecuaciones homogéneas de primer orden

Definición: Una ecuación diferencial de primer orden $\frac{dy}{dx}=f(x,y)$, se dice que es homogénea, cuando la función f(x, y) es homogénea de grado 'cero'. Observación. Si la ecuación viene dada de la forma:

M(x, y)dx+N(x, y)dy=0

Será homogénea cuando las funciones M(x, y) y N(x, y) sean homogéneas del mismo grado.

Resolución: Haciendo el cambio, y=u·x, y'=u'x+u, se reduce a una de variables separadas

Ejemplos:

1. 4x-3y+y'(2y-3x)=0

${\displaystyle \begin{array}{l}y\text{'}=\frac{3y-4x}{2y-3x}\{\begin{array}{l}y=ux\\ y\text{'}=u\text{'}x+u\end{array}\Rightarrow u\text{'}x+u=\frac{3ux-4x}{2ux-3x}=\frac{3u-4}{2u-3}\Rightarrow \\ u\text{'}x=\frac{3u-4}{2u-3}-u=\frac{-2u^2+6u-4}{2u-3}=\frac{-2(u^2-3u+2)}{2u-3}=\frac{-2(u-1)(u-2)}{2u-3}\\ \frac{2u-3}{(u-1)(u-2)}du=-2\frac{dx}{x}\\ \frac{2u-3}{(u-1)(u-2)}=\frac{A}{u-1}+\frac{B}{u-2}\Rightarrow \{\begin{array}{l}A=1\\ B=1\end{array}\\ \ln (u-1)+\ln (u-2)=-2\ln x+\ln C\Rightarrow \ln \left[(u-1)(u-2)\right]=\ln \frac{C}{x^2}\Rightarrow \\ \Rightarrow (u-1)(u-2)=\frac{C}{x^2}\end{array}}$

Deshaciendo el cambio (y-x)(y-2x)=C

2. ${\displaystyle 2x^{3}y\text{'}=y(y^2+3x^2)}$

${\displaystyle \begin{array}{l}u\text{'}x+u=\frac{ux(u^2{x}^2+3x^2)}{2x^3}=\frac{u^3+3u}{2}\Rightarrow u\text{'}x=\frac{u^3+3u-2u}{2}=\\ =\frac{u^3+u}{2}\Rightarrow 2\frac{du}{u^3+u}=\frac{dx}{x}\\ \frac{2}{u(u^2+1)}=\frac{A}{u}+\frac{Bu+C}{u^2+1}\Rightarrow \{\begin{array}{l}A(u^2+1)+(Bu+C)u=2\\ A=2,B=-2,C=0\end{array}\\ \ln x+\ln K=2\ln u-\ln (u^2+1)\Rightarrow \ln (xK)=\ln \frac{u^2}{u^2+1}\Rightarrow xK=\frac{u^2}{u^2+1}\end{array}}$

Deshaciendo el cambio

${\displaystyle xK=\frac{\frac{y^2}{x^2}}{\frac{y^2}{x^2}+1}\Rightarrow xK=\frac{y^2}{y^2+x^2}}$

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

${\displaystyle y\text{'}=\frac{xy-2y^2}{x^2-3xy}}$

Haciendo el cambio

${\displaystyle \begin{array}{l}\{\begin{array}{l}y=ux\\ y\text{'}=u\text{'}x+u\end{array}\Rightarrow u\text{'}x+u=\frac{u{x}^2-2u^2{x}^2}{x^2-3x^{2}u}=\frac{u-2u^2}{1-3u}\Rightarrow u\text{'}x=\frac{u-2u^2}{1-3u}-u=\\ \frac{u^2}{1-3u}\Rightarrow \frac{1-3u}{u^2}du=\frac{dx}{x}\Rightarrow {\displaystyle \int \left(\frac{1}{u^2}-\frac{3}{u}\right)}du={\displaystyle \int \frac{dx}{x}}\Rightarrow \\ -\frac{1}{u}-3\ln u=\ln x+\ln C\Rightarrow -\frac{1}{u}=\ln \left(u^{3}xC\right)\end{array}}$

Deshaciendo el cambio

${\displaystyle -\frac{x}{y}=\ln \left(\frac{y^3}{x^3}xC\right)\Rightarrow e^{-\frac{x}{y}}=\frac{C{y}^3}{x^2}}$

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

${\displaystyle \begin{array}{l}y\text{'}=\frac{y^2-3x{}^2-2xy}{x^2-2xy-3y^2}y=ux\Rightarrow u\text{'}x+u=\frac{u^2-3-2u}{1-2u-3u^2}\\ u\text{'}x=\frac{u^2-3-2u-u+2u^2+3u^3}{1-2u-3u^2}=\frac{3u^3+3u^2-3u-3}{1-2u-3u^2}\\ \Rightarrow \frac{dx}{x}=\frac{1-2u-3u^2}{3u^3+3u^2-3u-3}du\\ \frac{1}{3}{\displaystyle \int \frac{1-2u-3u^2}{{\left(u+1\right)}^2(u-1)}du\Rightarrow }\frac{1-2u-3u^2}{{\left(u+1\right)}^2(u-1)}=\frac{-3u+1}{(u+1)(u-1)}\\ =\frac{A}{u+1}+\frac{B}{u-1}\{\begin{array}{l}A=-2\\ B=-1\end{array}\\ \frac{1}{3}\left[-2\ln (u+1)-\ln (u-1)\right]=\ln x+\ln K\Rightarrow \frac{1}{{\left(u+1\right)}^2(u-1)}={\left(xk\right)}^3\Rightarrow \\ \Rightarrow \frac{x^3}{{(y+x)}^2(y-x)}=k^3{x}^3\Rightarrow {\left(y+x\right)}^2(y-x)=\frac{1}{k^3}=\text{cte}\end{array}}$

5. ${\displaystyle y\text{'}=\frac{2xy}{3x^2-y^2}}$

${\displaystyle \begin{array}{l}u\text{'}x+u=\frac{2u}{3-u^2}\Rightarrow u\text{'}x=\frac{2u}{3-u^2}-u=\frac{2u-3u+u^3}{3-u^2}=\frac{u^3-u}{3-u^2}\\ \frac{3-u^2}{u(u^2-1)}du=\frac{dx}{x}.\\ \frac{3-u^2}{u(u^2-1)}=\frac{A}{u}+\frac{B}{u-1}+\frac{C}{u+1}\Rightarrow \{\begin{array}{l}A=-3\\ B=1\\ C=1\end{array}\\ -3\ln u+\ln (u-1)+\ln (u+1)=\ln x+\ln K\Rightarrow \frac{u^2-1}{u^3}=Kx\\ \frac{x(y^2-x^2)}{y^3}=Kx\Rightarrow y^2-x^2=K{y}^3\end{array}}$

6. ${\displaystyle y\text{'}=\frac{4x^2+xy-3y^2}{5x^2-2xy-y^2}}$

${\displaystyle \begin{array}{l}u\text{'}x+u=\frac{4+u-3u^2}{5-2u-u^2}\Rightarrow u\text{'}x=\frac{4+u-3u^2}{5-2u-u^2}-u=\frac{u^3-u^2-4u+4}{5-2u-u^2}\\ \frac{5-2u-u^2}{u^3-u^2-4u+4}du=\frac{dx}{x}\\ \frac{5-2u-u^2}{(u-1)(u-2)(u+2)}=\frac{A}{u-1}+\frac{B}{u-2}+\frac{C}{u+2}\Rightarrow \{\begin{array}{l}A=-\frac{2}{3}\\ B=-\frac{3}{4}\\ C=\frac{5}{12}\end{array}\\ \ln Kx=-\frac{2}{3}\ln (u-1)-\frac{3}{4}\ln (u-2)+\frac{5}{12}\ln (u+2)=\\ =\frac{1}{12}\ln \frac{{\left(u+2\right)}^5}{{\left(u-1\right)}^8{\left(u-2\right)}^9}\Rightarrow x^{12}\text{cte}=\frac{{\left(y+2x\right)}^5{x}^{12}}{{\left(y-x\right)}^8{\left(y-2x\right)}^9}\Rightarrow \\ \Rightarrow {\left(y-x\right)}^8{\left(y-2x\right)}^9\text{cte}={\left(y+2x\right)}^5\end{array}}$

Ecuaciones reducibles a homogéneas

Ecuaciones de la forma:

${\displaystyle \frac{dy}{dx}=f\left(\frac{ax+by+c}{mx+ny+l}\right)}$

Evidentemente si c=l=0, es una ecuación homogénea, caso contrario se estudia el sistema:

${\displaystyle \{\begin{array}{l}ax+by+c=0\\ mx+ny+l=0\end{array}}$

Este sistema representa dos rectas en el plano y se tiene los siguientes casos:

1. Sistema compatible determinado, las rectas se cortan en un punto (x₀, y₀). Realizando el cambio de variables:

${\displaystyle \begin{array}{l}x=X+x_0,y=Y+y_0\Rightarrow \{\begin{array}{l}dx=dX\\ dy=dY\end{array}\Rightarrow y\text{'}=Y\text{'}\\ ax+by+c=a(X+x_0)+b(Y+y_0)+c=aX+bY+(a{x}_0+b{y}_0+c)=aX+bY+0\\ mx+ny+l=m(X+x_0)+n(Y+y_0)+l=mX+nY+(m{x}_0+n{y}_0+l)=mX+nY+0\end{array}}$

La ecuación se convierte en la homogénea:

${\displaystyle \frac{dY}{dX}=f\left(\frac{aX+bY}{mX+nY}\right)}$

Ejemplo: (3y-7x+7)dx-(3x-7y-3)dy=0

${\displaystyle y\text{'}=\frac{-7x+3y+7}{3x-7y-3}\Rightarrow \{\begin{array}{l}-7x+3y+7=0\\ 3x-7y-3=0\end{array}}$

Como

${\displaystyle \left|\begin{array}{cc}-7& 3\\ 3& -7\end{array}\right|\ne 0}$

solución única

${\displaystyle \begin{array}{l}{x}_0=\frac{\left|\begin{array}{cc}-7& 3\\ 3& -7\end{array}\right|}{\left|\begin{array}{cc}-7& 3\\ 3& -7\end{array}\right|}=1,y_0=0\Rightarrow \{\begin{array}{l}x=X+1\\ y=Y\end{array}\\ Y\text{'}=\frac{-7X+3Y}{3X-7Y}\text{\hspace{0.17em}homogénea\hspace{0.17em}}\{\begin{array}{l}Y=uX\\ Y\text{'}=u\text{'}X+u\end{array}\Rightarrow u\text{'}X+u=\frac{-7+3u}{3-7u}\\ u\text{'}X=\frac{7u^2-7}{3-7u}\Rightarrow \frac{3-7u}{u^2-1}du=7\frac{dX}{X}\\ \frac{3-7u}{u^2-1}=\frac{A}{u-1}+\frac{B}{u+1}\Rightarrow \{\begin{array}{l}B=-5\\ A=-2\end{array}\\ \ln K-2\ln (u-1)-5\ln (u+1)=7\ln X\Rightarrow \frac{K}{{\left(u-1\right)}^2{\left(u+1\right)}^5}=X^7\\ \Rightarrow K={\left(Y-X\right)}^2{\left(Y+X\right)}^5={\left(y-x+1\right)}^2{\left(y+x-1\right)}^5\end{array}}$

2. El sistema es compatible indeterminado (rectas coincidentes)

${\displaystyle \frac{a}{m}=\frac{b}{n}=\frac{c}{l}}$

Se realiza la división y la ecuación es de variables separadas

Ejemplo: (x-2y+5)dx+(-3x+6y-15)dy=0

${\displaystyle y\text{'}=-\frac{x-2y+5}{-3x+6y-15}=\frac{1}{3}\Rightarrow 3dy=dx\Rightarrow 3y=x+K}$

3. El sistema es incompatible (rectas paralelas)

${\displaystyle \frac{a}{m}=\frac{b}{n}\ne \frac{c}{l}}$

Al ser

${\displaystyle \frac{a}{m}=\frac{b}{n}=h\Rightarrow \{\begin{array}{l}a=m.h\\ b=n.h\end{array}\Rightarrow y\text{'}=f\left(\frac{mhx+nhy+c}{mx+ny+l}\right)=f\left(\frac{h(mx+ny)+c}{mx+ny+l}\right)}$

se hace el cambio mx+ny=z y pasa a variables separadas

Ejemplo: (x-2y-1)dx+(3x-6y+2)dy=0

${\displaystyle \begin{array}{l}y\text{'}=-\frac{x-2y-1}{3x-6y+2}\text{\hspace{0.17em}haciendo\hspace{0.17em}}\{\begin{array}{l}x-2y=z\\ 1-2y\text{'}=z\text{'}\end{array}\Rightarrow \frac{1-z\text{'}}{2}=\frac{-z+1}{3z+2}\\ -z\text{'}=\frac{2(-z+1)}{3z+2}-1=\frac{-5z}{3z+2}\Rightarrow z\text{'}=\frac{dz}{dx}=\frac{5z}{3z+2}\Rightarrow \frac{3z+2}{5z}dz=dx\\ \left(\frac{3}{5}+\frac{2}{5z}\right)dz=dx\Rightarrow \frac{3}{5}z+\frac{2}{5}\ln z+\ln K=x.\\ 2\ln z+\ln C=5x-3z\Rightarrow 2\ln (x-2y)+\ln C=2x+6y=2(x+3y)\\ \ln (x-2y)+\ln A=x+3y\Rightarrow A(x-2y)=e^{x+3y}\end{array}}$

Integrar las ecuaciones:

1. ${\displaystyle ydx-(x+y)dy=0}$

2. ${\displaystyle xdy-ydx=\sqrt{x^2+y^2}dx}$

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

4. ${\displaystyle xy\text{'}=\sqrt{x^2-y^2}+y}$

5. ${\displaystyle 4x^2-xy+y^2+y\text{'}(x^2-xy+4y^2)=0}$

6. ${\displaystyle y\text{'}=\frac{y^3+2x^{2}y}{2(x^3+x{y}^2)}.}$

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

8. ${\displaystyle ydx+\left(2\sqrt{xy}-x\right)dy=0}$

9. ${\displaystyle y\text{'}=2{\left(\frac{y+2}{x+y-1}\right)}^2}$

10. ${\displaystyle \left(x{e}^{\frac{y}{x}}+y\right)dx-xdy=0}$

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

12. ${\displaystyle y\text{'}=\frac{x+y-1}{3x-y+5}}$

13. ${\displaystyle (x+y+1)dx-(x+y-1)dy=0}$

14. ${\displaystyle y\text{'}=\frac{2x+3y-1}{5-4x-6y}}$

15. ${\displaystyle 3x+y-2+y\text{'}(x-1)=0}$

16. ${\displaystyle (5y+7x)dy+(8y+10x)dx=0}$