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Ecuaciones Diferenciales Exactas, por WikiMatematica.org
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Ecuaciones Diferenciales Exactas

De por WikiMatematica.org


http://youtu.be/UKttZ1xlQGE

Diferencial Total:

 z = f(x,y)

 dz = \frac {\partial f}{\partial x} dx + \frac {\partial f}{\partial y}dy

Cuando  f(x,y) = k Siendo  k una constante.

 dz = \frac {\partial f}{\partial x} dx + \frac {\partial f}{\partial y} dy = 0

Para ecuaciones que no se formaron por un diferencial total, no todos se pueden resolver.

Una Ecuación Diferencial Ordinaria es exacta si mediante operaciones algebraicas tiene la forma:  M(x,y)dx + N(x,y)dy = 0


Para verificar que la función es exacta debe cumplir con:

 \frac {\partial M} {\partial y} = \frac {\partial N} {\partial x}

Lo cual se conoce como "Propiedad de Clairaout".

Contenido

Ejemplo # 1

  •  2xy dx + (x^{2} - 1) dy = 0

Sea  2xy = M(x,y) y 2x= N(x,y) Podemos ver que es exacta.

 \frac {\partial M} {\partial y} = 2x

 \frac {\partial N} {\partial x} = 2x

Solución 1

 f(x,y) =?

 \frac {\partial f} {\partial x} = M(x,y)

 \frac {\partial f}{\partial x} = 2xy   \left \| \int ... dx

 f(x,y) = \int 2xy dx

 f(x,y) = x^{2} y + h(y) \left \|   \frac{\partial }{\partial y}

 \frac {\partial f} {\partial y} = x^{2} + h'(y) = x^{2} - 1

 h'(y) = -1 \left \|  \int ... dy

 h(y) = - \int dy

 h(y) = -y + c

 f(x,y) = x^{2}y - y + c = k

 x^{2}y - y = k

Solución 2
 f(x,y) =?

 \frac {\partial f} {\partial y} = N(x,y)

 \frac {\partial f} {\partial y} = x^{2} - 1 \left \| \int ... dy

 f(x,y) = x^{2}y - y + h(x) \left \| \frac {\partial} {\partial x}

 \frac {\partial f} {\partial x} = 2xy + h'(x) = 2xy

 h'(x) = 0 \left \| \int... dx

 h(x) = c

 f(x,y) = x^{2}y - y + c = k

x^{2}y - y = k

Ejemplo # 2

Resolver

  •  (e^{2y} - y \cos xy) dx + (2xe^{2y} - x\cos xy + 2y) dy = 0


Es una EDO Exacta, ya que:

 \frac {\partial M} {\partial y} = 2e^{2y} - \cos xy + xy \sin xy

 \frac {\partial N} {\partial x} = 2e^{2y} - \cos xy + xy \sin xy


Solución
 \frac {\partial f} {\partial x} = 2e^{2y} - \cos xy \left \| \int... dx

 f(x,y) = \int [e^{2y} - y \cos xy ] dx

 f(x,y)= xe^{2y} - \sin (xy) + h(y) \left \| \frac {\delta} {\delta y}

 \frac {\partial f}{\partial y} = 2xe^{2y} - x \cos xy + h'(y)= 2xe^{2y} - x\cos xy + 2y

 h'(y) = 2y \left \| \int ... dy

 h(y) = \int 2y dy

 h(y) = y^{2} + c

 f(x,y) = xe^{2y} - \sin xy + y^{2} + c = k

 xe^{2y} - \sin xy + y^{2} = k

Ejemplo # 3

  • Resolver :
            (2x-5y)dx + (-5x + 3y^{2})dy=0
            M(x,y)= 2x -5y 
            N(x,y)= -5x + 3y^{2}


probando:

\frac {\partial M}{\partial y} = -5

\frac {\partial N}{\partial x} = -5


solución a:

 \frac {\partial f}{\partial x} = M(x,y) = 2x -5y  \setminus \setminus\int... dx

f(x,y)=\int(2x - 5y) dx

f(x,y) = x^{2} - 5xy + h(y)  \setminus \setminus \frac{\partial f}{\partial y}

\frac {\partial f }{\partial y} = -5x + h'(y) = N(x,y) = -5x + 3y^{2}

h'(y) = 3y^{2} \setminus \setminus \int... dy

h(y)=y^{3} + c

f(x,y)= x^{2} -5xy +y^{3}+c1=c2

f(x,y)= x^{2} - 5xy + y^{3}=c

solución b:

 \frac {\partial f}{\partial x} = N(x,y) = -5x + 3y^{2}  \setminus \setminus\int... dy

f(x,y)=\int(-5x + 3y^{2}) dy

f(x,y) = - 5xy + y^{3}+  h(x)  \setminus \setminus \frac{\partial f}{\partial x}

\frac {\partial f }{\partial x} = -5y + h'(x) = M(x,y) = 2x - 5y

h'(x) = 2y \setminus \setminus \int... dx

h(x) =x^{2} + c

f(x,y)= x^{2} - 5xy + y^{3}=c

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