Sr Examen
Ecuación diferencial y'*y''-y^3=cos(x)y
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Solución
Ha introducido
[src]
2
3 d d
- y (x) + --(y(x))*---(y(x)) = cos(x)*y(x)
dx 2
dx
$$- y^{3}{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = y{\left(x \right)} \cos{\left(x \right)}$$
-y^3 + y'*y'' = y*cos(x)
Solución detallada
Tenemos la ecuación:
$$- y^{3}{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = y{\left(x \right)} \cos{\left(x \right)}$$
Esta ecuación diferencial tiene la forma:
donde
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(y' \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
$$\operatorname{g_{2}}{\left(y' \right)} = \frac{1}{\frac{d}{d x} y{\left(x \right)}}$$
Pasemos la ecuación a la forma:
Dividamos ambos miembros de la ecuación en g2(y')
$$\frac{1}{\frac{d}{d x} y{\left(x \right)}}$$
obtendremos
$$\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
Con esto hemos separado las variables x y y'.
Ahora multipliquemos las dos partes de la ecuación por dx,
entonces la ecuación será así
$$dx \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = dx \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
o
$$dy' \frac{d}{d x} y{\left(x \right)} = dx \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
Tomemos la integral de las dos partes de la ecuación:
- de la parte izquierda la integral por y',
- de la parte derecha la integral por x.
$$\int y'\, dy' = \int \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}\, dx$$
Tomemos estas integrales
$$\frac{y'^{2}}{2} = Const + \int \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}\, dx$$
Hemos recibido una ecuación ordinaria con la incógnica y'.
(Const - es una constante)
La solución:
$$\operatorname{y'1} = \operatorname{y'}{\left(x \right)} = - \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}$$
$$\operatorname{y'2} = \operatorname{y'}{\left(x \right)} = \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}$$
tomemos estas integrales
$$\operatorname{y_{1}} = \int \frac{d}{d x} y{\left(x \right)}\, dx = \int \left(- \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\right)\, dx$$ =
$$\operatorname{y_{1}} = y{\left(x \right)} = C_{3} - \sqrt{2} \int \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\, dx$$
$$\operatorname{y_{2}} = \int \frac{d}{d x} y{\left(x \right)}\, dx = \int \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\, dx$$ =
$$\operatorname{y_{2}} = y{\left(x \right)} = C_{4} + \sqrt{2} \int \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\, dx$$
$$- y^{3}{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = y{\left(x \right)} \cos{\left(x \right)}$$
Esta ecuación diferencial tiene la forma:
f1(x)*g1(y')*y'' = f2(x)*g2(y'),
donde
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(y' \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
$$\operatorname{g_{2}}{\left(y' \right)} = \frac{1}{\frac{d}{d x} y{\left(x \right)}}$$
Pasemos la ecuación a la forma:
g1(y')/g2(y')*y''= f2(x)/f1(x).
Dividamos ambos miembros de la ecuación en g2(y')
$$\frac{1}{\frac{d}{d x} y{\left(x \right)}}$$
obtendremos
$$\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
Con esto hemos separado las variables x y y'.
Ahora multipliquemos las dos partes de la ecuación por dx,
entonces la ecuación será así
$$dx \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = dx \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
o
$$dy' \frac{d}{d x} y{\left(x \right)} = dx \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}$$
Tomemos la integral de las dos partes de la ecuación:
- de la parte izquierda la integral por y',
- de la parte derecha la integral por x.
$$\int y'\, dy' = \int \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}\, dx$$
Tomemos estas integrales
$$\frac{y'^{2}}{2} = Const + \int \left(y^{2}{\left(x \right)} + \cos{\left(x \right)}\right) y{\left(x \right)}\, dx$$
Hemos recibido una ecuación ordinaria con la incógnica y'.
(Const - es una constante)
La solución:
$$\operatorname{y'1} = \operatorname{y'}{\left(x \right)} = - \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}$$
$$\operatorname{y'2} = \operatorname{y'}{\left(x \right)} = \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}$$
tomemos estas integrales
$$\operatorname{y_{1}} = \int \frac{d}{d x} y{\left(x \right)}\, dx = \int \left(- \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\right)\, dx$$ =
$$\operatorname{y_{1}} = y{\left(x \right)} = C_{3} - \sqrt{2} \int \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\, dx$$
$$\operatorname{y_{2}} = \int \frac{d}{d x} y{\left(x \right)}\, dx = \int \sqrt{2} \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\, dx$$ =
$$\operatorname{y_{2}} = y{\left(x \right)} = C_{4} + \sqrt{2} \int \sqrt{C_{1} + \int y{\left(x \right)} \cos{\left(x \right)}\, dx + \int y^{3}{\left(x \right)}\, dx}\, dx$$