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Ecuación diferencial (y'')^2=y'
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Solución
Ha introducido
[src]
2 / 2 \ | d | d |---(y(x))| = --(y(x)) | 2 | dx \dx /
$$\left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} = \frac{d}{d x} y{\left(x \right)}$$
y''^2 = y'
Solución detallada
Tenemos la ecuación:
$$- \frac{d}{d x} y{\left(x \right)} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} = 0$$
Esta ecuación diferencial tiene la forma:
Esta ecuación se resuelve con los pasos siguientes:
Pasemos la ecuación a la forma
En nuestro caso
$$\operatorname{f_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{f_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{d^{2}}{d x^{2}} y{\left(x \right)}$$
$$\operatorname{g_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{g_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{d}{d x} y{\left(x \right)}$$
es decir
$$\frac{\left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = 1$$
Multipliquemos las dos partes de la ecuación por dx
$$\frac{dx \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = dx$$
Como
entonces
entonces
$$dx \frac{\left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = \frac{dy}{\left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)}}$$
o
$$dx \left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)} \frac{\left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = dy$$
$$dx \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} = dy$$
$$\int \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}\, dx = \int 1\, dy$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
от правой части интеграл по y
$$\int 1\, dy = y{\left(x \right)}$$
es decir
$$\int \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}\, dx = C_{1} + y{\left(x \right)}$$
Resolvermos esta ecuación:
Hallemos y'
$$- \frac{d}{d x} y{\left(x \right)} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} = 0$$
Esta ecuación diferencial tiene la forma:
f1(y)*f2(y')*y'' = g1(y)*g2(y')
Esta ecuación se resuelve con los pasos siguientes:
Pasemos la ecuación a la forma
f2(y')/g2(y')*y'' = g1(y)/f1(y)
En nuestro caso
$$\operatorname{f_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{f_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{d^{2}}{d x^{2}} y{\left(x \right)}$$
$$\operatorname{g_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{g_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{d}{d x} y{\left(x \right)}$$
es decir
$$\frac{\left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = 1$$
Multipliquemos las dos partes de la ecuación por dx
$$\frac{dx \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = dx$$
Como
y'=dy/dx
entonces
dx=dy/y'
entonces
$$dx \frac{\left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = \frac{dy}{\left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)}}$$
o
$$dx \left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)} \frac{\left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}{\frac{d}{d x} y{\left(x \right)}} = dy$$
$$dx \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} = dy$$
$$\int \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}\, dx = \int 1\, dy$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
True
от правой части интеграл по y
$$\int 1\, dy = y{\left(x \right)}$$
es decir
$$\int \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}\, dx = C_{1} + y{\left(x \right)}$$
Resolvermos esta ecuación:
Hallemos y'
Respuesta
[src]
3 2 2
x C2*x x*C2
y(x) = C1 + -- - ----- + -----
12 4 4
$$y{\left(x \right)} = C_{1} + \frac{C_{2}^{2} x}{4} - \frac{C_{2} x^{2}}{4} + \frac{x^{3}}{12}$$
Clasificación
nth order reducible