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Ecuación diferencial 4*y^(3*y'')=y^4-1
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Solución
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2
d
3*---(y(x))
2
dx 4
4*(y(x)) = -1 + y (x)
$$4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)} = y^{4}{\left(x \right)} - 1$$
4*y^(3*y'') = y^4 - 1
Solución detallada
Tenemos la ecuación:
$$- y^{4}{\left(x \right)} + 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)} + 1 = 0$$
Esta ecuación diferencial tiene la forma:
Esta ecuación se resuelve con los pasos siguientes:
Pasemos la ecuación a la forma
En nuestro caso
$$\operatorname{f_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{f_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{1}{\frac{d^{2}}{d x^{2}} y{\left(x \right)}}$$
$$\operatorname{g_{1}}{\left(y{\left(x \right)} \right)} = y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}$$
$$\operatorname{g_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = 1$$
es decir
$$1 = y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}$$
Multipliquemos las dos partes de la ecuación por dx
$$dx = dx \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)$$
Como
entonces
entonces
$$dx = \frac{dy \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)}{\left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)}}$$
o
$$dx \left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)} = dy \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)$$
$$dx \frac{d}{d x} y{\left(x \right)} = dy \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)$$
$$\int \frac{d}{d x} y{\left(x \right)}\, dx = \int \left(y^{4} + \left(-4\right) y^{3 \frac{d^{2}}{d x^{2}} y}\right)\, dy$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
$$\int \frac{d}{d x} y{\left(x \right)}\, dx = y{\left(x \right)}$$
от правой части интеграл по y
$$\int \left(y^{4} + \left(-4\right) y^{3 \frac{d^{2}}{d x^{2}} y}\right)\, dy = \frac{y^{5}{\left(x \right)}}{5} - 4 y{\left(x \right)}$$
es decir
$$y{\left(x \right)} = C_{1} + \frac{y^{5}{\left(x \right)}}{5} - 4 y{\left(x \right)}$$
Resolvermos esta ecuación:
Hallemos y'
$$- y^{4}{\left(x \right)} + 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)} + 1 = 0$$
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
$$\operatorname{f_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{f_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{1}{\frac{d^{2}}{d x^{2}} y{\left(x \right)}}$$
$$\operatorname{g_{1}}{\left(y{\left(x \right)} \right)} = y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}$$
$$\operatorname{g_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = 1$$
es decir
$$1 = y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}$$
Multipliquemos las dos partes de la ecuación por dx
$$dx = dx \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)$$
Como
y'=dy/dx
entonces
dx=dy/y'
entonces
$$dx = \frac{dy \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)}{\left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)}}$$
o
$$dx \left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)} = dy \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)$$
$$dx \frac{d}{d x} y{\left(x \right)} = dy \left(y^{4}{\left(x \right)} - 4 y^{3 \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(x \right)}\right)$$
$$\int \frac{d}{d x} y{\left(x \right)}\, dx = \int \left(y^{4} + \left(-4\right) y^{3 \frac{d^{2}}{d x^{2}} y}\right)\, dy$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
$$\int \frac{d}{d x} y{\left(x \right)}\, dx = y{\left(x \right)}$$
от правой части интеграл по y
$$\int \left(y^{4} + \left(-4\right) y^{3 \frac{d^{2}}{d x^{2}} y}\right)\, dy = \frac{y^{5}{\left(x \right)}}{5} - 4 y{\left(x \right)}$$
es decir
$$y{\left(x \right)} = C_{1} + \frac{y^{5}{\left(x \right)}}{5} - 4 y{\left(x \right)}$$
Resolvermos esta ecuación:
Hallemos y'