Sr Examen
Ecuación diferencial yy'y"=(y')^3+(y")^2
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Solución
Ha introducido
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2
2 3 / 2 \
d d /d \ | d |
--(y(x))*---(y(x))*y(x) = |--(y(x))| + |---(y(x))|
dx 2 \dx / | 2 |
dx \dx /
$$y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} = \left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}$$
y*y'*y'' = y'^3 + y''^2
Solución detallada
Tenemos la ecuación:
$$y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} - \left(\frac{d}{d x} y{\left(x \right)}\right)^{3} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} = 0$$
Esta ecuación diferencial tiene la forma:
Esta ecuación se resuelve con los pasos siguientes:
Pasemos la ecuación a la forma
En nuestro caso
$$\operatorname{f_{1}}{\left(y{\left(x \right)} \right)} = y{\left(x \right)}$$
$$\operatorname{f_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{d}{d x} y{\left(x \right)}$$
$$\operatorname{g_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{g_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}$$
es decir
$$\frac{\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{1}{y{\left(x \right)}}$$
Multipliquemos las dos partes de la ecuación por dx
$$\frac{dx \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dx}{y{\left(x \right)}}$$
Como
entonces
entonces
$$dx \frac{\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dy \frac{1}{y{\left(x \right)}}}{\left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)}}$$
o
$$dx \left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)} \frac{\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dy}{y{\left(x \right)}}$$
$$dx \frac{\left(\frac{d}{d x} y{\left(x \right)}\right)^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dy}{y{\left(x \right)}}$$
$$\int \frac{\left(\frac{d}{d x} y{\left(x \right)}\right)^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}\, dx = \int \frac{1}{y}\, dy$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
$$\int \frac{\left(\frac{d}{d x} y{\left(x \right)}\right)^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}\, dx = \frac{\log{\left(\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} \right)}}{3}$$
от правой части интеграл по y
$$\int \frac{1}{y}\, dy = \log{\left(y{\left(x \right)} \right)}$$
es decir
$$\frac{\log{\left(\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} \right)}}{3} = C_{1} + \log{\left(y{\left(x \right)} \right)}$$
Resolvermos esta ecuación:
Hallemos y'
1) Recibimos la ecuación diferencial de 1 orden
$$\frac{d}{d x} y{\left(x \right)} = \frac{\left(-1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}{2}$$
Volvamos a aplicar
Obtenemos
$$dx = \frac{dy}{\frac{1}{2} \left(-1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
от правой части интеграл по y
$$x = \int \frac{2}{\left(-1 + \sqrt{3} i\right) \sqrt[3]{y^{3} e^{3 C_{1}} - \left(\frac{d^{2}}{d x^{2}} y\right)^{2}}}\, dy$$
$$x = C_{2} + \frac{2 \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}}}}\, dy}{-1 + \sqrt{3} i}$$
2) Recibimos la ecuación diferencial de 1 orden
$$\frac{d}{d x} y{\left(x \right)} = - \frac{\left(1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}{2}$$
Volvamos a aplicar
Obtenemos
$$dx = \frac{dy}{\left(-1\right) \frac{1}{2} \left(1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
от правой части интеграл по y
$$x = \int \left(- \frac{2}{\left(1 + \sqrt{3} i\right) \sqrt[3]{y^{3} e^{3 C_{1}} - \left(\frac{d^{2}}{d x^{2}} y\right)^{2}}}\right)\, dy$$
$$x = C_{2} - \frac{2 \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}}}}\, dy}{1 + \sqrt{3} i}$$
3) Recibimos la ecuación diferencial de 1 orden
$$\frac{d}{d x} y{\left(x \right)} = \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}$$
Volvamos a aplicar
Obtenemos
$$dx = \frac{dy}{\sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
от правой части интеграл по y
$$x = \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}} - \left(\frac{d^{2}}{d x^{2}} y\right)^{2}}}\, dy$$
$$x = C_{2} + \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}}}}\, dy$$
$$y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)} - \left(\frac{d}{d x} y{\left(x \right)}\right)^{3} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} = 0$$
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
$$\operatorname{f_{1}}{\left(y{\left(x \right)} \right)} = y{\left(x \right)}$$
$$\operatorname{f_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \frac{d}{d x} y{\left(x \right)}$$
$$\operatorname{g_{1}}{\left(y{\left(x \right)} \right)} = 1$$
$$\operatorname{g_{2}}{\left(\frac{d}{d x} y{\left(x \right)} \right)} = \left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}$$
es decir
$$\frac{\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{1}{y{\left(x \right)}}$$
Multipliquemos las dos partes de la ecuación por dx
$$\frac{dx \frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dx}{y{\left(x \right)}}$$
Como
y'=dy/dx
entonces
dx=dy/y'
entonces
$$dx \frac{\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dy \frac{1}{y{\left(x \right)}}}{\left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)}}$$
o
$$dx \left(\begin{cases} x & \text{for}\: 0 = 1 \\1 & \text{for}\: 1 = 1 \\0 & \text{otherwise} \end{cases}\right) \frac{d}{d x} y{\left(x \right)} \frac{\frac{d}{d x} y{\left(x \right)} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dy}{y{\left(x \right)}}$$
$$dx \frac{\left(\frac{d}{d x} y{\left(x \right)}\right)^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}} = \frac{dy}{y{\left(x \right)}}$$
$$\int \frac{\left(\frac{d}{d x} y{\left(x \right)}\right)^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}\, dx = \int \frac{1}{y}\, dy$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
$$\int \frac{\left(\frac{d}{d x} y{\left(x \right)}\right)^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)}}{\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}\, dx = \frac{\log{\left(\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} \right)}}{3}$$
от правой части интеграл по y
$$\int \frac{1}{y}\, dy = \log{\left(y{\left(x \right)} \right)}$$
es decir
$$\frac{\log{\left(\left(\frac{d}{d x} y{\left(x \right)}\right)^{3} + \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2} \right)}}{3} = C_{1} + \log{\left(y{\left(x \right)} \right)}$$
Resolvermos esta ecuación:
Hallemos y'
1) Recibimos la ecuación diferencial de 1 orden
$$\frac{d}{d x} y{\left(x \right)} = \frac{\left(-1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}{2}$$
Volvamos a aplicar
y'=dy/dx.
Obtenemos
$$dx = \frac{dy}{\frac{1}{2} \left(-1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
от правой части интеграл по y
$$x = \int \frac{2}{\left(-1 + \sqrt{3} i\right) \sqrt[3]{y^{3} e^{3 C_{1}} - \left(\frac{d^{2}}{d x^{2}} y\right)^{2}}}\, dy$$
$$x = C_{2} + \frac{2 \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}}}}\, dy}{-1 + \sqrt{3} i}$$
2) Recibimos la ecuación diferencial de 1 orden
$$\frac{d}{d x} y{\left(x \right)} = - \frac{\left(1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}{2}$$
Volvamos a aplicar
y'=dy/dx.
Obtenemos
$$dx = \frac{dy}{\left(-1\right) \frac{1}{2} \left(1 + \sqrt{3} i\right) \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
от правой части интеграл по y
$$x = \int \left(- \frac{2}{\left(1 + \sqrt{3} i\right) \sqrt[3]{y^{3} e^{3 C_{1}} - \left(\frac{d^{2}}{d x^{2}} y\right)^{2}}}\right)\, dy$$
$$x = C_{2} - \frac{2 \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}}}}\, dy}{1 + \sqrt{3} i}$$
3) Recibimos la ecuación diferencial de 1 orden
$$\frac{d}{d x} y{\left(x \right)} = \sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}$$
Volvamos a aplicar
y'=dy/dx.
Obtenemos
$$dx = \frac{dy}{\sqrt[3]{e^{3 \left(C_{1} + \log{\left(y{\left(x \right)} \right)}\right)} - \left(\frac{d^{2}}{d x^{2}} y{\left(x \right)}\right)^{2}}}$$
Tomemos la integral de las dos partes de la ecuación
от левой части интеграл по x
от правой части интеграл по y
$$x = \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}} - \left(\frac{d^{2}}{d x^{2}} y\right)^{2}}}\, dy$$
$$x = C_{2} + \int \frac{1}{\sqrt[3]{y^{3} e^{3 C_{1}}}}\, dy$$
Clasificación
factorable