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Ecuación diferencial x*y''=y'*(ln(y')-ln(x))

El profesor se sorprenderá mucho al ver tu solución correcta😉

v

Para el problema de Cauchy:

y() =
y'() =
y''() =
y'''() =
y''''() =

Gráfico:

interior superior

Solución

Ha introducido [src]
    2                                           
   d          /             /d       \\ d       
x*---(y(x)) = |-log(x) + log|--(y(x))||*--(y(x))
    2         \             \dx      // dx      
  dx                                            
$$x \frac{d^{2}}{d x^{2}} y{\left(x \right)} = \left(- \log{\left(x \right)} + \log{\left(\frac{d}{d x} y{\left(x \right)} \right)}\right) \frac{d}{d x} y{\left(x \right)}$$
x*y'' = (-log(x) + log(y'))*y'
Respuesta [src]
            //             1 + C1*x              \
            ||(-1 + C1*x)*e                2     |
            ||---------------------  for C1  != 0|
            ||           2                       |
            ||         C1                        |
y(x) = C1 + |<                                   |
            ||          2                        |
            ||         x                         |
            ||         --             otherwise  |
            ||         2                         |
            \\                                   /
$$y{\left(x \right)} = C_{1} + \begin{cases} \frac{\left(C_{1} x - 1\right) e^{C_{1} x + 1}}{C_{1}^{2}} & \text{for}\: C_{1}^{2} \neq 0 \\\frac{x^{2}}{2} & \text{otherwise} \end{cases}$$
Clasificación
nth order reducible