Sr Examen

Ecuación diferencial 2y'-y/(x-1)-1/y=0

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Para el problema de Cauchy:

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

Gráfico:

interior superior

Solución

Ha introducido [src]
   1       d           y(x)     
- ---- + 2*--(y(x)) - ------ = 0
  y(x)     dx         -1 + x    
$$2 \frac{d}{d x} y{\left(x \right)} - \frac{1}{y{\left(x \right)}} - \frac{y{\left(x \right)}}{x - 1} = 0$$
2*y' - 1/y - y/(x - 1) = 0
Respuesta [src]
          __________________________________________________________________
y(x) = -\/ -C1 - log(-1 + x) - 2*log(2) + C1*x + x*log(-1 + x) + 2*x*log(2) 
$$y{\left(x \right)} = - \sqrt{C_{1} x - C_{1} + x \log{\left(x - 1 \right)} + 2 x \log{\left(2 \right)} - \log{\left(x - 1 \right)} - 2 \log{\left(2 \right)}}$$
         __________________________________________________________________
y(x) = \/ -C1 - log(-1 + x) - 2*log(2) + C1*x + x*log(-1 + x) + 2*x*log(2) 
$$y{\left(x \right)} = \sqrt{C_{1} x - C_{1} + x \log{\left(x - 1 \right)} + 2 x \log{\left(2 \right)} - \log{\left(x - 1 \right)} - 2 \log{\left(2 \right)}}$$
Gráfico para el problema de Cauchy
Clasificación
Bernoulli
1st power series
lie group
Bernoulli Integral
Respuesta numérica [src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 1.5587694849527012)
(-5.555555555555555, 1.9308722443859792)
(-3.333333333333333, 2.0635752823502944)
(-1.1111111111111107, 1.8954483510104692)
(1.1111111111111107, 0.5417763871809098)
(3.333333333333334, 3.1933833808213433e-248)
(5.555555555555557, 1.5636038433718505e+185)
(7.777777777777779, 8.388243571828173e+296)
(10.0, 3.861029683e-315)
(10.0, 3.861029683e-315)