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Ecuación diferencial (x^2*ln^2x)y''-2xlnxy'+(lnx+2)y=ln^4x

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                                      
                     2    2     d              d                    4   
(2 + log(x))*y(x) + x *log (x)*---(y(x)) - 2*x*--(y(x))*log(x) = log (x)
                                 2             dx                       
                               dx                                       
$$x^{2} \log{\left(x \right)}^{2} \frac{d^{2}}{d x^{2}} y{\left(x \right)} - 2 x \log{\left(x \right)} \frac{d}{d x} y{\left(x \right)} + \left(\log{\left(x \right)} + 2\right) y{\left(x \right)} = \log{\left(x \right)}^{4}$$
x^2*log(x)^2*y'' - 2*x*log(x)*y' + (log(x) + 2)*y = log(x)^4