Sr Examen

Ecuación diferencial yy'=1-2x/y

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]
d                   2*x 
--(y(x))*y(x) = 1 - ----
dx                  y(x)
$$y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = - \frac{2 x}{y{\left(x \right)}} + 1$$
y*y' = -2*x/y + 1
Respuesta [src]
                                                  /             /     3 \\      /              /     3 \      /     9 \\        
                                                  |           5*|-8 - --||      |            5*|-8 - --|   10*|20 + --||        
                  2 /     1 \    3 /     3 \    4 |      30     \     C1/|    5 |      390     \     C1/      \     C1/|        
                 x *|-2 - --|   x *|10 + --|   x *|-24 - -- + -----------|   x *|496 + --- - ----------- + ------------|        
            x       \     C1/      \     C1/      \      C1        C1    /      \       C1        C1            C1     /    / 6\
y(x) = C1 + -- + ------------ + ------------ + --------------------------- + ------------------------------------------- + O\x /
            C1          2              4                       5                                     7                          
                    2*C1           6*C1                   24*C1                                120*C1                           
$$y{\left(x \right)} = \frac{x^{5} \left(496 - \frac{5 \left(-8 - \frac{3}{C_{1}}\right)}{C_{1}} + \frac{10 \left(20 + \frac{9}{C_{1}}\right)}{C_{1}} + \frac{390}{C_{1}}\right)}{120 C_{1}^{7}} + \frac{x^{4} \left(-24 + \frac{5 \left(-8 - \frac{3}{C_{1}}\right)}{C_{1}} - \frac{30}{C_{1}}\right)}{24 C_{1}^{5}} + \frac{x^{3} \left(10 + \frac{3}{C_{1}}\right)}{6 C_{1}^{4}} + \frac{x^{2} \left(-2 - \frac{1}{C_{1}}\right)}{2 C_{1}^{2}} + \frac{x}{C_{1}} + C_{1} + O\left(x^{6}\right)$$
Clasificación
1st power series
lie group