Sr Examen

Ecuación diferencial yy'=x√y²+1

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                         
--(y(x))*y(x) = 1 + x*y(x)
dx                        
$$y{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = x y{\left(x \right)} + 1$$
y*y' = x*y + 1
Respuesta [src]
                                                                  /            /      15\\        
                                                                  |          7*|-4 + ---||        
                                                                  |            |       3||        
                  2 /     1 \    3 /      3 \    4 /     15\    5 |     35     \     C1 /|        
                 x *|1 - ---|   x *|-1 + ---|   x *|7 - ---|   x *|6 - --- + ------------|        
                    |      3|      |       3|      |      3|      |      3         3     |        
            x       \    C1 /      \     C1 /      \    C1 /      \    C1        C1      /    / 6\
y(x) = C1 + -- + ------------ + ------------- + ------------ + --------------------------- + O\x /
            C1        2                 2               4                      3                  
                                    6*C1           24*C1                 120*C1                   
$$y{\left(x \right)} = \frac{x^{4} \left(7 - \frac{15}{C_{1}^{3}}\right)}{24 C_{1}^{4}} + \frac{x^{5} \left(6 + \frac{7 \left(-4 + \frac{15}{C_{1}^{3}}\right)}{C_{1}^{3}} - \frac{35}{C_{1}^{3}}\right)}{120 C_{1}^{3}} + \frac{x^{3} \left(-1 + \frac{3}{C_{1}^{3}}\right)}{6 C_{1}^{2}} + \frac{x}{C_{1}} + \frac{x^{2} \left(1 - \frac{1}{C_{1}^{3}}\right)}{2} + C_{1} + O\left(x^{6}\right)$$
Clasificación
1st power series
lie group