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Ecuaciones diferenciales/ dy/dx=y-x/(y+x)

Ecuación diferencial dy/dx=y-x/(y+x)

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Para el problema de Cauchy:

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

Gráfico:

interior superior
⌨

Solución

Ha introducido [src] 
d               x           
--(y(x)) = - -------- + y(x)
dx           x + y(x)
$$\frac{d}{d x} y{\left(x \right)} = - \frac{x}{x + y{\left(x \right)}} + y{\left(x \right)}$$ 
y' = -x/(x + y) + y
Respuesta [src] 
                                                               /                                        1 \      /               /                           1              \       12    27        1     4      /     1 \             \        
                                                               |                                   1 + ---|      |               |                      1 + ---     /    6 \|   1 + -- + ---   1 - --- - ---   2*|1 + ---|     /    6 \|        
                                                               |                                         2|      |               |                            2   2*|1 + --||       C1     2         2     3     |      2|   4*|1 + --||        
                                   3 / 2       /     1 \\    4 |   8     6       /     1     4 \       C1 |    5 | 12    50      |     1     8     26       C1      \    C1/|            C1        C1    C1      \    C1 /     \    C1/|        
                    2 /     1 \   x *|--- + C1*|1 + ---||   x *|- --- - --- + C1*|1 - --- - ---| - -------|   x *|--- + --- + C1*|1 - --- + --- + --- + ------- + ----------| + ------------ - ------------- + ----------- + ----------|        
                   x *|C1 - --|      |  2      |      2||      |    3     2      |      2     3|      C1  |      |  3     4      |      2     3     4       2          2    |        C1              C1              2            2    |        
                      \     C1/      \C1       \    C1 //      \  C1    C1       \    C1    C1 /          /      \C1    C1       \    C1    C1    C1      C1         C1     /                                      C1           C1     /    / 6\
y(x) = C1 + C1*x + ------------ + ----------------------- + ----------------------------------------------- + -------------------------------------------------------------------------------------------------------------------------- + O\x /
                        2                    6                                     24                                                                                    120
$$y{\left(x \right)} = \frac{x^{2} \left(C_{1} - \frac{1}{C_{1}}\right)}{2} + \frac{x^{3} \left(C_{1} \left(1 + \frac{1}{C_{1}^{2}}\right) + \frac{2}{C_{1}^{2}}\right)}{6} + \frac{x^{4} \left(C_{1} \left(1 - \frac{1}{C_{1}^{2}} - \frac{4}{C_{1}^{3}}\right) - \frac{1 + \frac{1}{C_{1}^{2}}}{C_{1}} - \frac{6}{C_{1}^{2}} - \frac{8}{C_{1}^{3}}\right)}{24} + \frac{x^{5} \left(C_{1} \left(1 + \frac{1 + \frac{1}{C_{1}^{2}}}{C_{1}^{2}} + \frac{2 \left(1 + \frac{6}{C_{1}}\right)}{C_{1}^{2}} - \frac{1}{C_{1}^{2}} + \frac{8}{C_{1}^{3}} + \frac{26}{C_{1}^{4}}\right) - \frac{1 - \frac{1}{C_{1}^{2}} - \frac{4}{C_{1}^{3}}}{C_{1}} + \frac{1 + \frac{12}{C_{1}} + \frac{27}{C_{1}^{2}}}{C_{1}} + \frac{2 \left(1 + \frac{1}{C_{1}^{2}}\right)}{C_{1}^{2}} + \frac{4 \left(1 + \frac{6}{C_{1}}\right)}{C_{1}^{2}} + \frac{12}{C_{1}^{3}} + \frac{50}{C_{1}^{4}}\right)}{120} + C_{1} + C_{1} x + O\left(x^{6}\right)$$
Trazar el gráfico con C=-1
Trazar el gráfico con C=0
Trazar el gráfico con C=1
Gráfico para el problema de Cauchy
Clasificación
1st power series
lie group
Respuesta numérica [src] 
(x, y):
(-10.0, 0.75)
(-7.777777777777778, -1.5772094674352533)
(-5.555555555555555, -19.73491298457441)
(-3.333333333333333, -183.10267671510678)
(-1.1111111111111107, -1689.7079639572219)
(1.1111111111111107, -15592.313742841869)
(3.333333333333334, -143882.98184382103)
(5.555555555555557, -1327725.4982134027)
(7.777777777777779, -12252004.902847927)
(10.0, -113059231.21580042)
(10.0, -113059231.21580042)
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