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Ecuación diferencial dy/dx=(xy+2y-x-2)/(xy-3y+x-3)

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]
d          -2 - x + 2*y(x) + x*y(x)
--(y(x)) = ------------------------
dx         -3 + x - 3*y(x) + x*y(x)
$$\frac{d}{d x} y{\left(x \right)} = \frac{x y{\left(x \right)} - x + 2 y{\left(x \right)} - 2}{x y{\left(x \right)} + x - 3 y{\left(x \right)} - 3}$$
y' = (x*y - x + 2*y - 2)/(x*y + x - 3*y - 3)
Respuesta [src]
              /    _________________________________________________________ \
              |   /    /        5        2       4       3        \  -1 + x  |
              |-\/  C1*\-243 + x  - 270*x  - 15*x  + 90*x  + 405*x/*e        |
y(x) = 1 + 2*W|--------------------------------------------------------------|
              \                              2                               /
$$y{\left(x \right)} = 2 W\left(- \frac{\sqrt{C_{1} \left(x^{5} - 15 x^{4} + 90 x^{3} - 270 x^{2} + 405 x - 243\right) e^{x - 1}}}{2}\right) + 1$$
              /   _________________________________________________________\
              |  /    /        5        2       4       3        \  -1 + x |
              |\/  C1*\-243 + x  - 270*x  - 15*x  + 90*x  + 405*x/*e       |
y(x) = 1 + 2*W|------------------------------------------------------------|
              \                             2                              /
$$y{\left(x \right)} = 2 W\left(\frac{\sqrt{C_{1} \left(x^{5} - 15 x^{4} + 90 x^{3} - 270 x^{2} + 405 x - 243\right) e^{x - 1}}}{2}\right) + 1$$
Clasificación
separable
1st exact
1st power series
lie group
separable Integral
1st exact Integral