Sr Examen
Ecuación diferencial dy/dx+y/(x-2)=5(x-2)y^(1/2)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
y(x) d ______ ------ + --(y(x)) = \/ y(x) *(-10 + 5*x) -2 + x dx
$$\frac{d}{d x} y{\left(x \right)} + \frac{y{\left(x \right)}}{x - 2} = \left(5 x - 10\right) \sqrt{y{\left(x \right)}}$$
y' + y/(x - 2) = (5*x - 10)*sqrt(y)
Respuesta
[src]
/ 2 5 2 4 3 2
| 32 C1 x 80*x 10*x 8*C1 40*x 80*x 8*C1*x 2*C1*x
| - ------ + ------ + ------ - ------ - ------ + ---------- + ------ + ------ - ---------- + ---------- for |-2 + x| < 1
| -2 + x -2 + x -2 + x -2 + x -2 + x ________ -2 + x -2 + x ________ ________
| \/ -2 + x \/ -2 + x \/ -2 + x
|
y(x) = < 2 2
| __1, 1 / 1 7/2 | \ __0, 2 /7/2, 1 | \ __1, 1 / 1 7/2 | \ __0, 2 /7/2, 1 | \ __1, 1 / 1 7/2 | \ __0, 2 /7/2, 1 | \
| 2 25*/__ | | -2 + x| 25*/__ | | -2 + x| C1*/__ | | -2 + x| C1*/__ | | -2 + x| 25*/__ | | -2 + x|*/__ | | -2 + x|
| C1 \_|2, 2 \5/2 0 | / \_|2, 2 \ 5/2, 0 | / \_|2, 2 \5/2 0 | / \_|2, 2 \ 5/2, 0 | / \_|2, 2 \5/2 0 | / \_|2, 2 \ 5/2, 0 | /
|------ + ------------------------------- + ------------------------------------- + ------------------------------ + ------------------------------------ + ---------------------------------------------------------------- otherwise
|-2 + x 4*(-2 + x) 4*(-2 + x) -2 + x -2 + x 2*(-2 + x)
\
$$y{\left(x \right)} = \begin{cases} \frac{C_{1}^{2}}{x - 2} + \frac{2 C_{1} x^{2}}{\sqrt{x - 2}} - \frac{8 C_{1} x}{\sqrt{x - 2}} + \frac{8 C_{1}}{\sqrt{x - 2}} + \frac{x^{5}}{x - 2} - \frac{10 x^{4}}{x - 2} + \frac{40 x^{3}}{x - 2} - \frac{80 x^{2}}{x - 2} + \frac{80 x}{x - 2} - \frac{32}{x - 2} & \text{for}\: \left|{x - 2}\right| < 1 \\\frac{C_{1}^{2}}{x - 2} + \frac{C_{1} {G_{2, 2}^{1, 1}\left(\begin{matrix} 1 & \frac{7}{2} \\\frac{5}{2} & 0 \end{matrix} \middle| {x - 2} \right)}}{x - 2} + \frac{C_{1} {G_{2, 2}^{0, 2}\left(\begin{matrix} \frac{7}{2}, 1 & \\ & \frac{5}{2}, 0 \end{matrix} \middle| {x - 2} \right)}}{x - 2} + \frac{25 {{G_{2, 2}^{1, 1}\left(\begin{matrix} 1 & \frac{7}{2} \\\frac{5}{2} & 0 \end{matrix} \middle| {x - 2} \right)}}^{2}}{4 \left(x - 2\right)} + \frac{25 {G_{2, 2}^{1, 1}\left(\begin{matrix} 1 & \frac{7}{2} \\\frac{5}{2} & 0 \end{matrix} \middle| {x - 2} \right)} {G_{2, 2}^{0, 2}\left(\begin{matrix} \frac{7}{2}, 1 & \\ & \frac{5}{2}, 0 \end{matrix} \middle| {x - 2} \right)}}{2 \left(x - 2\right)} + \frac{25 {{G_{2, 2}^{0, 2}\left(\begin{matrix} \frac{7}{2}, 1 & \\ & \frac{5}{2}, 0 \end{matrix} \middle| {x - 2} \right)}}^{2}}{4 \left(x - 2\right)} & \text{otherwise} \end{cases}$$
Gráfico para el problema de Cauchy
Clasificación
Bernoulli
1st power series
lie group
Bernoulli Integral
Respuesta numérica
[src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, nan)
(-5.555555555555555, nan)
(-3.333333333333333, nan)
(-1.1111111111111107, nan)
(1.1111111111111107, nan)
(3.333333333333334, nan)
(5.555555555555557, nan)
(7.777777777777779, nan)
(10.0, nan)
(10.0, nan)