Sr Examen
Ecuación diferencial dy/dt=y^t
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
d t --(y(t)) = y (t) dt
$$\frac{d}{d t} y{\left(t \right)} = y^{t}{\left(t \right)}$$
y' = y^t
Respuesta
[src]
/ / 1 \ / 1 \ \
| 2 2*|- -- + 2*log(C1)|*log(C1) 2*|- -- + log(C1)|*log(C1)|
4 / 3 3 9*log(C1)\ 5 | 4 8 20 20*log(C1) 22*log (C1) \ C1 / \ C1 / |
3 / 2 2 \ t *|log (C1) - --- + ---------| t *|log (C1) + --- + --- - ---------- + ----------- + ---------------------------- + --------------------------|
2 t *|log (C1) + --| | 2 C1 | | 3 2 2 C1 C1 C1 |
t *log(C1) \ C1/ \ C1 / \ C1 C1 C1 / / 6\
y(t) = C1 + t + ---------- + ------------------ + ------------------------------- + ---------------------------------------------------------------------------------------------------------------- + O\t /
2 6 24 120
$$y{\left(t \right)} = t + \frac{t^{2} \log{\left(C_{1} \right)}}{2} + \frac{t^{3} \left(\log{\left(C_{1} \right)}^{2} + \frac{2}{C_{1}}\right)}{6} + \frac{t^{4} \left(\log{\left(C_{1} \right)}^{3} + \frac{9 \log{\left(C_{1} \right)}}{C_{1}} - \frac{3}{C_{1}^{2}}\right)}{24} + \frac{t^{5} \left(\log{\left(C_{1} \right)}^{4} + \frac{2 \left(\log{\left(C_{1} \right)} - \frac{1}{C_{1}}\right) \log{\left(C_{1} \right)}}{C_{1}} + \frac{2 \left(2 \log{\left(C_{1} \right)} - \frac{1}{C_{1}}\right) \log{\left(C_{1} \right)}}{C_{1}} + \frac{22 \log{\left(C_{1} \right)}^{2}}{C_{1}} - \frac{20 \log{\left(C_{1} \right)}}{C_{1}^{2}} + \frac{20}{C_{1}^{2}} + \frac{8}{C_{1}^{3}}\right)}{120} + C_{1} + O\left(t^{6}\right)$$
Clasificación
1st power series
lie group