Al Resolver la ecuación diferencial dy/sen(x-y+1)=dx; si y(0)=π-1, el valor de la constante c corresponde a:
1. -1
2. 0
3. 1
4. 2
CarlosMath:
sugiero el cambio de variable x=(u+v)/2 , y = (v-u+2)/2
Respuestas
Respuesta dada por:
0
Sea
y
entonces
y ![y=\dfrac{v-u+2}{2} y=\dfrac{v-u+2}{2}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7Bv-u%2B2%7D%7B2%7D)
Así la ecuación se transforma en
![\dfrac{d\left(\frac{v-u+2}{2}\right)}{\sin u}=d\left(\dfrac{u+v}{2}\right)\\ \\ \\
\text{Desarrollando}\\ \\ \\
\dfrac{dv-du}{\sin u}=du+dv\\ \\ \\
dv-du=(du+dv)\sin u\\ \\ \\
(1-\sin u )\,dv=(1+\sin u)\, du\\ \\ \\
dv=\dfrac{1+\sin u}{1-\sin u}\, du\\ \\ \\
\displaystyle
v=\int \dfrac{1+\sin u}{1-\sin u}\, du \\ \\ \\
v=2\tan\left(\dfrac{u}{2}+\dfrac{\pi}{4}\right)-u+C \\ \\ \\ \dfrac{d\left(\frac{v-u+2}{2}\right)}{\sin u}=d\left(\dfrac{u+v}{2}\right)\\ \\ \\
\text{Desarrollando}\\ \\ \\
\dfrac{dv-du}{\sin u}=du+dv\\ \\ \\
dv-du=(du+dv)\sin u\\ \\ \\
(1-\sin u )\,dv=(1+\sin u)\, du\\ \\ \\
dv=\dfrac{1+\sin u}{1-\sin u}\, du\\ \\ \\
\displaystyle
v=\int \dfrac{1+\sin u}{1-\sin u}\, du \\ \\ \\
v=2\tan\left(\dfrac{u}{2}+\dfrac{\pi}{4}\right)-u+C \\ \\ \\](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5Cleft%28%5Cfrac%7Bv-u%2B2%7D%7B2%7D%5Cright%29%7D%7B%5Csin+u%7D%3Dd%5Cleft%28%5Cdfrac%7Bu%2Bv%7D%7B2%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0A%5Ctext%7BDesarrollando%7D%5C%5C+%5C%5C+%5C%5C%0A%5Cdfrac%7Bdv-du%7D%7B%5Csin+u%7D%3Ddu%2Bdv%5C%5C+%5C%5C+%5C%5C%0Adv-du%3D%28du%2Bdv%29%5Csin+u%5C%5C+%5C%5C+%5C%5C%0A%281-%5Csin+u+%29%5C%2Cdv%3D%281%2B%5Csin+u%29%5C%2C+du%5C%5C+%5C%5C+%5C%5C%0Adv%3D%5Cdfrac%7B1%2B%5Csin+u%7D%7B1-%5Csin+u%7D%5C%2C+du%5C%5C+%5C%5C+%5C%5C%0A%5Cdisplaystyle%0Av%3D%5Cint+%5Cdfrac%7B1%2B%5Csin+u%7D%7B1-%5Csin+u%7D%5C%2C+du+%5C%5C+%5C%5C+%5C%5C%0Av%3D2%5Ctan%5Cleft%28%5Cdfrac%7Bu%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29-u%2BC+%5C%5C+%5C%5C+%5C%5C)
![x+y-1=2\tan\left(\dfrac{x-y+1}{2}+\dfrac{\pi}{4}\right)-x+y-1+C \\ \\ \\
x=\tan\left(\dfrac{x-y+1}{2}+\dfrac{\pi}{4}\right)+C'\\ \\ \\
\text{Reemplazando valores:}\\ \\
0=\tan\left(\dfrac{0-\pi+1+1}{2}+\dfrac{\pi}{4}\right)+C'\\ \\ \\
0=\tan\left(\dfrac{4-\pi}{4}\right)+C'\\ \\ \\
C'=\tan\left(\dfrac{\pi-4}{4}\right)\\ \\ \\
\text{Soluci\'on de la EDO}\\ \\ \\
x=\tan\left(\dfrac{x-y+1}{2}+\dfrac{\pi}{4}\right)+\tan\left(\dfrac{\pi-4}{4}\right) x+y-1=2\tan\left(\dfrac{x-y+1}{2}+\dfrac{\pi}{4}\right)-x+y-1+C \\ \\ \\
x=\tan\left(\dfrac{x-y+1}{2}+\dfrac{\pi}{4}\right)+C'\\ \\ \\
\text{Reemplazando valores:}\\ \\
0=\tan\left(\dfrac{0-\pi+1+1}{2}+\dfrac{\pi}{4}\right)+C'\\ \\ \\
0=\tan\left(\dfrac{4-\pi}{4}\right)+C'\\ \\ \\
C'=\tan\left(\dfrac{\pi-4}{4}\right)\\ \\ \\
\text{Soluci\'on de la EDO}\\ \\ \\
x=\tan\left(\dfrac{x-y+1}{2}+\dfrac{\pi}{4}\right)+\tan\left(\dfrac{\pi-4}{4}\right)](https://tex.z-dn.net/?f=x%2By-1%3D2%5Ctan%5Cleft%28%5Cdfrac%7Bx-y%2B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29-x%2By-1%2BC+%5C%5C+%5C%5C+%5C%5C%0Ax%3D%5Ctan%5Cleft%28%5Cdfrac%7Bx-y%2B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29%2BC%27%5C%5C+%5C%5C+%5C%5C%0A%5Ctext%7BReemplazando+valores%3A%7D%5C%5C+%5C%5C%0A0%3D%5Ctan%5Cleft%28%5Cdfrac%7B0-%5Cpi%2B1%2B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29%2BC%27%5C%5C+%5C%5C+%5C%5C%0A0%3D%5Ctan%5Cleft%28%5Cdfrac%7B4-%5Cpi%7D%7B4%7D%5Cright%29%2BC%27%5C%5C+%5C%5C+%5C%5C%0AC%27%3D%5Ctan%5Cleft%28%5Cdfrac%7B%5Cpi-4%7D%7B4%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0A%5Ctext%7BSoluci%5C%27on+de+la+EDO%7D%5C%5C+%5C%5C+%5C%5C%0Ax%3D%5Ctan%5Cleft%28%5Cdfrac%7Bx-y%2B1%7D%7B2%7D%2B%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Cright%29%2B%5Ctan%5Cleft%28%5Cdfrac%7B%5Cpi-4%7D%7B4%7D%5Cright%29)
Así la ecuación se transforma en
Preguntas similares
hace 6 años
hace 9 años
hace 9 años
hace 9 años
hace 9 años
hace 9 años