ecuaciones

1. MATEMATICAS III
Leidy Roxana Rivadeneira Aguas
04 June 2020
DEBER 1
Resolver mediante el m´etodo de variables separables las siguiente ecuaciones:
1. dy
dx = sen(x + y)
2. dy
dx = 2 +
√
y − 2x + 3
3. dy
dx = (x + y − 1)2
4. 1 + ey−x+5
Solucion
1 Soluci´on Ejercicio 1
dy
dx = sen(x + y)
dy
dx = sen(x + y) u = ax + bx + C
dy = sen(x + y)dx u = x + y
du − dx = sen(u)dx du = dx + dy
du − dx − sen(u)dx = 0
du − dx(1 + sen(u)) = 0
1

2. du = (1 + sen(u))dx
du
sen(u)+1 = dx
1
sen(u)+1 du = dx
1
sen(u)+1 du = x + c u = tan(u
2 )
1
2u
1+u2 +1
· 2
1+u2 du = x + c sen(u) = 2
1+u2
1
2u+1+u2
1+u2
· 2
1+u2 du = x + c du = 2
1+u2 du
1+u2
2u+1+u2 · 2
1+u2 du = x + c
2
2u+1+u2 du = x + c
2 2
2u+1+u2 du = x + c
2 2
(1+u)2 du = x + C v = u + 1
2 2
v2 dv = x + C dv = du
2 · v−2+1
−2+1 = x + C
2 · (u+1)−2+1
−2+1 = x + C
2 ·
(tan( v
2 )+1)−2+1
−2+1 = x + C
2 · (− 1
(tan( v
2 )+1 ) = x + C
− 2
tan( v
2 )+1 = x + C
−2 = (x + C)(tan(v
2 ) + 1)
−2 = xtan(v
2 ) + x + ctan(v
2 ) + C
−2 − x − C = xtan(v
2 ) + ctan(v
2 )
−2 − x − C = tan(v
2 ) · (x + C) tan(x) = a
−2−x−C
x+c = tan(v
2 ) x = arctan(a) + πn
v
2 = arctan(−2−x−C
x+C ) + πn
v = 2 · (arctan(−2−x−C
x+C ) + πn
2

3. x + y = 2 · arctan(−2−x−C
x+C ) + 2πn
y = 2 · arctan(−2−x−C
x+C ) + 2πn − x
2 Soluci´on Ejercicio 2
∗ dx
dy = 2 +
√
y − 2x + 3
– dx
dy = 2 +
√
y − 2x + 3
– dy
dx + 2 = 2 +
√
u u =
√
y − 2x + 3
– du√
u
= dx du = dy − 2dx
– u− 1
2
+1
1
2 +1
= dx du
dx = dy
dx − 2
– 2u
1
2 = x du
dx + 2 = dy
dx
– 2
√
y − 2x + 3 = x + C
3 Soluci´on Ejercicio 3
∗ dx
dy = (x + y + 1)2
– dy
dx = (x + y + 1)2
u = x + y + 1
– dy = (x + y + 1)2
dx
– du = dx + dy
– du − dx = (u2
)dx
– du = (u2
dx) + dx
3

4. – du = (u2
+ 1)dx
– du
u2+1 = dx
– 1
v2+1 du = dx
– arctan(u) = x + C
– 1
1+x2 dx = arctan(x)
– arctan(x + y + 1) = x + c
– x + y + 1 = tan(x + C)
– y = tan(x + C) − x − 1
4 Soluci´on Ejercicio 4
1 + ey−x+5
dy
dx = 1 + ey−x+5
v = y − x + 5
dy = (1 + ey−x+5
)dx du = dy − dx
dy = (1 + ev
)dx
du + dx = (1 + eu
)dx
du = (1 + eu
)dx − dx
du = (1 + eu
− 1)dx
du = (eu
)dx
du
eu = dx
4

5. 1
eu du = dx
1
ev du = x + C u = eu
1
u · du
u = x + C du = eu
du
1
u2 du = x + C du
eu = du
u
u−2
du = x + C
u−2+1
−2+1 = x + C
(eu
)−2+1
−2+1 = x + C
− 1
eu = x + C
−e−u
= x + C
ln(e−u
) = ln(−x − C)
ln(e−u
) = ln(−x − C)
−u = ln(−x − C)
u = −ln(−x − C)
y − x + 5 = ln(−x − C)
y = ln(−x − C) + x − 5
5