Question

Verify that y=tan(x+c) is a one parameter family of solutions of the differential equation y'=1+y^2

Answer 1

y=tan(x+c)
y'=sec^2(x+c)

Answer 2

Answer:
No real solutions.

Answer 3

keep going @Mateaus

Answer 4

y'=sec^2(x+c)
y'=tan^2(x+c)+1
that's how far I can simplify it but I am unsure if there is a possible way to get y'=1+y^2

Answer 5

??
you have done it,

you were given : y'=1+y^2

with: y=tan(x+c)

you showed that : y'=sec^2(x+c) [= y^2 +1]

Answer 6

oh! derp... I see. Did the same mistake for another problem lol

Answer 7

Thanks!