Differential Equations HELP!!!!?

Question

y' = sqrt(x+y+1)

We've normally been substituting vx for y, and vdx + xdv for dy, but i'm so lost from the start :(

THANK YOU VERY VERY MUCH IN ADVANCE!!!

Answer 1

Try the substitution v=x+y+1. Then v' = 1+y', so we have:

v'-1 = √v

This is separable, and easily solved:

v' = √v + 1
1/(√v+1) v' = 1
∫1/(√v+1) dv/dx dx = ∫1 dx
∫1/(√v+1) dv = x + C

Let u=√v+1, du=1/(2√v) dv = 1/(2u-2) dv, dv = (2u-2) du

∫(2u-2)/u du = x + C
∫2-2/u du = x + C
2u - 2 ln u = x + C
2(√v+1) - 2 ln (√v+1) = x+C
2√(x+y+1) + 2 - 2 ln (√(x+y+1) + 1) = x+C

It doesn't appear that much further simplification is possible, so we'll just leave this as an implicit solution. So we are done.

Answer 2

Are you asking for the derivative of sqrt(x+y+1)? If so, you get the derivative by using the chain rule.

dy = (1/2) / sqrt(x+y+1).