Question

checking my answer:
Solve y y' =1. Is y(x) = (2x+C)^(1/2) where C is arbitrary constants asolution y y' = 1!
y(0) = 4

Answer 1

\[y'=\frac{1}{\sqrt{2x+C}}\]

Answer 2

so it looks like \(yy'=1\) no matter what

Answer 3

y (dy/dx) = 1
y dy = dx
(because y(0) = 4)
int y dy (4 from y) = x dx (0 from x)
(y^2/2) - (4^2/2) = x
((y^2)/2) - 8 = x
y^2 = (x+8)2
y^2 = 2x + 16

Answer 4

checking (y(0) = 4)
y^2 = 2x + 16
y^2 = 2(0) + 16
y^2 = 16
y = sqrt(16)
y = 4