Seperable Differential equation?

Question

I have (x^2+1)y'=xy

The answer I get is y = Ksqrt(ln|x^2+1|)

But the answer in the book is y = Ksqrt(x^2+1)

Why is the natural LOG not there?

I used the udu substitution for the integral x/(x^2+1)

Answer 1

y'/y = x/(x²+1)

ln|y|=½ln|x²+1| + C = ln{K√(x²+1)}
Thus y = K√(x²+1)

Answer 2

You did the calculus right. This is what you need to remember:

1/2*ln(x^2 + 1) = ln( x^2 + 1)^1/2) or ln((sqrt(x^2 + 1))

so exp(ln( x^2 + 1)^1/2)) just leaves sqrt(x^2 + 1)

You just forgot a log exponent rule thing is all. It happens. (-:

Answer 3

(x^2 + 1)y' = xy
(x^2 + 1)dy = xydx
dy/y = xdx/(x^2 + 1)
let u = x^2 + 1
du = 2xdx
xdx = (1/2)du
dy/y = (1/2)du/u
lny = (1/2) lnu + lnC
lny = (1/2) lnCu
lny = ln[(Cu)^(1/2)]
y = (Cu)^(1/2)
y = (C(x^2 + 1))^(1/2)
let K = C^(1/2)
y = K(x^2 + 1)^(1/2)

Since both sides are logs, the arguments must be equal, and the ln can be dropped.

Answer 4

the answer of the book is ok

look at the right hand side of the equal to

dy/y = xdx/x^2+1
integrating both sides
2logy = log (x^2+1) + log K
therefore,

y= k sqrt(x^2+1)