What is the integral of dy/dx=√(x/y)?

Question

I have to solve for y, using that equation, but every time I try I keep getting it wrong. I can't seem to figure out how to solve for y. I end up getting (2/3)y^(3/2)+C=(2/3)x^(3/2)+C, but those are the exact same equations.

Answer 1

Seperable diff. eq

sqrt(y) dy = sqrt(x) dx

y^(1/2) dy = x^(1/2) dx

Integrating...

(2/3) y^(3/2) = (2/3)x^(3/2) + c

y^(3/2) = x^(3/2) + C <--- multiplies both sides by 3/2; constant of integration gets tweaked

y = [x^(3/2) + C ]^2/3 <--- raises both sides to the power 2/3

This is the alleged solution.

check:

sqrt(y) = sqrt( [x^(3/2) + C ]^2/3 ) = [x^(3/2) + C ]^1/3

SO the right side is sqrt(x) / sqrt(y) = sqrt(x) / [x^(3/2) + C ]^1/3

Now the derivative (of the alleged solition) is 2/3 [ x^(3/2)+C]^(-1/3) [ (3/2)x ^(1/2) ]

= x^(1/2) / [ x^(3/2) + c]^(1/3)

which agrees EXACTLY with the right hand side.

So the solution in bold above is correct.