# Math Help - separation of variables (calculus)

1. ## separation of variables (calculus)

ok, so how does one attempt this using separation of variables?:

y' = the square root of ty

so far, what i have done is set y' equal to t^(1/2) * y^(1/2) since the square root of two variables is just the square root of each individual variable.

i'm not really sure what to do after that so i would be grateful for any help!

2. Originally Posted by clockingly
ok, so how does one attempt this using separation of variables?:

y' = the square root of ty

so far, what i have done is set y' equal to t^(1/2) * y^(1/2) since the square root of two variables is just the square root of each individual variable.

i'm not really sure what to do after that so i would be grateful for any help!
you were right so far. Now just divide both sides to obtain

y'/y^(1/2) = t^(1/2) or equivalently y^(/1/2) dy = t^(1/2) dt

now integrate both sides

2y^(1/2) = (2/3)t^(3/2) + C
=> y = [{(2/3)t^(3/2) + C}^2]/2

3. Originally Posted by Jhevon
you were right so far. Now just divide both sides to obtain

y'/y^(1/2) = t^(1/2) or equivalently y^(/1/2) dy = t^(1/2) dt

now integrate both sides

2y^(1/2) = (2/3)t^(3/2) + C
=> y = [{(2/3)t^(3/2) + C}^2]/2
(Do not worry many make that mistake when solving differencial equations. I do not even think professors mention this for some reason).

4. Originally Posted by ThePerfectHacker
(Do not worry many make that mistake when solving differencial equations. I do not even think professors mention this for some reason).
Yeah, professors usually don't mention that. Perhaps the possibility of y=0 exists within the constant itself though

5. Originally Posted by Jhevon
Yeah, professors usually don't mention that. Perhaps the possibility of y=0 exists within the constant itself though
There is actually a stranger situation, that might occur. I addressed it on this site some time ago, when y=0 at some point on the interval. My belief is that y=0 either everywhere on the interval or y not = 0. Meaning y=0 at some points yes and some points no, cannot happen. I never been able to formally prove it, but my idea is that when a curve y not = 0 for some point, then around that neighborhood of points we have a solution (which is of the form you mentioned it) now if it is zero at some other point and you connect the non-trivial solution (of the form you mentioned) with a curve, we get a "rough" point, you know a sharp turn, and therefore it is non-differenciable at that point. But that cannot happen because by definition a solution to some differencial equation must be differenciable. This tells us a very useful fact (assuming it is true), only consider the case y=0 before you divide.