separation of variables (calculus)

• Feb 27th 2007, 06:15 PM
clockingly
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!
• Feb 27th 2007, 06:23 PM
Jhevon
Quote:

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
• Feb 27th 2007, 06:55 PM
ThePerfectHacker
Quote:

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).
• Feb 27th 2007, 07:00 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker