- #1
cheeee
- 15
- 0
2nd order diff Eq with t missing
I am trying to find y as a function of t
and y'' - y = 0
The two IV given are y(0) = 7, and y(1) = 5 .. Remark: the initial condition involves values at two points.
Well since y = {y,y''} and the independent variable t does not appear, I went about it by setting z = y' and trying to reduce it to a first order DE.
z = dy/dt = y'
z * dz/dy - y = 0
z(dz/dy) = y
Separating the variables and integrating...
zdz = ydy
(z^2)/2 = (y^2)/2 + c
z^2 = (y^2) + C (where C = 2c)
z = sqrt (y^2 + C)
Normally here I would use the IVs to determine C before..however both there is no IV involving y'(t)...
Anyways...
z = sqrt (y^2 + C)
y' = sqrt (y^2 + C)
dy/dt = sqrt (y^2 + C)
dy/sqrt(y^2 + C) = dt
So do I just integrate from here? Isn't this a composite function on the left hand side...so the variable y would appear as well as y^2 ?
How do I solve this problem and get an answer in the form of y(t) = _________________
Any help would be appreciated.
I am trying to find y as a function of t
and y'' - y = 0
The two IV given are y(0) = 7, and y(1) = 5 .. Remark: the initial condition involves values at two points.
Well since y = {y,y''} and the independent variable t does not appear, I went about it by setting z = y' and trying to reduce it to a first order DE.
z = dy/dt = y'
z * dz/dy - y = 0
z(dz/dy) = y
Separating the variables and integrating...
zdz = ydy
(z^2)/2 = (y^2)/2 + c
z^2 = (y^2) + C (where C = 2c)
z = sqrt (y^2 + C)
Normally here I would use the IVs to determine C before..however both there is no IV involving y'(t)...
Anyways...
z = sqrt (y^2 + C)
y' = sqrt (y^2 + C)
dy/dt = sqrt (y^2 + C)
dy/sqrt(y^2 + C) = dt
So do I just integrate from here? Isn't this a composite function on the left hand side...so the variable y would appear as well as y^2 ?
How do I solve this problem and get an answer in the form of y(t) = _________________
Any help would be appreciated.