- #1
MathewsMD
- 433
- 7
Hi,
I'm currently finding a solution (i.e. y) for y′′ +t(y′)2 = 0.
The answer is:
y=(1/k)ln|(k−t)/(k+t)|+c2 if c1 =k^2 >0; y=(2/k)arctan(t/k)+c2 if c1 =−k^2 <0; y=−2t−1 +c2 if c1 =0; also y=c
(NOTE: c1 or c2 is just c_1 or c_2)
I've gotten the solutions as y = 0 or y=(2/k)arctan(t/k)+c2, depending on the initial conditions, but don't seem to understand how the first solution (i.e. y=(1/k)ln|(k−t)/(k+t)|+c2) provided in the answer was derived. Also, I don't quite understand why the restrictions are being stated (i.e. if c1 > k^2). If anyone could please explain why there are multiple restrictions and different solutions for each, and also how the first solution was derived, that would be extremely helpful.
Thank you!
I'm currently finding a solution (i.e. y) for y′′ +t(y′)2 = 0.
The answer is:
y=(1/k)ln|(k−t)/(k+t)|+c2 if c1 =k^2 >0; y=(2/k)arctan(t/k)+c2 if c1 =−k^2 <0; y=−2t−1 +c2 if c1 =0; also y=c
(NOTE: c1 or c2 is just c_1 or c_2)
I've gotten the solutions as y = 0 or y=(2/k)arctan(t/k)+c2, depending on the initial conditions, but don't seem to understand how the first solution (i.e. y=(1/k)ln|(k−t)/(k+t)|+c2) provided in the answer was derived. Also, I don't quite understand why the restrictions are being stated (i.e. if c1 > k^2). If anyone could please explain why there are multiple restrictions and different solutions for each, and also how the first solution was derived, that would be extremely helpful.
Thank you!