Solve the following initial value problem:
cos(t)^2 (dy/dt) = 1
with y( 4 ) = tan(4).
(Find y as a function of t.)
I found that y = tan(t)
but am unable to find on what interval is the solution valid.
? < t < ? How can I find this??
cos(t)^2 (dy/dt) = 1
with y( 4 ) = tan(4).
(Find y as a function of t.)
I found that y = tan(t)
but am unable to find on what interval is the solution valid.
? < t < ? How can I find this??
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without the side condition you would state that t is in ( nπ - π / 2 , nπ + π / 2 )...
but the side condition needs 4 to be in the interval { dy /dt existing means y is continuous }
so you choose ( π / 2 , 3π / 2 )
but the side condition needs 4 to be in the interval { dy /dt existing means y is continuous }
so you choose ( π / 2 , 3π / 2 )
-
0 < t < pi/2 + pi*n
pi/2 + pi*n < t < 2pi*n
Where n is an integer greater than or equal to 0.
At t = pi/2 + pi*n, the solution is not valid because then you'd have division by 0.
pi/2 + pi*n < t < 2pi*n
Where n is an integer greater than or equal to 0.
At t = pi/2 + pi*n, the solution is not valid because then you'd have division by 0.