initial value problem

Question

initial value problem
dv/dt = 7/(1+t^2) +(sect)^2

v(0)=3

v=?

Answer 1

dv/dt = 7/(1+t^2) +(sect)^2

 

dv = 7dt/(1+t^2) +(sect)^2 dt

 

v = 7∫dt/(1+t^2) +∫(sect)^2 dt

 

The first integral can be solved with the arctan trig substitution. The second integral can be solved recalling that tan(t) is the antiderivative of sec²(t).

 

v(t) = 7arctan(t) + tan(t) + v0

 

Apply the initial condition, v(0) = 3. Let t = 0, then

7arctan(0) + tan(0) + v0 = 3, which implies that v0 = 3.

 

Thus, we have

 

v(t) = 7arctan(t) + tan(t) + 3

Answer 2

Integrate both sides:

 

v(t)=7ArcTan(t)+Tan(t)+C, C some constant

v(0)=3, 

So

3=7ArcTan(0)+Tan(0)+C, C=3

 

v(t)=7ArcTan(t)+Tan(t)+3