What is a particular solution to the differential equation #(du)/dt=(2t+sec^2t)/(2u)# and #u(0)=-5#?

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Answer 1 · 2016-07-02T21:12:30

#u^2= t^2+tan t + 25#

#(du)/dt=(2t+sec^2t)/(2u)#

#2u (du)/dt=2t+sec^2t#

#int du qquad 2 u = int dt qquad 2t+sec^2t #

#u^2= t^2+tan t + C#

applying the IV

#(-5)^2 = 2(0) + tan (0) + C#

#implies C = 25#

#u^2= t^2+tan t + 25#

Answer 2 · 2016-07-02T21:17:04

#u^2=t^2+tant+25#

Start by multiplying both sides by #2u# and #dt# to separate the differential equation:
#2udu=2t+sec^2tdt#

Now integrate:
#int2udu=int2t+sec^2tdt#

These integrals aren't too complicated, but if you have any questions on them do not be afraid to ask. They evaluate to:
#u^2+C=t^2+C+tan t+C#

We can combine all the #C#s to make one general constant:
#u^2=t^2+tant+C#

We are given the initial condition #u(0)=-5# so:
#(-5)^2=(0)^2+tan(0)+C#
#25=C#

Thus the solution is #u^2=t^2+tant+25#

Answer 3 · 2016-07-02T21:20:33

#u(t) = -sqrt(t^2+tan(t)+25)#

Grouping variables

#2 u du = (2t+sec^2(t))dt#

Integrating both sides

#u^2=t^2+tan (t) + C#

#u(t) = pm sqrt(t^2+tan (t) + C)#

but considering the initial conditions

#u(0) = -sqrt(C) = -5->C = 25#

and finally

#u(t) = -sqrt(t^2+tan(t)+25)#