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

3 Answers
Jul 2, 2016

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

Explanation:

(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

Jul 2, 2016

u^2=t^2+tant+25

Explanation:

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 Cs 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

Jul 2, 2016

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

Explanation:

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)