What is a particular solution to the differential equation dudt=2t+sec2t2u and u(0)=-5?

dudt=2t+sec2t2u2ududt=2t+sec2t∫du2u=∫dt2t+sec2tu2=t2+tan⁡t+C(−5)2=2(0)+tan⁡(0)+C⇒C=25u2=t2+tan⁡t+25

Resolved: "What is a particular solution to the differential..."

Bradley Barron

2022-04-17

What is a particular solution to the differential equation

\frac{d u}{d t} = \frac{2 t + \sec^{2} t}{2 u}

and u(0)=-5?

gil001q4wq

Beginner2022-04-18Added 11 answers

\frac{d u}{d t} = \frac{2 t + \sec^{2} t}{2 u}

2 u \frac{d u}{d t} = 2 t + \sec^{2} t

\int d u 2 u = \int d t 2 t + \sec^{2} t

u^{2} = t^{2} + \tan t + C

{(- 5)}^{2} = 2 (0) + \tan (0) + C

\Rightarrow C = 25

u^{2} = t^{2} + \tan t + 25

gsmckibbenx7ga

Beginner2022-04-19Added 17 answers

Start by multiplying both sides by 2u and dt to separate the differential equation:

2 u d u = 2 t + \sec^{2} t d t

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} + \tan t + 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} + \tan t + 25

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