Question

Find the solution of the differential equation that satisfies the given initial condition. du dt = 2t + sec2(t) 2u , u(0) = â2

Answer 1

It looks like the equation is

`(du)/(dt) = 2t + \frac12 \sec^2(t) u`

with initial value
`u(0) = \frac\pi2`.

Rearrange the equation to

`(du)/(dt) - \frac12 \sec^2(t) u = 2t`

Multiply both sides by the integrating factor

`\displaystyle \mu = \exp\left( \int -\frac12 \sec^2(t)\,dt\right) = \exp\left(-\frac12\tan(t)\right)`

and solve for
`u`.

`\implies e^(-\frac12 \tan(t)) (du)/(dt) - \frac12 \sec^2(t) e^(-\frac12 \tan(t)) u = 2t e^(-\frac12 \tan(t))`

`\implies (d)/(dt)\left[e^(-\frac12 \tan(t)) u\right] = 2t e^(-\frac12 \tan(t))`

By the fundamental theorem of calculus, integrating both sides yields

`e^(-\frac12\tan(t)) u = e^(-\frac12\tan(t)) u \bigg|_(t=0) + \displaystyle \int_(\xi=0)^(\xi=t) 2\xi e^(-\frac12 \tan(\xi)) \, d\xi`

`\implies e^(-\frac12\tan(t)) u = 1*\frac\pi2 + \displaystyle \int_(0)^(t) 2\xi e^(-\frac12 \tan(\xi)) \, d\xi`

`\implies \boxed{\displaystyle u = \frac\pi2 e^(\frac12\tan(t)) + 2e^(\frac12\tan(t)) \int_0^t \xi e^(-\frac12 \tan(\xi)) \, d\xi}`