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

$\dfrac{du}{dt} = 2t + \dfrac12 \sec^2(t) u$

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

Rearrange the equation to

$\dfrac{du}{dt} - \dfrac12 \sec^2(t) u = 2t$

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 $u$.

$\implies e^{-\frac12 \tan(t)} \dfrac{du}{dt} - \dfrac12 \sec^2(t) e^{-\frac12 \tan(t)} u = 2t e^{-\frac12 \tan(t)}$

$\implies \dfrac{d}{dt}\left[e^{-\frac12 \tan(t)} u\right] = 2t e^{-\frac12 \tan(t)}$

By the fundamental theorem of calculus, integrating both sides yields

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

$\implies e^{-\frac12\tan(t)} u = 1\times\dfrac\pi2 + \displaystyle \int_{0}^{t} 2\xi e^{-\frac12 \tan(\xi)} \, d\xi$

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