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

Jul 2, 2016

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

#### Explanation:

$\frac{\mathrm{du}}{\mathrm{dt}} = \frac{2 t + {\sec}^{2} t}{2 u}$

$2 u \frac{\mathrm{du}}{\mathrm{dt}} = 2 t + {\sec}^{2} t$

$\int \mathrm{du} q \quad 2 u = \int \mathrm{dt} q \quad 2 t + {\sec}^{2} t$

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

applying the IV

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

$\implies C = 25$

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

Jul 2, 2016

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

#### Explanation:

Start by multiplying both sides by $2 u$ and $\mathrm{dt}$ to separate the differential equation:
$2 u \mathrm{du} = 2 t + {\sec}^{2} t \mathrm{dt}$

Now integrate:
$\int 2 u \mathrm{du} = \int 2 t + {\sec}^{2} t \mathrm{dt}$

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

We are given the initial condition $u \left(0\right) = - 5$ so:
${\left(- 5\right)}^{2} = {\left(0\right)}^{2} + \tan \left(0\right) + C$
$25 = C$

Thus the solution is ${u}^{2} = {t}^{2} + \tan t + 25$

Jul 2, 2016

$u \left(t\right) = - \sqrt{{t}^{2} + \tan \left(t\right) + 25}$

#### Explanation:

Grouping variables

$2 u \mathrm{du} = \left(2 t + {\sec}^{2} \left(t\right)\right) \mathrm{dt}$

Integrating both sides

${u}^{2} = {t}^{2} + \tan \left(t\right) + C$

$u \left(t\right) = \pm \sqrt{{t}^{2} + \tan \left(t\right) + C}$

but considering the initial conditions

$u \left(0\right) = - \sqrt{C} = - 5 \to C = 25$

and finally

$u \left(t\right) = - \sqrt{{t}^{2} + \tan \left(t\right) + 25}$