Question

\(\;{\bf{9}} - {\bf{14}}\) Find the solution of the differential equation that satisfies the given initial condition.

\(\frac{{du}}{{dt}} = \frac{{2t + {{\sec }^2}t}}{{2u}};u(0) = - 5\)

Short answer

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

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Separate variables & integrate

Consider the following differential equation\(\frac{{du}}{{dt}} = \frac{{2t + {{\sec }^2}t}}{{2u}}\)

\(\begin{aligned}2udu &= \left( {2t + {{\sec }^2}t} \right)dt\\\int 2 udu &= \int {\left( {2t + {{\sec }^2}t} \right)} dt\\2\frac{{{u^2}}}{2} &= 2\frac{{{t^2}}}{2} + \tan t + C{\rm{ }}\,\,\,\,\,{\rm{Using }}\int {{{\sec }^2}} tdt &= \tan t + C\\{u^2} &= {t^2} + \tan t + C\\u &= \sqrt {{t^2} + \tan t + C} \end{aligned}\)

Apply initial condition

Apply the condition\(u(0) = - 5\).

Put \(t = 0\) in\(u = \sqrt {{t^2} + \tan t + C} \)

\(\begin{aligned}- 5 &= \sqrt C \\C &= {( - 5)^2}\end{aligned}\)

Therefore, the solution of the initial value problem is\(u = - \sqrt {{t^2} + \tan t + 25} \).