calculus

Subtract 2t from both sides: du/dt-2t = sec2t/2u Integrate both sides: ∫du/u = ∫sec2t/2 dt ln|u| = tan2t + c Take exponent of both sides and solve for u: u = e^(tan2t + c) Insert initial value: −5 = e^0 + c Solve for c: c = -5 Therefore, the solution to the initial value problem is u(t) = e^(tan2t - 5). The solution to this initial value problem was found by separating the variables, then integrating both sides. The solution can then be found when the initial value is substituted into the equation and solved for c.