Answer to Question #216370 in Differential Equations for Ohida

Question

Solve the initial-value problem: (dx/dt)+(tant)x=(cost)^2, x(0)=-1.

Answer 1

Substitution:

$x(t) = u(t)v(t) \Rightarrow \frac{{dx}}{{dt}} = x' = u'v + uv'$

Then

$u'v + uv' + uv\tan t = {\cos ^2}t$

$u'v + u\left( {v' + v\tan t} \right) = {\cos ^2}t$

Let

$v' + v\tan t = 0 \Rightarrow \frac{{dv}}{{dt}} = - v\tan t \Rightarrow \frac{{dv}}{v} = - \frac{{\sin t}}{{\cos t}}dt = \frac{{d\cos t}}{{\cos t}} \Rightarrow \ln v = \ln \cos t \Rightarrow v = \cos t$

Then

$u'\cos t = {\cos ^2}t \Rightarrow u' = \cos t \Rightarrow u = \sin t + C$

Then

$x = uv = \left( {\sin t + C} \right)\cos t$

$x(0) = - 1 \Rightarrow (\sin 0 + C)\cos 0 = - 1 \Rightarrow C = - 1$

Answer: $x = \left( {\sin t - 1} \right)\cos t$