Solve the initial value problem \[ 9\left(\sin (t) \frac{d y}{d t}+\cos (t) y\right)=\cos (t) \sin ^{2}(t) \] for \( 0

Ask by O'Quinn Lang. in the United States

Mar 20,2025

To solve the given initial value problem, we first need to rewrite the equation in standard form, which is typically written as: \[ \frac{dy}{dt} + P(t) y = Q(t). \] Starting with the original equation: \[ 9\left(\sin (t) \frac{d y}{d t}+\cos (t) y\right)=\cos (t) \sin ^{2}(t). \] We can expand and rearrange it: \[ 9 \sin(t) \frac{dy}{dt} + 9 \cos(t) y = \cos(t) \sin^2(t). \] Next, divide the entire equation by \(9 \sin(t)\) (assuming \(t \neq 0\) as we are in the interval \(0 < t < \pi\)): \[ \frac{dy}{dt} + \frac{9 \cos(t)}{9 \sin(t)} y = \frac{\cos(t) \sin^2(t)}{9 \sin(t)}. \] This simplifies to: \[ \frac{dy}{dt} + \cot(t) y = \frac{1}{9} \sin(t) \cos(t). \] Now we have the equation in standard form, where: - \(P(t) = \cot(t)\), - \(Q(t) = \frac{1}{9} \sin(t) \cos(t)\). Next, we will also note the initial condition given in the problem: \[ y\left(\frac{\pi}{2}\right) = 16. \] This provides us with the necessary information to solve the equation once we use an integrating factor or an appropriate method for solving first-order linear differential equations.