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
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Bonus Knowledge
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.