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 Perkins Johnson. in the United States

Mar 20,2025

To solve the initial value problem, we first put it in standard form. We start with the equation given: \[ 9\left(\sin(t) \frac{dy}{dt} + \cos(t) y\right) = \cos(t) \sin^2(t) \] Dividing through by \( 9\sin(t) \), we obtain: \[ \frac{dy}{dt} + \frac{\cos(t)}{\sin(t)} y = \frac{\cos(t) \sin(t)}{9} \] Now, we have it in the standard linear form \( \frac{dy}{dt} + P(t) y = Q(t) \), where: - \( P(t) = \frac{\cos(t)}{\sin(t)} \) - \( Q(t) = \frac{\cos(t) \sin(t)}{9} \) Next, we find the integrating factor \( \rho(t) \). The integrating factor is calculated as: \[ \rho(t) = e^{\int P(t) \, dt} = e^{\int \frac{\cos(t)}{\sin(t)} \, dt} \] To compute this integral, we realize that: \[ \int \frac{\cos(t)}{\sin(t)} \, dt = \int \cot(t) \, dt = \ln|\sin(t)| \] Thus, the integrating factor becomes: \[ \rho(t) = e^{\ln|\sin(t)|} = |\sin(t)| \] Since \( \sin(t) > 0 \) for \( 0 < t < \pi \): \[ \rho(t) = \sin(t) \] Now, we multiply through the first equation by \( \rho(t) \): \[ \sin(t) \frac{dy}{dt} + \cos(t) y = \frac{\sin(t) \cos(t) \sin(t)}{9} \] This becomes: \[ \sin(t) \frac{dy}{dt} + \cos(t) y = \frac{\cos(t) \sin^2(t)}{9} \] The left side can be written as the derivative of a product: \[ \frac{d}{dt}(\sin(t) y) = \frac{\cos(t) \sin^2(t)}{9} \] Now, we can integrate both sides: \[ \int \frac{d}{dt}(\sin(t) y) \, dt = \int \frac{\cos(t) \sin^2(t)}{9} \, dt \] This gives us: \[ \sin(t) y = \frac{1}{9} \int \cos(t) \sin^2(t) \, dt + C \] To solve the integral on the right, we can use integration by parts or a substitution. Here, we'll use substitution: Let \( u = \sin(t) \) so \( du = \cos(t) \, dt \), and when \( t = 0, u = 0 \); when \( t = \pi, u = 0 \): \[ \frac{1}{9} \int u^2 \, du = \frac{1}{9} \cdot \frac{u^3}{3} = \frac{u^3}{27} = \frac{\sin^3(t)}{27} \] Therefore, we have: \[ \sin(t) y = \frac{\sin^3(t)}{27} + C \] Now we solve for \( y \): \[ y = \frac{\sin^2(t)}{27} + \frac{C}{\sin(t)} \] To find \( C \), we use the initial condition \( y(\frac{\pi}{2})=16 \): When \( t = \frac{\pi}{2} \): \[ \sin\left(\frac{\pi}{2}\right) = 1 \implies 16 = \frac{1^2}{27} + C \implies 16 = \frac{1}{27} + C \] \[ C = 16 - \frac{1}{27} = \frac{432 - 1}{27} = \frac{431}{27} \] So the solution becomes: \[ y(t) = \frac{\sin^2(t)}{27} + \frac{431/27}{\sin(t)} = \frac{\sin^2(t) + 431}{27 \sin(t)} \] Therefore, the final solution is: \[ y(t) = \frac{\sin^2(t) + 431}{27 \sin(t)} \]