Question

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

Upstudy AI Solution

Tutor-Verified Answer

Answer

The equation in standard form is: \[ \frac{d y}{d t} + \frac{\cos(t)}{9\sin(t)} y = \frac{\cos(t) \sin(t)}{9} \] with the initial condition \( y\left(\frac{\pi}{2}\right) = 16 \).

Solution

To solve the initial value problem, we first need to rewrite the given equation in standard form. The given equation is: \[ 9\left(\sin (t) \frac{d y}{d t}+\cos (t) y\right)=\cos (t) \sin ^{2}(t) \] We can start by distributing the 9: \[ 9\sin(t) \frac{d y}{d t} + 9\cos(t) y = \cos(t) \sin^2(t) \] Next, we can isolate the term involving \(\frac{d y}{d t}\): \[ 9\sin(t) \frac{d y}{d t} = \cos(t) \sin^2(t) - 9\cos(t) y \] Now, we can divide through by \(9\sin(t)\) (noting that \(\sin(t) \neq 0\) for \(0 < t < \pi\)) to put it in standard form: \[ \frac{d y}{d t} = \frac{1}{9\sin(t)}\left(\cos(t) \sin^2(t) - 9\cos(t) y\right) \] This simplifies to: \[ \frac{d y}{d t} = \frac{\cos(t)}{9\sin(t)}\left(\sin^2(t) - 9y\right) \] Now, we can express this in standard form: \[ \frac{d y}{d t} + \frac{1}{\sin(t)}\cos(t) y = \frac{1}{9\sin(t)}\cos(t) \sin^2(t) \] Thus, the initial value problem in standard form is: \[ \frac{d y}{d t} + \frac{\cos(t)}{9\sin(t)} y = \frac{\cos(t) \sin(t)}{9} \] with the initial condition \(y\left(\frac{\pi}{2}\right) = 16\).

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

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.

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy