Question

solve the initial value problem.

\theta \frac { d y } { d \theta } + y = 3 sin \theta , \theta > 0 , y left( \frac { pi } { 2 }
ight) = 1

y =

Explanation:

Step1: Rewrite the equation in standard linear form

The given equation \(\theta\frac{dy}{d\theta}+y = 3\sin\theta\) can be rewritten as \(\frac{dy}{d\theta}+\frac{1}{\theta}y=\frac{3\sin\theta}{\theta}\).
The standard form of a first - order linear differential equation is \(\frac{dy}{d\theta}+P(\theta)y = Q(\theta)\), where \(P(\theta)=\frac{1}{\theta}\) and \(Q(\theta)=\frac{3\sin\theta}{\theta}\).

Step2: Find the integrating factor

The integrating factor \(\mu(\theta)=e^{\int P(\theta)d\theta}\).
Since \(\int\frac{1}{\theta}d\theta=\ln\theta\), then \(\mu(\theta)=e^{\ln\theta}=\theta\).

Step3: Multiply the equation by the integrating factor

Multiply \(\frac{dy}{d\theta}+\frac{1}{\theta}y=\frac{3\sin\theta}{\theta}\) by \(\theta\).
We get \(\theta\frac{dy}{d\theta}+y = 3\sin\theta\).
The left - hand side is the derivative of the product \(\theta y\) with respect to \(\theta\), i.e., \(\frac{d(\theta y)}{d\theta}=3\sin\theta\).

Step4: Integrate both sides

Integrate \(\frac{d(\theta y)}{d\theta}=3\sin\theta\) with respect to \(\theta\).
\(\theta y=\int3\sin\theta d\theta\).
Since \(\int\sin\theta d\theta=-\cos\theta + C\), then \(\theta y=- 3\cos\theta + C\).

Step5: Solve for \(y\)

\(y =-\frac{3\cos\theta}{\theta}+\frac{C}{\theta}\).

Step6: Use the initial condition \(y(\frac{\pi}{2}) = 1\)

Substitute \(\theta=\frac{\pi}{2}\) and \(y = 1\) into \(y=-\frac{3\cos\theta}{\theta}+\frac{C}{\theta}\).
\(1=-\frac{3\cos(\frac{\pi}{2})}{\frac{\pi}{2}}+\frac{C}{\frac{\pi}{2}}\).
Since \(\cos(\frac{\pi}{2}) = 0\), then \(1=\frac{2C}{\pi}\), so \(C=\frac{\pi}{2}\).

Answer:

\(y=\frac{-3\cos\theta+\frac{\pi}{2}}{\theta}\)