Question

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.θ dy/dθ + y = sin θ, θ \u003e 0, y(π/2) = 1

Accepted Answer

Rewrite the given differential equation $$	heta \frac{dy}{d	heta} + y = \sin 	heta in $$standard linear form by dividing both sides by $$	heta ($$noting that $$	heta \u003e 0$$):

$$\frac{dy}{d	heta} + \frac{1}{	heta} y = \frac{\sin 	heta}{	heta}$$ Identify the integrating factor $$\mu(	heta)$$, which is given by:

$$\mu(	heta) = e$$^{\int P(	heta) \, d	heta}$$,

where P$$(	heta) = \frac{1}{	heta}$$. So,

$$\mu(	heta) = $$e^{\int \frac{1}$${	heta} \, d	heta} = $$e^{\ln |	heta|}$$ = 	heta$$ (since $$	heta \u003e 0$$). Multiply the entire differential equation by the integrating factor $$	heta$$ to get:

$$	heta \frac{dy}{d	heta} + y = \sin 	heta$$,

which matches the original equation, confirming the integrating factor is correct. This allows us to write the left side as the derivative of a product:

$$\frac{d}{d	heta} (	heta y) = \sin 	heta$$. Integrate both sides with respect to $$	heta$$:

$$\int \frac{d}{d	heta} (	heta y) \, d	heta = \int \sin 	heta \, d	heta$$,

which simplifies to:

$$	heta y = -\cos 	heta + C$$,

where C$ is the constant of integration. Use the initial condition y$$\left(\frac{\pi}{2}ight) = 1$$ to solve for C$:

Substitute $$	heta = \frac{\pi}{2}$$ and y = 1$ into the equation:

$$\left(\frac{\pi}{2}ight)(1) = -\cos \left(\frac{\pi}{2}ight) + C$$.

Solve this equation for C$ to find the particular solution.

Solving Initial Value Problems
Solve the initial value problems in Exercises 15–20.

θ dy/dθ + y = sin θ, θ > 0, y(π/2) = 1

Verified video answer for a similar problem:

Key Concepts

First-Order Linear Differential Equations

Integrating Factor Method

Initial Value Problems (IVP)

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.θ dy/dθ + y = sin θ, θ > 0, y(π/2) = 1