Question

Solve \(y^{\prime}-\frac{1}{t} y=\sin t, y(1)=3 .\) Express your answer in terms of the sine integral, \(\operatorname{Si}(t)\), where \(\operatorname{Si}(t)=\int_{0}^{t} \frac{\sin s}{s} d s . \quad\left[\right.\) Note that \(\left.\operatorname{Si}(t)=\operatorname{Si}(1)+\int_{1}^{t} \frac{\sin s}{s} d s .\right]\)

Short answer

Question: Solve the first-order linear differential equation \(y^{\prime}-\frac{1}{t} y=\sin t\) with the initial condition \(y(1)=3\) and express the answer in terms of the sine integral, \(\operatorname{Si}(t)\). Answer: \(y(t) = t\left(\operatorname{Si}(t)+ 3 - \operatorname{Si}(1)\right)\).

Step-by-step solution

Identify the differential equation and the initial condition

The given differential equation is \(y^{\prime}-\frac{1}{t} y=\sin t\) with the initial condition \(y(1)=3\).

Find the integrating factor

The integrating factor is given by \(I(t)=\exp(\int(-\frac{1}{t}) dt)=t^{-1}\) (using properties of exponents in the final step).

Multiply the differential equation by the integrating factor

Multiply the differential equation by the integrating factor \(t^{-1}\): \(\frac{1}{t}y^{\prime}-\frac{1}{t^2}y=t^{-1}\sin t\).

Write as a derivative of the product

Observe that the left side of the above equation can be written as the derivative of the product \(y(t)I(t)\): \((y(t)t^{-1})^{\prime}=t^{-1}\sin t\).

Integrate both sides

Integrate both sides with respect to \(t\): \(\int (y(t)t^{-1})^{\prime} dt = \int t^{-1}\sin t dt\), which simplifies to \(y(t)t^{-1}=\int\frac{\sin t}{t}dt+C\).

Apply the initial condition

Apply the initial condition \(y(1)=3\): \(3\cdot 1^{-1}=\int\frac{\sin 1}{1} dt+C\). The integral can be replaced by \(\operatorname{Si}(1)\) according to the given definition, giving us \(3=\operatorname{Si}(1)+C\).

Solve for C and rewrite the expression for y(t)

Solve for \(C\): \(C=3-\operatorname{Si}(1)\). Rewrite the expression for \(y(t)\): \(y(t)=t\left(\int\frac{\sin t}{t} dt + 3 - \operatorname{Si}(1)\right)\).

Write the answer in terms of the sine integral

According to the given definition, we can rewrite the integral and arrive at the final answer: \(y(t) = t\left(\operatorname{Si}(t)+ 3 - \operatorname{Si}(1)\right)\).

Key concepts

Integrating Factor

The integrating factor is a crucial tool for solving linear first-order differential equations. It is particularly useful when the equation is not exact, and we need to manipulate it into a form that is easier to integrate. In our exercise, the differential equation given is \( y' - \frac{1}{t}y = \sin t \). To make this equation solvable, we find an integrating factor, which turns the left-hand side into a simple derivative.
To determine the integrating factor \( I(t) \), we calculate \( \exp\left(\int -\frac{1}{t} dt\right) \). Computing this integral gives \( -\ln|t| \), hence \( I(t) = e^{-\ln|t|} = t^{-1} \). This transformation helps in converting the differential equation into a form where its left-hand side can be directly integrated.
Once we multiply the entire differential equation by this integrating factor, it becomes easier to integrate, eventually yielding a simpler solution process. Thus, identifying and applying the integrating factor is a foundational step in transforming and solving linear first-order differential equations.

Sine Integral

A sine integral often appears in problems involving periodic functions, like the sine function, especially when divided by a variable. In this case, the integral \( \int \frac{\sin t}{t} dt \) is not straightforward to solve using elementary functions. As a result, it's common to express the solution in terms of special functions like the sine integral, \( \operatorname{Si}(t) \).
The sine integral is defined as \( \operatorname{Si}(t) = \int_{0}^{t} \frac{\sin s}{s} ds \). This integral accounts for the oscillatory nature of the sine function over a unit interval, providing a means to express its cumulative effect up to any given point \( t \).
In our differential equation context, this means rather than solving \( \int \frac{\sin t}{t} dt \) directly, we can use \( \operatorname{Si}(t) \) for a concise expression. Hence, the exercise solution becomes expressible in terms of \( \operatorname{Si}(t) \), simplifying the computation and representation of the solution.

Initial Value Problems

Initial value problems are a fundamental concept in differential equations, where we seek a solution that satisfies both the differential equation and an initial condition. This condition specifies the value of the solution at a particular point, like \( y(1) = 3 \) in our example.
Initial conditions help ensure that a unique solution is obtained, as many differential equations can have infinite solutions without such constraints. The initial value guides the process of finding the constant of integration, which arises after integrating the equation.
In our problem, after integrating and rewriting the expression in terms of the sine integral, we apply the initial condition \( y(1) = 3 \). This helps in determining the constant \( C \) by substituting into the expression: \( 3 = \operatorname{Si}(1) + C \). Solving this yields \( C = 3 - \operatorname{Si}(1) \), allowing us to express \( y(t) \) fully. Hence, the initial condition not only assists in solving the equation but also in constraining and verifying the correctness of the solution.