In Exercises 5–8, show that each function is a solution of the given initial value problem.
7. Differential Equation: xy' + y = -sin(x), x>0
Initial condition: y(π/2) = 0
Solution candidate: y = cos(x)/x

Verified step by step guidance

1. Start by identifying the given differential equation: \(x y' + y = -\sin(x)\), with \(x > 0\), and the initial condition \(y\left(\frac{\pi}{2}\right) = 0\).
2. Given the candidate solution \(y = \frac{\cos(x)}{x}\), compute its derivative \(y'\) using the quotient rule: \(y' = \frac{d}{dx} \left( \frac{\cos(x)}{x} \right)\).
3. Apply the quotient rule: \(y' = \frac{-\sin(x) \cdot x - \cos(x) \cdot 1}{x^2} = \frac{-x \sin(x) - \cos(x)}{x^2}\).
4. Substitute \(y\) and \(y'\) back into the differential equation: \(x y' + y = x \cdot \frac{-x \sin(x) - \cos(x)}{x^2} + \frac{\cos(x)}{x}\).
5. Simplify the expression step-by-step to verify if it equals \(-\sin(x)\). Then, check the initial condition by evaluating \(y\left(\frac{\pi}{2}\right)\) to confirm it equals zero.

Key Concepts

Initial Value Problem (IVP)

An initial value problem consists of a differential equation paired with a specific condition that the solution must satisfy at a given point. This condition, called the initial condition, ensures a unique solution. In this problem, y(π/2) = 0 specifies the value of the solution at x = π/2.

Verification of a Solution to a Differential Equation

To verify a candidate solution, substitute it and its derivative into the differential equation. If both sides are equal for all x in the domain, the function is a solution. This process confirms that the candidate satisfies the equation's requirements.

Differentiation of Functions Involving Quotients

Differentiating functions like y = cos(x)/x requires applying the quotient rule, which states that the derivative of u/v is (v*u' - u*v')/v². Correct differentiation is essential to substitute y' accurately into the differential equation.