Another second-order equation Consider the differential equation y''(t) + k²y(t) = 0, where k is a positive real number.
c. Give the general solution of the equation for arbitrary k > 0 and verify your conjecture.

Verified step by step guidance

1. Recognize that the given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \(y''(t) + k^{2} y(t) = 0\), where \(k > 0\).
2. Write the characteristic equation associated with the differential equation by replacing \(y(t)\) with \(e^{\lambda t}\), which gives: \(\lambda^{2} + k^{2} = 0\).
3. Solve the characteristic equation for \(\lambda\): \(\lambda^{2} = -k^{2}\), so \(\lambda = \pm i k\), where \(i\) is the imaginary unit.
4. Since the roots are purely imaginary, the general solution to the differential equation is a linear combination of sine and cosine functions: \(y(t) = C_{1} \cos(k t) + C_{2} \sin(k t)\), where \(C_{1}\) and \(C_{2}\) are arbitrary constants.
5. Verify the solution by differentiating \(y(t)\) twice to find \(y''(t)\), then substitute \(y(t)\) and \(y''(t)\) back into the original equation \(y''(t) + k^{2} y(t) = 0\) to confirm that it holds true for all \(t\).

Key Concepts

Second-Order Linear Homogeneous Differential Equations

These are differential equations involving the second derivative of a function, with no terms independent of the function or its derivatives. The general form is y'' + p(t)y' + q(t)y = 0. Solutions depend on the characteristic equation derived from constant coefficients, which determines the behavior of the solution.

Characteristic Equation and Roots

For constant coefficient equations like y'' + k²y = 0, the characteristic equation is r² + k² = 0. Solving this yields complex roots ±ik, indicating oscillatory solutions. The nature of these roots guides the form of the general solution using sine and cosine functions.

General Solution of Homogeneous Equations with Complex Roots

When the characteristic roots are complex conjugates α ± βi, the general solution is y(t) = e^{αt}(C₁ cos(βt) + C₂ sin(βt)). For purely imaginary roots (α=0), the solution simplifies to y(t) = C₁ cos(kt) + C₂ sin(kt), representing oscillations with frequency k.