A second-order equation Consider the differential equation y''(t) - k²y(t) = 0 where k > 0 is a real number.


a. Verify by substitution that when k = 1, a solution of the equation is y(t) = C₁eᵗ + C₂e⁻ᵗ. You may assume this function is the general solution.

Verified step by step guidance

1. Start by writing down the given differential equation for the case when \( k = 1 \): \[ y''(t) - y(t) = 0 \]
2. Write the proposed solution for \( k = 1 \): \[ y(t) = C_1 e^{t} + C_2 e^{-t} \]
3. Compute the first derivative \( y'(t) \) of the proposed solution: \[ y'(t) = C_1 e^{t} - C_2 e^{-t} \]
4. Compute the second derivative \( y''(t) \) of the proposed solution: \[ y''(t) = C_1 e^{t} + C_2 e^{-t} \]
5. Substitute \( y(t) \) and \( y''(t) \) back into the differential equation and simplify: \[ y''(t) - y(t) = (C_1 e^{t} + C_2 e^{-t}) - (C_1 e^{t} + C_2 e^{-t}) = 0 \] Since this holds true for all \( t \), the proposed function is indeed a solution.

Key Concepts

Second-Order Linear Differential Equations

These are differential equations involving the second derivative of a function. They often have the form y'' + p(t)y' + q(t)y = 0. Solutions typically involve characteristic equations that help find general solutions based on roots.

Characteristic Equation and Its Roots

For constant-coefficient linear differential equations like y'' - k²y = 0, the characteristic equation is r² - k² = 0. Solving this quadratic gives roots that determine the form of the general solution, such as exponential functions when roots are real.

Verification by Substitution

To verify a proposed solution, substitute it and its derivatives into the original differential equation. If the equation holds true for all t, the function is a valid solution. This method confirms the correctness of the general solution.