Find the general solution of the equation y'' - 3y' + 2y = 0.
Practice Questions
Q1
Find the general solution of the equation y'' - 3y' + 2y = 0.
y = C1 e^(2x) + C2 e^(x)
y = C1 e^(x) + C2 e^(2x)
y = C1 e^(3x) + C2 e^(0)
y = C1 e^(0) + C2 e^(3x)
Questions & Step-by-Step Solutions
Find the general solution of the equation y'' - 3y' + 2y = 0.
Step 1: Write down the given differential equation: y'' - 3y' + 2y = 0.
Step 2: Identify the characteristic equation by replacing y'' with r^2, y' with r, and y with 1. This gives us r^2 - 3r + 2 = 0.
Step 3: Factor the characteristic equation. Look for two numbers that multiply to 2 and add to -3. The factors are (r - 1)(r - 2) = 0.
Step 4: Set each factor equal to zero to find the roots: r - 1 = 0 gives r = 1, and r - 2 = 0 gives r = 2.
Step 5: Write the general solution using the roots found. The general solution is y = C1 e^(1x) + C2 e^(2x), where C1 and C2 are constants.
Homogeneous Linear Differential Equations – The question tests the ability to solve second-order homogeneous linear differential equations with constant coefficients.
Characteristic Equation – It assesses the understanding of deriving the characteristic equation from the differential equation and solving it for roots.
General Solution – The question evaluates the ability to construct the general solution from the roots of the characteristic equation.
