What is the solution to the differential equation y'' + 4y = 0?
Practice Questions
Q1
What is the solution to the differential equation y'' + 4y = 0?
y = C1 cos(2x) + C2 sin(2x)
y = C1 e^(2x) + C2 e^(-2x)
y = C1 e^(x) + C2 e^(-x)
y = C1 sin(2x) + C2 cos(2x)
Questions & Step-by-Step Solutions
What is the solution to the differential equation y'' + 4y = 0?
Step 1: Identify the type of equation. The equation y'' + 4y = 0 is a second-order linear homogeneous differential equation.
Step 2: Write the characteristic equation. Replace y'' with r^2 and y with 1 to get the characteristic equation: r^2 + 4 = 0.
Step 3: Solve the characteristic equation for r. Rearranging gives r^2 = -4, so r = ±2i.
Step 4: Use the roots to write the general solution. Since the roots are complex (±2i), the solution is of the form y = C1 cos(2x) + C2 sin(2x), where C1 and C2 are constants.
