Find the solution of the differential equation y'' + 4y = 0.
Practice Questions
1 question
Q1
Find the solution of 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)
This is a second-order linear homogeneous differential equation. The characteristic equation has roots ±2i.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the solution of the differential equation y'' + 4y = 0.
Solution: This is a second-order linear homogeneous differential equation. The characteristic equation has roots ±2i.
Steps: 5
Step 1: Identify the type of differential equation. This is a second-order linear homogeneous differential equation because it involves the second derivative of y and has no external forcing term.
Step 2: Write the differential equation in standard form: y'' + 4y = 0.
Step 3: Find the characteristic equation by replacing y'' with r^2 and y with 1: r^2 + 4 = 0.
Step 4: Solve the characteristic equation for r: r^2 = -4, which gives r = ±2i.
Step 5: Since the roots are complex (±2i), the general solution of the differential equation is: y(t) = C1 * cos(2t) + C2 * sin(2t), where C1 and C2 are constants determined by initial conditions.
