My Learning Cart
?
Categories
Account

What is the solution to the differential equation y'' + 4y = 0?

₹0.0
Login to Download
  • 📥 Instant PDF Download
  • ♾ Lifetime Access
  • 🛡 Secure & Original Content

What’s inside this PDF?

Question: What is the solution to the differential equation y\'\' + 4y = 0?

Options:

  1. y = C1 cos(2x) + C2 sin(2x)
  2. y = C1 e^(2x) + C2 e^(-2x)
  3. y = C1 e^(x) + C2 e^(-x)
  4. y = C1 sin(2x) + C2 cos(2x)

Correct Answer: y = C1 cos(2x) + C2 sin(2x)

Solution: 

This is a second-order linear homogeneous differential equation. The characteristic equation has roots ±2i, leading to the solution y = C1 cos(2x) + C2 sin(2x).

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?
  1. y = C1 cos(2x) + C2 sin(2x)
  2. y = C1 e^(2x) + C2 e^(-2x)
  3. y = C1 e^(x) + C2 e^(-x)
  4. 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.
No concepts available.
Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely
Home Practice Performance eBooks