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

Practice Questions

Q1
What is the solution of the equation y' + 4y = 0?
  1. y = Ce^(-4x)
  2. y = Ce^(4x)
  3. y = 4Ce^x
  4. y = Ce^(x/4)

Questions & Step-by-Step Solutions

What is the solution of the equation y' + 4y = 0?
  • Step 1: Identify the equation. We have y' + 4y = 0, where y' is the derivative of y with respect to x.
  • Step 2: Rewrite the equation. We can write it as y' = -4y.
  • Step 3: Recognize that this is a separable equation. We can separate the variables y and x.
  • Step 4: Rewrite the equation in a separable form: dy/y = -4 dx.
  • Step 5: Integrate both sides. The left side becomes ∫(1/y) dy = ln|y|, and the right side becomes ∫-4 dx = -4x + C, where C is the constant of integration.
  • Step 6: Combine the results from the integration. We have ln|y| = -4x + C.
  • Step 7: Exponentiate both sides to solve for y. This gives us |y| = e^(-4x + C) = e^C * e^(-4x).
  • Step 8: Let C' = e^C, which is also a constant. So, |y| = C' * e^(-4x).
  • Step 9: Remove the absolute value. We can write y = ±C' * e^(-4x). We can just call it y = Ce^(-4x), where C can be any constant (positive or negative).
  • Step 10: Final solution: The solution to the equation y' + 4y = 0 is y = Ce^(-4x).
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