What is the particular solution of the equation dy/dx = 2y with y(0) = 5?

Practice Questions

Q1
What is the particular solution of the equation dy/dx = 2y with y(0) = 5?
  1. y = 5e^(2x)
  2. y = 2e^(2x)
  3. y = 5e^(-2x)
  4. y = 5 + 2x

Questions & Step-by-Step Solutions

What is the particular solution of the equation dy/dx = 2y with y(0) = 5?
  • Step 1: Start with the differential equation dy/dx = 2y.
  • Step 2: Recognize that this is a separable equation, meaning we can separate y and x.
  • Step 3: Rewrite the equation as dy/y = 2 dx.
  • Step 4: Integrate both sides. The left side becomes ln|y| and the right side becomes 2x + C (where C is a constant).
  • Step 5: Write the result of the integration: ln|y| = 2x + C.
  • Step 6: Exponentiate both sides to solve for y: y = e^(2x + C).
  • Step 7: Rewrite e^(2x + C) as y = e^(2x) * e^C. Let C' = e^C, so y = C'e^(2x).
  • Step 8: The general solution is y = Ce^(2x), where C is a constant.
  • Step 9: Use the initial condition y(0) = 5 to find the specific value of C.
  • Step 10: Substitute x = 0 into the general solution: y(0) = Ce^(2*0) = C.
  • Step 11: Set C equal to 5 since y(0) = 5, so C = 5.
  • Step 12: The particular solution is y = 5e^(2x).
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