What is the particular solution of the equation dy/dx = 2y with the initial cond

Practice Questions

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

Questions & Step-by-Step Solutions

What is the particular solution of the equation dy/dx = 2y with the initial condition y(0) = 1?
  • Step 1: Start with the differential equation dy/dx = 2y.
  • Step 2: Recognize that this is a separable equation, which means we can separate the variables 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: After integrating, we have ln|y| = 2x + C.
  • Step 6: Exponentiate both sides to solve for y. This gives us 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: We can rename C' as C, so the general solution is y = Ce^(2x).
  • Step 9: Now, use the initial condition y(0) = 1 to find the value of C.
  • Step 10: Substitute x = 0 into the general solution: y(0) = Ce^(2*0) = C * 1 = C.
  • Step 11: Set C equal to 1 because y(0) = 1.
  • Step 12: Therefore, the particular solution is y = e^(2x).
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