What is the particular solution of the equation dy/dx = 2y with the initial condition y(0) = 1?

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Q: What is the particular solution of the equation dy/dx = 2y with the initial condition y(0) = 1?

Solution: The general solution is y = Ce^(2x). Using the initial condition y(0) = 1, we find C = 1.

Steps: 12

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).