What is the solution to the initial value problem dy/dx = 4y, y(1) = 2?

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Question: What is the solution to the initial value problem dy/dx = 4y, y(1) = 2?

Options:

Correct Answer: y = 2e^(4x-4)

Solution:

The general solution is y = Ce^(4x). Using y(1) = 2, we find C = 2e^(-4).

What is the solution to the initial value problem dy/dx = 4y, y(1) = 2?

What is the solution to the initial value problem dy/dx = 4y, y(1) = 2?

Step 1: Start with the differential equation dy/dx = 4y.
Step 2: Recognize that this is a separable equation, meaning we can separate y and x.
Step 3: Rewrite the equation as dy/y = 4 dx.
Step 4: Integrate both sides. The left side becomes ln|y| and the right side becomes 4x + C (where C is a constant).
Step 5: After integrating, we have ln|y| = 4x + C.
Step 6: Exponentiate both sides to solve for y. This gives us y = e^(4x + C) = e^(4x) * e^C.
Step 7: Let e^C be a new constant, which we can call C. So, we have y = Ce^(4x).
Step 8: Now, we use the initial condition y(1) = 2 to find the value of C.
Step 9: Substitute x = 1 into the equation: 2 = Ce^(4*1) = Ce^4.
Step 10: Solve for C: C = 2/e^4.
Step 11: Therefore, the solution to the initial value problem is y = (2/e^4)e^(4x) = 2e^(4x - 4).

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