What is the solution of the equation y' = -ky, where k is a constant?

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Question: What is the solution of the equation y\' = -ky, where k is a constant?

Options:

Correct Answer: y = Ce^(-kt)

Solution:

This is a separable equation. Integrating gives y = Ce^(-kt).

What is the solution of the equation y' = -ky, where k is a constant?

What is the solution of the equation y' = -ky, where k is a constant?

Step 1: Recognize that the equation y' = -ky is a first-order differential equation.
Step 2: Rewrite y' as dy/dt to clarify that we are dealing with a function of t.
Step 3: Rearrange the equation to separate variables: dy/y = -k dt.
Step 4: Integrate both sides: ∫(1/y) dy = ∫(-k) dt.
Step 5: The left side integrates to ln|y| and the right side integrates to -kt + C, where C is the constant of integration.
Step 6: Write the equation as ln|y| = -kt + C.
Step 7: Exponentiate both sides to solve for y: |y| = e^(-kt + C).
Step 8: Rewrite e^C as a new constant, say C', so |y| = C'e^(-kt).
Step 9: Since y can be positive or negative, we can drop the absolute value and write y = Ce^(-kt), where C can be any constant.

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