What is the solution to the equation y' + 3y = 0?

What is the solution to the equation y' + 3y = 0?

Step 1: Identify the equation. The equation is y' + 3y = 0.
Step 2: Recognize that this is a first-order linear differential equation.
Step 3: Rearrange the equation to isolate y'. This gives us y' = -3y.
Step 4: Use the method of separation of variables. Rewrite it as dy/dx = -3y.
Step 5: Separate the variables by dividing both sides by y and multiplying both sides by dx. This gives us (1/y) dy = -3 dx.
Step 6: Integrate both sides. The left side becomes ln|y| and the right side becomes -3x + C, where C is the constant of integration.
Step 7: Write the result of the integration: ln|y| = -3x + C.
Step 8: Exponentiate both sides to solve for y. This gives us |y| = e^(-3x + C).
Step 9: Rewrite e^C as a new constant, which we can call C. So, |y| = Ce^(-3x).
Step 10: Since y can be positive or negative, we can drop the absolute value, giving us y = Ce^(-3x).

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