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Find the solution of the equation y' + 2y = 0.
Find the solution of the equation y' + 2y = 0.
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Practice Questions
1 question
Q1
Find the solution of the equation y' + 2y = 0.
y = Ce^(-2x)
y = Ce^(2x)
y = 2Ce^x
y = Ce^x
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This is a first-order linear differential equation. The solution is y = Ce^(-2x).
Questions & Step-by-step Solutions
1 item
Q
Q: Find the solution of the equation y' + 2y = 0.
Solution:
This is a first-order linear differential equation. The solution is y = Ce^(-2x).
Steps: 10
Show Steps
Step 1: Identify the equation. The equation is y' + 2y = 0.
Step 2: Recognize that this is a first-order linear differential equation.
Step 3: Rewrite the equation in standard form: y' = -2y.
Step 4: Separate the variables by dividing both sides by y: dy/y = -2 dx.
Step 5: Integrate both sides. The left side becomes ln|y| and the right side becomes -2x + C (where C is the constant of integration).
Step 6: Write the result of the integration: ln|y| = -2x + C.
Step 7: Exponentiate both sides to solve for y: |y| = e^(-2x + C).
Step 8: Simplify the right side: |y| = e^C * e^(-2x).
Step 9: Let C' = e^C (a new constant), so |y| = C' * e^(-2x).
Step 10: Remove the absolute value by allowing C' to be any real number (positive or negative): y = Ce^(-2x), where C is a constant.
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