What is the solution of the differential equation y' = 2y + 3?
Practice Questions
Q1
What is the solution of the differential equation y' = 2y + 3?
y = Ce^(2x) - 3/2
y = Ce^(2x) + 3/2
y = 3e^(2x)
y = 2e^(x) + C
Questions & Step-by-Step Solutions
What is the solution of the differential equation y' = 2y + 3?
Step 1: Write the differential equation in standard form: y' - 2y = 3.
Step 2: Identify the coefficient of y, which is -2.
Step 3: Calculate the integrating factor, which is e^(∫-2dx) = e^(-2x).
Step 4: Multiply the entire differential equation by the integrating factor: e^(-2x)y' - 2e^(-2x)y = 3e^(-2x).
Step 5: Recognize that the left side is the derivative of (e^(-2x)y). So, rewrite it: d/dx(e^(-2x)y) = 3e^(-2x).
Step 6: Integrate both sides with respect to x: ∫d/dx(e^(-2x)y)dx = ∫3e^(-2x)dx.
Step 7: The left side simplifies to e^(-2x)y. For the right side, use integration by parts or a table to find the integral: ∫3e^(-2x)dx = -3/2 e^(-2x) + C.
Step 8: Set the two sides equal: e^(-2x)y = -3/2 e^(-2x) + C.
Step 9: Multiply both sides by e^(2x) to solve for y: y = Ce^(2x) - 3/2.
Step 10: Rearrange the equation to get the final solution: y = Ce^(2x) + 3/2.
