Find the general solution of the equation y' = 3y + 2.

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Q: Find the general solution of the equation y' = 3y + 2.

Solution: This is a first-order linear differential equation. The integrating factor is e^(-3x).

Steps: 11

Step 1: Identify the equation. We have y' = 3y + 2, which is a first-order linear differential equation.
Step 2: Rewrite the equation in standard form. This means we want it to look like y' - 3y = 2.
Step 3: Find the integrating factor. The integrating factor is e^(∫-3dx) = e^(-3x).
Step 4: Multiply the entire equation by the integrating factor. This gives us e^(-3x)y' - 3e^(-3x)y = 2e^(-3x).
Step 5: Recognize the left side as the derivative of a product. The left side can be written as d/dx(e^(-3x)y).
Step 6: Integrate both sides. ∫d/dx(e^(-3x)y) dx = ∫2e^(-3x) dx.
Step 7: Solve the left side. This gives us e^(-3x)y = ∫2e^(-3x) dx.
Step 8: Calculate the integral on the right side. The integral ∫2e^(-3x) dx = -2/3 e^(-3x) + C, where C is the constant of integration.
Step 9: Combine the results. We have e^(-3x)y = -2/3 e^(-3x) + C.
Step 10: Solve for y. Multiply both sides by e^(3x) to isolate y: y = -2/3 + Ce^(3x).
Step 11: Write the final general solution. The general solution is y = -2/3 + Ce^(3x), where C is a constant.