Find the particular solution of dy/dx = x + y, given y(0) = 1.

Find the particular solution of dy/dx = x + y, given y(0) = 1.

Step 1: Start with the differential equation dy/dx = x + y.
Step 2: Rearrange the equation to isolate dy/dx on one side: dy/dx - y = x.
Step 3: This is a first-order linear differential equation. We can solve it using an integrating factor.
Step 4: The integrating factor is e^(-x) because it is derived from the coefficient of y.
Step 5: Multiply the entire equation by the integrating factor e^(-x): e^(-x) * dy/dx - e^(-x) * y = e^(-x) * x.
Step 6: The left side can be rewritten as the derivative of (e^(-x) * y).
Step 7: So, we have d/dx(e^(-x) * y) = e^(-x) * x.
Step 8: Integrate both sides with respect to x: ∫d/dx(e^(-x) * y) dx = ∫e^(-x) * x dx.
Step 9: The left side simplifies to e^(-x) * y. For the right side, use integration by parts to find the integral.
Step 10: After integrating, we find e^(-x) * y = -e^(-x) * (x + 1) + C.
Step 11: Multiply both sides by e^(x) to solve for y: y = - (x + 1) + C * e^(x).
Step 12: This simplifies to y = e^(x) + C - 1.
Step 13: The general solution is y = e^(x) + C.
Step 14: Now, use the initial condition y(0) = 1 to find C: 1 = e^(0) + C, which gives C = 1.
Step 15: Substitute C back into the general solution: y = e^(x) + 1.

First-Order Linear Differential Equations – The question tests the ability to solve a first-order linear differential equation using an integrating factor and apply initial conditions to find a particular solution.
Initial Value Problems – The question assesses the understanding of how to apply initial conditions to determine the constant in the general solution.