Solving the FirstOrder Linear Differential Equation: (x+1)dy/dx + (x+2)y = 2xe^x
This article will guide you through the process of solving the firstorder linear differential equation: (x+1)dy/dx + (x+2)y = 2xe^x.
Understanding the Equation
The equation is a firstorder linear differential equation because:
 It involves the first derivative of the dependent variable y (dy/dx).
 The dependent variable y and its derivative appear linearly (not raised to any power other than 1).
Solving the Equation
The standard form of a firstorder linear differential equation is:
dy/dx + P(x)y = Q(x)
To solve our given equation, we need to rewrite it in this standard form:

Divide both sides by (x+1): dy/dx + (x+2)/(x+1)y = 2xe^x / (x+1)

Identify P(x) and Q(x): P(x) = (x+2)/(x+1) Q(x) = 2xe^x / (x+1)
Now we can use the integrating factor method to solve the equation.

Calculate the integrating factor: The integrating factor is given by: exp(∫P(x)dx)
Therefore, the integrating factor for our equation is: exp(∫(x+2)/(x+1)dx) = exp(x + lnx+1) = (x+1)e^x

Multiply both sides of the equation by the integrating factor: (x+1)e^x dy/dx + (x+1)e^x * (x+2)/(x+1)y = 2xe^x * (x+1)e^x / (x+1)
Simplifying, we get: d/dx [(x+1)e^x y] = 2xe^0

Integrate both sides with respect to x: (x+1)e^x y = ∫2x dx (x+1)e^x y = x^2 + C

Solve for y: y = (x^2 + C) / [(x+1)e^x]
The General Solution
The general solution to the differential equation is: y = (x^2 + C) / [(x+1)e^x], where C is an arbitrary constant.
Conclusion
By following the steps above, we successfully solved the firstorder linear differential equation (x+1)dy/dx + (x+2)y = 2xe^x. The solution represents a family of curves, each corresponding to a different value of the constant C.