(x+1)dy/dx+(x+2)y=2xe^-x

4 min read Jun 16, 2024

Solving the First-Order Linear Differential Equation: (x+1)dy/dx + (x+2)y = 2xe^-x

This article will guide you through the process of solving the first-order linear differential equation: (x+1)dy/dx + (x+2)y = 2xe^-x.

Understanding the Equation

The equation is a first-order 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 first-order 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 + ln|x+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 first-order 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.