(1 + X) Dy Dx − Xy = X + X2

Jasonbradley

4 min read

·

Jun 16, 2024

27

Share

Solving the Differential Equation: (1 + x) dy/dx - xy = x + x^2

This article will guide you through the process of solving the first-order linear differential equation:

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

1. Standard Form

First, we need to rewrite the equation in standard form:

dy/dx + P(x)y = Q(x)

To achieve this, divide both sides of the equation by (1 + x):

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

Now, we can identify:

• P(x) = -x/(1+x)
• Q(x) = (x + x^2)/(1+x)

2. Integrating Factor

The next step involves finding the integrating factor, which is defined as:

μ(x) = exp(∫P(x) dx)

Let's calculate the integrating factor:

μ(x) = exp(∫(-x/(1+x)) dx)

We can solve the integral using substitution (u = 1 + x):

μ(x) = exp(-∫(u-1)/u du) μ(x) = exp(-∫(1 - 1/u) du) μ(x) = exp(-(u - ln|u|)) μ(x) = exp(-(1+x - ln|1+x|)) μ(x) = (1+x)^(-1) * e^(-1-x)

3. Solving the Equation

Multiply both sides of the standard form equation by the integrating factor:

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

Notice that the left side of the equation is the derivative of the product of y and the integrating factor:

d/dx [y * (1+x)^(-1) * e^(-1-x)] = xe^(-1-x)

Now, integrate both sides with respect to x:

y * (1+x)^(-1) * e^(-1-x) = ∫xe^(-1-x) dx

Solve the integral on the right side using integration by parts (u = x, dv = e^(-1-x) dx):

y * (1+x)^(-1) * e^(-1-x) = -xe^(-1-x) - e^(-1-x) + C

Finally, solve for y to obtain the general solution:

y = -(1+x)xe^(-1-x) - (1+x)e^(-1-x) + C(1+x)e^(-1-x)

4. General Solution

Therefore, the general solution to the differential equation (1 + x) dy/dx - xy = x + x^2 is:

y = -(1+x)xe^(-1-x) - (1+x)e^(-1-x) + C(1+x)e^(-1-x)

where C is an arbitrary constant.