(x + Yey/x) Dx − Xey/x Dy = 0 Y(1) = 0

4 min read Jun 16, 2024

Solving the Differential Equation (x + yey/x) dx − xey/x dy = 0 with Initial Condition y(1) = 0

This article will guide you through the process of solving the given differential equation and applying the initial condition to find a particular solution.

1. Identifying the Type of Differential Equation

The equation (x + yey/x) dx − xey/x dy = 0 is a first-order homogeneous differential equation. This is because it can be rewritten in the form:

dy/dx = f(y/x)

where f(y/x) = (x + yey/x) / (xey/x).

2. Solving the Homogeneous Equation

To solve this type of equation, we use the substitution u = y/x. This leads to:

y = ux
dy = udx + xdu

Substituting these expressions into the original differential equation and simplifying, we get:

(x + uex)dx - xuex(udx + xdu) = 0

Rearranging and factoring out dx:

dx(x + uex - u^2x^2ex) - x^2uexdu = 0

Now we can separate the variables:

dx/(x + uex - u^2x^2ex) = x^2uexdu

Integrating both sides, we obtain:

∫dx/(x + uex - u^2x^2ex) = ∫x^2uexdu

The left-hand side integral is a bit complex. It can be solved using partial fraction decomposition, but for simplicity, let's focus on the right-hand side for now. We can integrate the right-hand side using integration by parts.

3. Integration by Parts

Let:

u = u
dv = x^2ex du

Then:

du = du
v = x^2ex - 2xex + 2ex

Applying integration by parts:

∫x^2uexdu = uv - ∫vdu
           = u(x^2ex - 2xex + 2ex) - ∫(x^2ex - 2xex + 2ex)du
           = u(x^2ex - 2xex + 2ex) - (x^2ex - 2xex + 2ex) + C

4. Back Substitution and Solving for the Constant

Now we substitute u = y/x back into the solution and solve for C using the initial condition y(1) = 0:

(y/x)(x^2ex - 2xex + 2ex) - (x^2ex - 2xex + 2ex) + C = ∫dx/(x + uex - u^2x^2ex)

Applying y(1) = 0:

0 - (e - 2e + 2e) + C = ∫dx/(x + uex - u^2x^2ex)
C = 3e

5. The Final Solution

Therefore, the particular solution to the given differential equation with the initial condition is:

(y/x)(x^2ex - 2xex + 2ex) - (x^2ex - 2xex + 2ex) + 3e = ∫dx/(x + uex - u^2x^2ex)

While this solution may look complex, it represents the implicit relationship between x and y that satisfies the differential equation and the initial condition.

Note: The left-hand side integral is still required to complete the solution explicitly. Its integration can be challenging, and the solution might involve special functions.