(x+2y)dx+ydy=0

4 min read Jun 16, 2024
(x+2y)dx+ydy=0

Solving the Differential Equation (x + 2y)dx + ydy = 0

This article will explore the solution of the differential equation (x + 2y)dx + ydy = 0. This type of equation is classified as a first-order homogeneous differential equation, where the coefficients of dx and dy are functions of x and y with the same degree.

Homogeneous Equation Recognition

Firstly, let's confirm that the equation is indeed homogeneous. We can rewrite the equation as:

M(x, y)dx + N(x, y)dy = 0 

Where:

  • M(x, y) = x + 2y
  • N(x, y) = y

Now, we check if the equation is homogeneous by examining the degree of M(x, y) and N(x, y). Both terms have a degree of 1 since the powers of x and y are 1. This confirms that the equation is homogeneous.

Solving the Homogeneous Equation

To solve this equation, we use the substitution:

y = vx

Differentiating both sides with respect to x, we get:

dy/dx = v + x(dv/dx)

Now, we substitute y and dy/dx in the original equation:

(x + 2vx)dx + vx(v + x(dv/dx))dx = 0

Simplifying the equation:

(1 + 2v)dx + v^2 dx + vx^2 (dv/dx)dx = 0

(1 + 2v + v^2)dx + vx^2 dv = 0

This is a separable differential equation, where we can separate the variables x and v:

dx / (vx^2) = -dv / (1 + 2v + v^2)

Integrating both sides:

∫(1/x^2)dx - ∫(v/(1+2v+v^2))dv = C

The integral on the left side can be solved directly, while the integral on the right side requires a partial fraction decomposition. The result after integration is:

-1/x + 1/2 * ln(v^2 + 2v + 1) - 1/2 * ln(1) = C

Now, we need to substitute back v = y/x:

-1/x + 1/2 * ln((y/x)^2 + 2(y/x) + 1) = C

Finally, we can rearrange and simplify the equation to obtain the general solution of the differential equation:

ln((y + x)^2) = 2C + 2/x

Conclusion

This article demonstrates how to solve a first-order homogeneous differential equation using substitution and integration techniques. The solution process involves recognizing the homogeneous nature of the equation, applying a suitable substitution, and performing integration to obtain the general solution. This technique is widely applicable for solving similar types of differential equations.

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