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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.