dx%2B(2x%2By)dy%3D0.jpeg "(x-2y)dx+(2x+y)dy=0")
Solving the Differential Equation (x-2y)dx + (2x+y)dy = 0
This article will explore the solution process for the given differential equation:
(x-2y)dx + (2x+y)dy = 0
This equation is a first-order homogeneous differential equation. Let's break down the steps to find its solution.
1. Identifying Homogeneity
A differential equation is considered homogeneous if it can be written in the form:
M(x,y)dx + N(x,y)dy = 0
Where M(x,y) and N(x,y) are homogeneous functions of the same degree. This means that if we replace x and y with tx and ty respectively, the function remains unchanged except for a factor of t raised to the degree of homogeneity.
In our equation, we can see that both (x-2y) and (2x+y) are homogeneous functions of degree 1. Therefore, the given equation is a homogeneous differential equation.
2. Substitution and Simplification
To solve homogeneous equations, we introduce a substitution:
y = vx
Differentiating both sides with respect to x, we get:
dy/dx = v + x(dv/dx)
Now, we can substitute y and dy/dx in the original equation:
(x - 2vx)dx + (2x + vx)(v + x(dv/dx))dy = 0
Simplifying the equation:
x(1 - 2v)dx + x(2 + v)(v + x(dv/dx)) = 0
Dividing both sides by x and rearranging terms:
(1 - 2v)dx + (2 + v)(v + x(dv/dx)) = 0
(1 - 2v)dx + (2v + v^2 + 2x + vx(dv/dx)) = 0
dx + (2v + v^2 + 2x + vx(dv/dx)) = 0
3. Separating Variables
Now, we can separate the variables x and v:
dx + (2v + v^2)dx + vx(dv/dx) = 0
(1 + 2v + v^2)dx + vx(dv/dx) = 0
(1 + 2v + v^2)dx = -vx(dv/dx)
(-1/v)(1 + 2v + v^2)dx = dv
-(1/v)(1 + v)^2 dx = dv
4. Integration
We can now integrate both sides of the equation:
∫(-1/v)(1 + v)^2 dx = ∫dv
∫(-1/v)dv + ∫(-2/v)dv + ∫(-1)dv = ∫dv
-ln|v| - 2ln|v| - v = v + C
-3ln|v| - 2v = C
5. Back Substitution
Substituting back v = y/x, we get:
-3ln|y/x| - 2(y/x) = C
This is the general solution to the given differential equation. We can write it in a more compact form:
ln|y/x|^3 + 2y/x = -C
Conclusion
We successfully solved the first-order homogeneous differential equation (x-2y)dx + (2x+y)dy = 0 using the substitution method and integration. The general solution to this equation is ln|y/x|^3 + 2y/x = -C, where C is an arbitrary constant.