(2y^2-6xy)dx+(3xy-4x^2)dy=0

4 min read Jun 16, 2024

Solving the Differential Equation (2y^2 - 6xy)dx + (3xy - 4x^2)dy = 0

This article will explore the solution process for the given differential equation.

The given equation is a first-order homogeneous differential equation. This can be identified because:

It is a first-order equation: The highest derivative present is the first derivative (dy/dx).
It is homogeneous: The equation can be rewritten in the form M(x,y)dx + N(x,y)dy = 0, where M and N are homogeneous functions of the same degree. In this case, both M = 2y^2 - 6xy and N = 3xy - 4x^2 are homogeneous of degree 2.

To solve this type of equation, we follow these steps:

Substitution: Introduce a new variable, say v, such that y = vx. This implies dy = vdx + xdv.
Substitution into the Equation: Substitute y and dy in terms of v and x into the original equation.
Simplification: The equation will become separable in terms of x and v.
Integration: Integrate both sides of the equation.
Substitution Back: Replace v with y/x to obtain the solution in terms of x and y.

Let's apply these steps to our equation:

Substitution: y = vx, dy = vdx + xdv
Substitution into the Equation:

(2(vx)^2 - 6x(vx))dx + (3x(vx) - 4x^2)(vdx + xdv) = 0
Simplification:

(2v^2x^2 - 6v^2x^2)dx + (3v^2x^2 - 4x^2)(vdx + xdv) = 0 -4v^2x^2dx + (3v^3x^2 - 4v^2x^2)dx + (3v^3x^3 - 4v^2x^3)dv = 0 (3v^3x^3 - 4v^2x^3)dv = -4v^2x^2dx (3v - 4)dv = -4(dx/x)
Integration:

∫(3v - 4)dv = ∫(-4/x)dx (3/2)v^2 - 4v = -4ln|x| + C
Substitution Back:

(3/2)(y/x)^2 - 4(y/x) = -4ln|x| + C

The solution to the differential equation (2y^2 - 6xy)dx + (3xy - 4x^2)dy = 0 is given by:

3y^2 - 8xy + 8x^2ln|x| = Cx^2

where C is an arbitrary constant.

This solution represents a family of curves that satisfy the original differential equation.