Solving the Differential Equation 3xy - y^2 dx x^2 - xy dy 0

In this article, we will solve the given differential equation and explore the methods and techniques involved. The differential equation provided is:

3xy - y^2 dx x^2 - xy dy 0

Step 1: Rewrite the Equation

First, let's rewrite the given differential equation in a more familiar form:

(3xy - y^2) dx (x^2 - xy) dy 0

We can then identify the functions (M(x, y)) and (N(x, y)) as follows:

(M(x, y) 3xy - y^2) and (N(x, y) x^2 - xy)

Now, let's check if the equation is exact:

(My 3x - 2y) and (Nx 2x - y)

Since (My) is not equal to (Nx), the equation is not exact.

Step 2: Finding the Integrating Factor

Next, we need to find an integrating factor (mu(x, y)) such that the equation becomes exact. We can use the following condition:

(frac{My - Nx}{N}) is a function of (x) only.

(frac{My - Nx}{N} frac{(3x - 2y) - (2x - y)}{x^2 - xy} frac{x - y}{x(x - y)} frac{1}{x})

Thus, the integrating factor (mu(x, y)) is (x).

Step 3: Multiply by the Integrating Factor

Multiplying the original equation by (mu(x, y) x), we get:

x(3xy - y^2) dx x(x^2 - xy) dy 0

This simplifies to:

(3x^2y - xy^2) dx (x^3 - x^2y) dy 0

Define new functions (P(x, y)) and (Q(x, y)) as:

(P(x, y) 3x^2y - xy^2) and (Q(x, y) x^3 - x^2y)

Now, check for exactness:

(Py 6x^2 - 2xy) and (Qx 3x^2 - 2xy Py)

Since (Py Qx), the new equation is exact.

Step 4: Solve the Exact Equation

Integrate (P(x, y)) with respect to (x):

(int (3x^2y - xy^2) dx x^3y - frac{x^2y^2}{2} h(y))

Now, differentiate this with respect to (y):

(x^3 h'(y) x^3 - x^2y)

This gives us:

(h'(y) -x^2y)

Integrate (h(y)) with respect to (y):

(h(y) -frac{x^2y^2}{2} C)

The general solution to the exact equation is:

(F(x, y) x^3y - frac{x^2y^2}{2} C 0)

This can be written as:

(2yx^3 - x^2y^2 C)

Step 5: Alternative Method Using Substitution

We can also solve the differential equation using the substitution (y vx). Then:

(frac{dy}{dx} v xfrac{dv}{dx})

Substitute (y vx):

(v xfrac{dv}{dx} frac{v^2x^2 - 3vx^2}{x^2 - vx^2})

(v xfrac{dv}{dx} frac{v^2 - 3v}{1 - v})

(xfrac{dv}{dx} frac{v^2 - 3v}{1 - v} - v)

(xfrac{dv}{dx} frac{2v^2 - 4v}{1 - v})

(int frac{v - 1}{v^2 - 2v} dv -2 int frac{dx}{x})

Using partial fraction decomposition:

(frac{v - 1}{v^2 - 2v} frac{1}{2v} frac{1}{2v - 2})

(int frac{1}{2v} frac{1}{2v - 2} dv -2 int frac{dx}{x})

(ln|v| - ln|v - 2| -4 ln|x| ln|C_0|)

(ln|v(v - 2)| ln|C_0 x^{-4}|)

(v(v - 2) C_0 x^{-4})

(left(frac{y}{x}right) left(frac{y}{x} - 2right) C_0 x^{-4})

(frac{y^2}{x^2} - 2frac{y}{x} C_0 x^{-4})

(frac{y^2 - 2xy}{x^2} C_0 x^{-4})

(y^2 - 2xy C_0 x^{-2})

(y^2 2xy frac{C_0}{x^2})

(y^2 - 2xy x^2 x^2 frac{C_0}{x^2})

((y - x)^2 x^2 frac{C_0}{x^2})

(y - x pm sqrt{x^2 frac{C_0}{x^2}})

(y x pm sqrt{x^2 frac{C_0}{x^2}})

Therefore, the general solution to the differential equation is:

(2yx^3 - x^2y^2 C)

and

(y x pm sqrt{x^2 frac{C_0}{x^2}})