Homogeneous differential equation help please.?

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

My = 2y, Nx = 6y

not exact,

so I'll gonna use Integrating Factor

..My - Nx

----------------

......N

....2y - 6y

=---------------

.......6xy

.... - 4y

= ----------

....6xy

.......2

=- -------

......3x

If = e^[-2/3∫dx/x] = e^[-2/3ln(x)] = e^[ln(x)^(-2/3)] = x^(-2/3)

when µ = x^(-2/3)

then we multiply x^(-2/3) both side of the equation we get:

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

.................y^2

[x^(4/3) + ------------]dx + 6x^(1/3)ydy = 0

................x^(2/3)

...........2y

My = ------------

..........x^(2/3)

..............................2y

Nx = 2x^(-2/3)y = --------------

..............................x^(2/3)

My = Nx Exact!

......................y^2

Ψ = x^(4/3) + -------------

......................x^(2/3)

∫ Ψxdx = ∫ [x^(4/3) + x^(-2/3)y^2]dx + f(y)

.......x^(7/3)........x^(1/3)y^2

Ψ =------------- + ------------------- + f(y)

........ 7/3............1/3

Ψ = 3/7x^(7/3) + 3x^(1/3)y^2 + f(y)

∂Ψ/∂y = 6x^(1/3)y + f'(y) = 6x^(1/3)y

f'(y) = 6x^(1/3)y - 6x^(1/3)y

f'(y) = 0

f(y) = C

Ψ(x.y) = 3/7x^(7/3) + 3x^(1/3)y^2 + C

3/7x^(7/3) + 3x^(1/3)y^2 = C answer//

check by using implicit differentiation

x^(4/3) + 3[y^2* 1/3x^(-2/3) + x^(1/3) * 2ydy/dx] = 0

x^(4/3) + 3[1/3x^(-2/3)y^2 + 2x^(1/3)ydy/dx] = 0

x^(4/3) + x^(-2/3)y^2 + 6x^(1/3)ydy/dx = 0

.................y^2

[x^(4/3) + -----------]dx + 6x^(1/3)ydy = 0 prove!!!

................x^(2/3)

There is no need to change the letter used for the constant as the above answerer has done, since the constant is arbitrary it can be modified in anyway you want.

Make a substitution to form a new and simpler differential equation:

(x² + y²) dx + 6xy dy = 0

x² + y² + 6xy(dy / dx) = 0

6xy(dy / dx) = -(x² + y²)

dy / dx = -(x² + y²) / (6xy)

dy / dx = -(1 + y² / x²) / (6y / x)

dy / dx = -[1 + (y / x)²] / [6(y / x)]

Let u = y / x,

y = ux

dy / dx = x(du / dx) + u

x(du / dx) + u = -(1 + u²) / (6u)

Solve this differential equation by separating the variables then integrating:

x(du / dx) = -(1 + u²) / (6u) - u

x(du / dx) = -(1 + u²) / (6u) - 6u² / (6u)

x(du / dx) = -(1 + 7u²) / (6u)

6u / (1 + 7u²) du = -1 / x dx

3 ∫ 14u / (1 + 7u²) du / 7 = - ∫ 1 / x dx

3ln(1 + 7u²) / 7 = -ln|x| + C

3ln(1 + 7u²) / 7 = C - ln|x|

ln(1 + 7u²) = C - 7ln|x| / 3

1 + 7u² = ℮^(C - 7ln|x| / 3)

1 + 7u² = ℮^(C - ln|∛x⁷|)

1 + 7u² = ℮ᶜ / ∛x⁷

1 + 7u² = C / ∛x⁷

7u² = C / ∛x⁷ - 1

u² = C - ∛x⁷ - 1 / 7

u = ±√(C - ∛x⁷ - 1 / 7)

Find the solution by substituting back for the previous variables:

Since u = y / x,

y / x = ±√(C - ∛x⁷ - 1 / 7)

y = ±x√(C / ∛x⁷ - 1 / 7)

dy/dx = -(x^2 + y^2)/(6xy) = -(1 + u^2)/6u

Where u(x) = y(x)/x, and dy/dx = d(ux)/dx = x du/dx + u

x du/dx + u = -(1+u^2)/6u

x du/dx = -(1+u^2+6u^2)/6u = -(1+7u^2)/6u

6u/(1+7u^2) du = -dx/x

Integrating both sides, we have

3 ln(1+7u^2) / 7 = -ln x + C

1 + 7u^2 = Ae(-7lnx/3) = Ax^(-7/3)

u = sqrt( (Ax^(7/3)-1) / 7 )

y = x sqrt( (Ax^(7/3)-1)/7 )

In your answer substitute v = y/x and simplify using laws of logarithm.