How do I solve this D.E. using integrating factor: y(8x-9y)dx + 2x(x-3y)dy= 0 ans:x^3(2x-3y)y=c but solution is missing? tnx?

Let M = y(8x-9y) = 8xy - 9y^2 , N = 2x(x-3y) = 2x^2 - 6xy
∂M/∂y = 8x - 18y , ∂N/∂x = 4x - 6y
∂M/∂y ≠ ∂N/∂x, so the eq. is not an exact eq.

To find the integrating factor, we use the following rule:
M(x,y) dx + N(x,y) dy = 0 is not an exact eq.
Let U = ∂M/∂y - ∂N/∂x
(1) If U/N = f(x) , then I = I(x) = e^( ∫ f(x)dx )
(2) If U/M = g(y) , then I = I(y) = e^( - ∫ g(y)dy )

Now,
U / N
= ( ∂M/∂y - ∂N/∂x ) / ( 2x^2 - 6xy )
= ( 4x - 12y ) / ( 2x^2 - 6xy )
= [ 4(x-3y) ] / [ 2x(x-3y) ]
= 2 / x
≡ f(x)

by part (1) of the rule, the integrating factor is
I
= I(x)
= e^( ∫ f(x)dx )
= e^( ∫ ( 2 / x ) dx )
= e^( 2 ln x )
= e^( ln x^2 )
= x^2

Multiply the eq. by x^2 , we have
( x^2 )*( 8xy - 9y^2 ) dx + ( x^2 )*( 2x^2 - 6xy ) dy = 0
( 8x^3 y - 9x^2 y^2 ) dx + ( 2x^4 - 6x^3 y ) dy = 0

Checking if the new eq. is exact :
Now, M = 8x^3 y - 9x^2 y^2 , N = 2x^4 - 6x^3 y
∂M/∂y = 8x^3 - 18x^2 y
∂N/∂x = 8x^3 - 18x^2 y
∂M/∂y = ∂N/∂x, so the new eq. is exact.

Suppose the solution is Φ(x,y) = c1 , then
∂Φ/∂x = 8x^3 y - 9x^2 y^2 , ∂Φ/∂y = 2x^4 - 6x^3 y
we start with ∂Φ/∂y = 2x^4 - 6x^3 y , so
Φ = 2x^4 y - 3x^3 y^2 + h(x)

∂Φ/∂x
= 8x^3 y - 9x^2 y^2 + h'(x)
= 8x^3 y - 9x^2 y^2

h'(x) = 0 , so h(x) = c2 , and hence
Φ = 2x^4 y - 3x^3 y^2 + h(x) = 2x^4 y - 3x^3 y^2 + c2
and Φ(x,y) = c1 , thus
Φ = 2x^4 y - 3x^3 y^2 + c2 = c1
2x^4 y - 3x^3 y^2 + c2 = c1 - c2 = c
x^3 * y * ( 2x - 3y ) = c

Ans: x^3( 2x - 3y )y = c