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is (2xy^4e^y+2xy^3+y)dx+(x^2y^4e^y-x^2y^2-3x)dy=0 exact, if yes solve it or else find the intergrating factor which makes the equation exact

The DE is,

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

Or,

M.dx + N.dy = 0.

Where M(x,y) = 2xy^4e^y + 2xy^3 + y, and N(x,y) = x^2y^4e^y - x^2y^2 - 3x

δM/δy = 8xy^3.e^y + 2xy^4.e^y + 6xy^2 + 1 ,      δN/δx = 2xy^4.e^y – 2xy^2 – 3

δM/δy =2xy^3.e^y(4 + y) + 6xy^2 + 1 ,    δN/δx = 2xy^2(y^2.e^y – 1) – 3

Since δM/δy ≠ δN/δx, then the DE is not exact.

We need to use an integrating factor. Let that be the function λ(x, y). Multiply all terms of the exact differential by λ(x, y) to give.

P.dx + Q.dy = 0

 

Where P(x, y) = λ(x, y).M(x, y) and Q(x, y) = λ(x, y).N(x, y)

To be exact we now need δP/δy = δQ/δx. i.e.

λ. δM/δy + M. δλ/δy = λ. δN/δx + N. δλ/δx

 

Assume λ(x, y) = λ(x) => δλ/δy = 0, then

δλ/δx  = λ( δM/δy – δN/δx)/N.

Substituting for δM/δy, δN/δx and N,

δλ/δx  = λ(8xy^3.e^y + 2xy^4.e^y + 6xy^2 + 1  – 2xy^4.e^y + 2xy^2 + 3)/( x^2y^4e^y - x^2y^2 - 3x)

δλ/δx  = λ(8xy^3.e^y + 8xy^2 + 4)/( x^2y^4e^y - x^2y^2 - 3x)

The above rational function does nor simplify, meaning λ() cannot be a function of x-only.

 

Assume λ(x, y) = λ(y) => δλ/δx = 0, then

δλ/δy  = λ( δN/δx – δM/δy)/M.

Substituting for δM/δy, δN/δx and M,

δλ/δy  = -λ(8xy^3.e^y + 8xy^2 + 4)/( 2xy^4e^y + 2xy^3 + y)

δλ/δy  = -λ4(2xy^3.e^y + 2xy^2 + 1)/[y( 2xy^3e^y + 2xy^2 + 1)]

δλ/δy  = -4λ/y

dλ/dy  = -4λ/y

dλ/ λ  = -4dy/y

integrating,

ln(λ) = -4.ln(y)

λ = y^(-4)

i.e. λ(x,y) = λ(y) = y^(-4)

This gives us,

P = λ.M = y^(-4)(2xy^4e^y+2xy^3+y) = 2xe^y + 2x/y + 1/y^3 = δU/δx

and Q = λ.N = y^(-4)( x^2y^4e^y-x^2y^2-3x) = x^2e^y - x^2/y^2 - 3x/y^4 = δU/δy

Integrating partially,

(from P): U(x,y) = x^2.e^y + x^2/y^2 + x/y^3 + f(y)

(from Q): U(x,y) = x^2.e^y + x^2/y^2 + x/y^3 + g(x)

By comparison, f(y) = g(x) = 0.

Our solution then is: U(x,y) = x^2.e^y + x^2/y^2 + x/y^3 = const

by Level 11 User (81.5k points)

I was coming back to continue, once I discovered it wasn't an exact DE. I was looking for an integrating factor, but you beat me to it. Thanks for your solution. I don't need to follow up mine, so instead I'll study your solution with interest!

Ah well, Small comfort is: some time was saved!

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