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