Answer:
[tex]\mathbf{6xe^xy+y^2e^x = C}[/tex] which implies that C is the integrating factor
Step-by-step explanation:
The correct format for the equation given is:
[tex]y(6x+y +6)dx +(6x +2y)dy=0[/tex]
By the application of the general differential equation:
⇒ Mdx + Ndy = 0
where:
M = 6xy+y²+6y
[tex]\dfrac{\partial M}{\partial y}= 6x+2y+6[/tex]
and
N = 6x +2y
[tex]\dfrac{\partial N}{\partial x}= 6[/tex]
∴
[tex]f(x) = \dfrac{1}{N}\Big(\dfrac{\partial M}{\partial y}- \dfrac{\partial N}{\partial x} \Big)[/tex]
[tex]f(x) = \dfrac{1}{6x+2y}(6x+2y+6-6)[/tex]
[tex]f(x) = \dfrac{1}{6x+2y}(6x+2y)[/tex]
f(x) = 1
Now, the integrating factor can be computed as:
[tex]\implies e^{\int fxdx}[/tex]
[tex]\implies e^{\int (1)dx}[/tex]
the integrating factor = [tex]e^x[/tex]
From the given equation:
[tex]y(6x+y +6)dx +(6x +2y)dy=0[/tex]
Let us multiply the above given equation by the integrating factor:
i.e.
[tex](6xy+y^2 +6y)dx +(6x +2y)dy=0[/tex]
[tex](6xe^xy+y^2 +6e^xy)dx +(6xe^x +2e^xy)dy=0[/tex]
[tex]6xe^xydx+6e^xydx+y^2e^xdx +6xe^xdy +2ye^xdy=0[/tex]
By rearrangement:
[tex]6xe^xydx+6e^xydx+6xe^xdy +y^2e^xdx +2ye^xdy=0[/tex]
Let assume that:
[tex]6xe^xydx+6e^xydx+6xe^xdy = d(6xe^xy)[/tex]
and:
[tex]y^2e^xdx +e^x2ydy=d(y^2e^x)[/tex]
Then:
[tex]d(6xe^xy)+d(y^2e^x) = 0[/tex]
[tex]6d (xe^xy) + d(y^2e^x) = 0[/tex]
By integration:
[tex]\mathbf{6xe^xy+y^2e^x = C}[/tex] which implies that C is the integrating factor
