Question

Solve the given differential equation by finding, as in Example 4 from Section 2.4, an appropriate integrating factor. y(6x y 6) dx (6x 2y) dy

Answer 1

`\mathbf{6xe^xy+y^2e^x = C}` which implies that C is the integrating factor

Explanation:

The correct format for the equation given is:

`y(6x+y +6)dx +(6x +2y)dy=0`

By the application of the general differential equation:

⇒ Mdx + Ndy = 0

where:

M = 6xy+y²+6y

`(\partial M)/(\partial y)= 6x+2y+6`

and

N = 6x +2y

`(\partial N)/(\partial x)= 6`

∴

`f(x) = (1)/(N)\Big((\partial M)/(\partial y)- (\partial N)/(\partial x) \Big)`

`f(x) = (1)/(6x+2y)(6x+2y+6-6)`

`f(x) = (1)/(6x+2y)(6x+2y)`

f(x) = 1

Now, the integrating factor can be computed as:

`\implies e^(\int fxdx)`

`\implies e^(\int (1)dx)`

the integrating factor =
`e^x`

From the given equation:

`y(6x+y +6)dx +(6x +2y)dy=0`

Let us multiply the above given equation by the integrating factor:

i.e.

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

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

`6xe^xydx+6e^xydx+y^2e^xdx +6xe^xdy +2ye^xdy=0`

By rearrangement:

`6xe^xydx+6e^xydx+6xe^xdy +y^2e^xdx +2ye^xdy=0`

Let assume that:

`6xe^xydx+6e^xydx+6xe^xdy = d(6xe^xy)`

and:

`y^2e^xdx +e^x2ydy=d(y^2e^x)`

Then:

`d(6xe^xy)+d(y^2e^x) = 0`

`6d (xe^xy) + d(y^2e^x) = 0`

By integration:

`\mathbf{6xe^xy+y^2e^x = C}` which implies that C is the integrating factor