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

Answer:

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

Step-by-step 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

$\dfrac{\partial M}{\partial y}= 6x+2y+6$

and

N = 6x +2y

$\dfrac{\partial N}{\partial x}= 6$

∴

$f(x) = \dfrac{1}{N}\Big(\dfrac{\partial M}{\partial y}- \dfrac{\partial N}{\partial x} \Big)$

$f(x) = \dfrac{1}{6x+2y}(6x+2y+6-6)$

$f(x) = \dfrac{1}{6x+2y}(6x+2y)$

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