Question

In Problems 13 and 14, find an integrating factor of the form and solve the equation.

$\left(\mathbf{12}\mathbf{+}\mathbf{5}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{6}{\mathbf{xy}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given equation is $3{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^5{\mathbf{y}}^3\mathbf{=}\mathbf{C}$.

Step-by-step solution

General form of special integrating factors

If $\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}}$is continuous and depends only on x, then

$\mu \left(\mathbf{x}\right)\mathbf{=}\mathbf{exp}\left[\int \left(\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}}\right)\mathbf{dx}\right]$is an integrating factor for the equation.

If $\frac{\left(\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\right)}{\mathbf{M}}$is continuous and depends only on y, then

is an integrating factor for the equation.

Find integrating factor of the form xnym

Given:

$\left(\mathbf{12}\mathbf{+}\mathbf{5}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{6}{\mathbf{xy}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(1\right)$

To find integrating factor of the form${\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}$${\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}$.

Multiply${\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}$on equation (1).

$\begin{array}{c} {\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\left(\mathbf{12}\mathbf{+}\mathbf{5}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\left(\mathbf{6}{\mathbf{xy}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left(\mathbf{12}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\mathbf{+}\mathbf{5}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{6}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\right)\mathbf{=}\mathbf{0}······\left(2\right)\end{array}$

Let $\mathbf{M}\mathbf{=}\mathbf{12}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\mathbf{+}\mathbf{5}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{6}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}$.

$\mathbf{M}\mathbf{=}\mathbf{12}{\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\mathbf{+}\mathbf{5}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{1}\mathbf{+}\mathbf{m}}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{6}{\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}$

Now let us consider the found equation is exact. Then, $\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$.

$\begin{array}{c}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{12}{\mathbf{mx}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{5}\left(\mathbf{1}\mathbf{+}\mathbf{m}\right){\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{6}\left(\mathbf{1}\mathbf{+}\mathbf{n}\right){\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}\left(\mathbf{2}\mathbf{+}\mathbf{n}\right){\mathbf{x}}^{\mathbf{1}\mathbf{+}\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\end{array}$

So, we can equalise the coefficients.

$\begin{array}{l}\mathbf{12}\mathbf{m}\mathbf{=}\mathbf{6}\left(\mathbf{1}\mathbf{+}\mathbf{n}\right)\\ \mathbf{12}\mathbf{m}\mathbf{=}\mathbf{6}\mathbf{+}\mathbf{6}\mathbf{n}\\ \mathbf{12}\mathbf{m}\mathbf{-}\mathbf{6}\mathbf{n}\mathbf{=}\mathbf{6}\\ 5\left(\mathbf{1}\mathbf{+}\mathbf{m}\right)\mathbf{=}3\left(\mathbf{2}\mathbf{+}\mathbf{n}\right)\\ 5\mathbf{+}5\mathbf{m}\mathbf{=}6\mathbf{+}3\mathbf{n}\\ 5\mathbf{m}\mathbf{-}3\mathbf{n}\mathbf{=}1\end{array}$

Solve the founded equation to get the value of m and n.

$m=2$

$\begin{array}{c}\mathbf{24}\mathbf{-}\mathbf{6}\mathbf{n}\mathbf{=}\mathbf{6}\\ \mathbf{-}\mathbf{6}\mathbf{n}\mathbf{=}\mathbf{-}\mathbf{18}\\ \mathbf{n}\mathbf{=}\mathbf{3}\end{array}$

The value of m = 2 and n = 3.

Then,

${\mathbf{x}}^{\mathbf{n}}{\mathbf{y}}^{\mathbf{m}}\mathbf{=}{\mathbf{x}}^3{\mathbf{y}}^2$

So, the integrating factor is found.

Simplifying method

Substitute m and n in equation (2)

$\begin{array}{l}\left(\mathbf{12}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}\mathbf{5}{\mathbf{x}}^{\mathbf{1}\mathbf{+}3}{\mathbf{y}}^{\mathbf{1}\mathbf{+}2}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{6}{\mathbf{x}}^{\mathbf{1}\mathbf{+}3}{\mathbf{y}}^{2\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}\mathbf{+}3}{\mathbf{y}}^2\right)\mathbf{=}\mathbf{0}\\ \left(\mathbf{12}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}\mathbf{5}{\mathbf{x}}^4{\mathbf{y}}^3\right)\mathbf{dx}\mathbf{+}\left(\mathbf{6}{\mathbf{x}}^4\mathbf{y}\mathbf{+}\mathbf{3}{\mathbf{x}}^5{\mathbf{y}}^2\right)\mathbf{=}\mathbf{0}······\left(3\right)\end{array}$

Solve the equation (3).

$\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{12}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}\mathbf{5}{\mathbf{x}}^4{\mathbf{y}}^3$

Now integrate the value to find F.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{12}{\mathbf{x}}^3{\mathbf{y}}^2\mathbf{+}\mathbf{5}{\mathbf{x}}^4{\mathbf{y}}^3\right)\mathbf{dx}\\ \mathbf{=}3{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^5{\mathbf{y}}^3\mathbf{+}\mathbf{g}\left(\mathbf{y}\right)\end{array}$

Differentiate F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\mathbf{6}{\mathbf{x}}^{\mathbf{4}}\mathbf{y}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\\ \mathbf{=}\mathbf{N}\end{array}$

Then equalise the N values.

$\begin{array}{c}\mathbf{6}{\mathbf{x}}^{\mathbf{4}}\mathbf{y}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\mathbf{6}{\mathbf{x}}^{\mathbf{4}}\mathbf{y}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}{\mathbf{y}}^{\mathbf{2}}\\ \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}0\end{array}$

Integrate on both sides with respect to y.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\int 0\mathbf{dy}\\ \mathbf{g}\left(\mathbf{y}\right)\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\end{array}$

Substitute in F.

$\begin{array}{c}3{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^5{\mathbf{y}}^3\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\\ 3{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^5{\mathbf{y}}^3\mathbf{=}\mathbf{C}\end{array}$

Hence the solution is$3{\mathbf{x}}^4{\mathbf{y}}^2\mathbf{+}{\mathbf{x}}^5{\mathbf{y}}^3\mathbf{=}\mathbf{C}$