Question

$\mathbf{x}\mathbf{+}\mathbf{y}$(a) Prove that $\mathbf{Mdx}\mathbf{+}\mathbf{Ndy}\mathbf{=}\mathbf{0}$has an integrating factor that depends only on the sum x + y if and only if the expression $\frac{\left(\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\right)}{\left(\mathbf{M}\mathbf{-}\mathbf{N}\right)}$depends only on .

(b) Use part (a) to solve the equation $\left(\mathbf{3}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{xy}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$.

Short answer

(a) Proved

(b)$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\left(\mathbf{3}\mathbf{+}\mathbf{xy}\right)\mathbf{+}\mathbf{C}$

Step-by-step solution

(a): Find the integrating factor of the form  

Given,$\mathbf{M}\mathbf{dx}\mathbf{+}\mathbf{N}\mathbf{dy}\mathbf{=}\mathbf{0} ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Has the integrating factor of the form $\mu \left(\mathbf{xy}\right)$.

To prove:$\frac{\left(\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\right)}{\left(\mathbf{M}\mathbf{-}\mathbf{N}\right)}\mathbf{=}\mathbf{H}\left(\mathbf{x}\mathbf{+}\mathbf{y}\right) ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Let us assume be the function of x + y. That is, role="math" localid="1664184972832" $\frac{\partial }{\partial \mathbf{y}}\left(\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\right)\mathbf{=}\frac{\partial }{\partial \mathbf{x}}\left(\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\right)$.

Since t is a function ofx + y, one has$\frac{\partial \mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}{\partial \mathbf{y}}\mathbf{=}\mathbf{t}\text{'}\left(\mathbf{x}\mathbf{+}\mathbf{y}\right)$

Multiply t on both sides in equation (1).

$\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{Mdx}\mathbf{+}\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{Ndy}\mathbf{=}\mathbf{0}$

Let us take $\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{M}$.

Differentiate with respect to y on both sides.

$\frac{\partial }{\partial \mathbf{y}}\left(\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\right)\mathbf{=}\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{+}\mathbf{M}\frac{\partial \mathbf{t}}{\partial \mathbf{y}}$

Similarly, $\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{N}\to \frac{\partial }{\partial \mathbf{x}}\left(\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\right)\mathbf{=}\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{+}\mathbf{N}\frac{\partial \mathbf{t}}{\partial \mathbf{x}}$.

Simplifying method

Now use the exactness condition.

That is,

$\begin{array}{c}\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{+}\mathbf{M}\frac{\partial \mathbf{t}}{\partial \mathbf{y}}\mathbf{=}\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{+}\mathbf{N}\frac{\partial \mathbf{t}}{\partial \mathbf{x}}\\ \mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)\mathbf{=}\mathbf{t}\text{'}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\left(\mathbf{N}\mathbf{-}\mathbf{M}\right)\\ \frac{\mathbf{t}\text{'}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}{\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}\mathbf{=}\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}\mathbf{-}\mathbf{M}}\\ \frac{\mathbf{t}\text{'}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}{\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}\mathbf{=}\mathbf{H}\left(\mathbf{x}\mathbf{+}\mathbf{y}\right)\end{array}$

Let $\mathbf{v}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}$. Then,

$\frac{\mathbf{t}\text{'}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}{\mathbf{t}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}\mathbf{=}\mathbf{H}\left(\mathbf{v}\right)$

Integrate on both sides.

$\mathbf{t}\mathbf{=}{\mathbf{e}}^{\int \mathbf{H}\left(\mathbf{v}\right)\mathbf{dv}} ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

So, the general form of localid="1664186265139" $\mathit{\mu }\left(x+y\right)\mathbf{=}{\mathbf{e}}^{\mathbf{\int }\mathbf{H}\left(x+y\right)\mathbf{d}\left(x+y\right)}$.

4(b): Using part (a) to solve the equation

Given, $\left(\mathbf{3}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{xy}\right)\mathbf{dy}\mathbf{=}\mathbf{0} ......\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Solve the equation (4).

$\mathbf{M}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{xy}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{xy}$

Let .

$\mathbf{M}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{xy}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{xy}$

Then,

$\begin{array}{l}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{y}\\ \frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}1\mathbf{+}\mathbf{x}\end{array}$

Substitute the values in equation (2).

$\begin{array}{c}\mathbf{H}\left(\mathbf{x}\mathbf{+}\mathbf{y}\right)\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{x}}{\mathbf{3}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{xy}\mathbf{-}\mathbf{3}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{xy}}\\ \mathbf{=}\frac{\mathbf{y}\mathbf{-}\mathbf{x}}{\mathbf{y}\mathbf{-}\mathbf{x}}\\ \mathbf{=}\mathbf{1}\end{array}$

Now substitute it in equation (3)

$\begin{array}{c}\mathbf{t}\mathbf{=}{\mathbf{e}}^{\int \mathbf{H}\left(\mathbf{v}\right)\mathbf{dv}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\end{array}$

Now multiply the value of t on equation (4).

$\begin{array}{c}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\left(\mathbf{3}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{xy}\right)\mathbf{dx}\mathbf{+}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\left(\mathbf{3}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{xy}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left(\mathbf{3}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{ye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{3}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xe}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\end{array}$

So, the founded equation is exact. Then,

$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}\left(\mathbf{3}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{ye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\right)\mathbf{dx}$

Integrate with respect toy.

$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{y}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{ye}}^{\mathbf{y}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{ye}}^{\mathbf{y}}\left({\mathbf{xe}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{x}}\right)\mathbf{+}\mathbf{g}\left(\mathbf{y}\right)$

Differentiate F with respect to y.

$\begin{array}{c}\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xe}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\\ \mathbf{=}\mathbf{N}\end{array}$

Then equalise the N values.

$\begin{array}{c}\mathbf{3}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xe}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xe}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\mathbf{+}{\mathbf{xye}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\\ \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}0\end{array}$

Integrate on both sides with respect to y.

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

Substitute in F.

$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\left(\mathbf{3}\mathbf{+}\mathbf{xy}\right)\mathbf{+}\mathbf{C}$

Therefore, the solution of the given equation is role="math" localid="1664186240117" $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}\left(\mathbf{3}\mathbf{+}\mathbf{xy}\right)\mathbf{+}\mathbf{C}$.