Question

(a) Find a condition on M and N that is necessary and sufficient for $\mathbf{Mdx}\mathbf{+}\mathbf{Ndy}\mathbf{=}\mathbf{0}$to have an integrating factor that depends only on the product${\mathbf{x}}^{\mathbf{2}}\mathbf{y}$.

(b) Use part (a) to solve the equation$\left(\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$.

Short answer

(a) Proved

(b)${\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\right)\mathbf{=}\mathbf{C}$

Step-by-step solution

(a): Find the integrating factor of the form  

Given,$\mathbf{Mdx}\mathbf{+}\mathbf{Ndy}\mathbf{=}\mathbf{0} ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Has the integrating factor of the form ${\mathbf{x}}^{\mathbf{2}}\mathbf{y}$will be written as $\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)$. Then, equation (1) becomes $\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\mathbf{Mdx}\mathbf{+}\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\mathbf{Ndy}\mathbf{=}\mathbf{0}$.

That is,$\frac{\partial }{\partial \mathbf{y}}\left(\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\mathbf{M}\right)\mathbf{=}\frac{\partial }{\partial \mathbf{x}}\left(\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\mathbf{N}\right)$.

Differentiating on both sides,

$\begin{array}{l}\mathbf{f}\text{'}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right){\mathbf{Mx}}^{\mathbf{2}}\mathbf{+}\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{f}\mathbf{\text{'}}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\mathbf{N}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\\ \mathbf{f}\text{'}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right){\mathbf{Mx}}^{\mathbf{2}}\mathbf{-}\mathbf{f}\text{'}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\mathbf{N}\mathbf{2}\mathbf{xy}\mathbf{=}\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\\ \mathbf{xf}\text{'}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\left(\mathbf{Mx}\mathbf{-}\mathbf{2}\mathbf{Ny}\right)\mathbf{=}\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\left(\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\right)\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\mathbf{xf}\text{'}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\left(\mathbf{Mx}\mathbf{-}\mathbf{2}\mathbf{Ny}\right)}{\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)}\\ \frac{\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}}{\mathbf{x}\left(\mathbf{Mx}\mathbf{-}\mathbf{2}\mathbf{Ny}\right)}=\frac{\mathbf{f}\text{'}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)}{\mathbf{f}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)}\\ \frac{\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}}{\mathbf{x}\left(\mathbf{Mx}\mathbf{-}\mathbf{2}\mathbf{Ny}\right)}=\mathbf{H}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{y}\right)\end{array}$

Let$\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}$. Then,$\frac{\mathbf{f}\text{'}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}{\mathbf{f}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)}\mathbf{=}\mathbf{H}\left(\mathbf{v}\right)$

$\mathbf{f}\mathbf{=}{\mathbf{e}}^{\int \mathbf{H}\left(\mathbf{v}\right)\mathbf{dv}} ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Integrate on both sides.

$\mathbf{f}\mathbf{=}{\mathbf{e}}^{\int \mathbf{H}\left(\mathbf{v}\right)\mathbf{dv}} ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

So, the general form of $\mu \left({\mathbf{x}}^2\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{H}\left({\mathbf{x}}^2\mathbf{y}\right)\mathbf{d}\left({\mathbf{x}}^2\mathbf{y}\right)}$.

(b): Use part (a) to solve the equation 2x+2y+2x3y+4x2y2dx+2x+x4+2x3ydy=0.

Given,

$\left(\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0} ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Solve the equation (3).

Let.$\mathbf{M}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}$

Then,

$\begin{array}{l}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\\ \frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{8}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\end{array}$

Substitute the values in equation (2).

$\begin{array}{c}\mathbf{H}\left({\mathbf{x}}^2\mathbf{y}\right)\mathbf{=}\frac{\mathbf{2}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{-}\mathbf{2}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{8}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}{\mathbf{x}\left(\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{4}}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{xy}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{4}}\mathbf{y}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\right)}\\ \mathbf{=}\frac{\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}{\mathbf{x}\left(\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{xy}\right)}\\ \mathbf{=}\frac{\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}{\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\\ \mathbf{=}\mathbf{1}\end{array}$$\begin{array}{l}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\\ \frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{8}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\end{array}$

Now substitute it in equation (3)

$\begin{array}{c}\mathbf{f}\mathbf{=}{\mathbf{e}}^{\int \mathbf{H}\left(\mathbf{v}\right)\mathbf{dv}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{v}}\\ \mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\end{array}$

Now multiply the value of f on equation (3).

${\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left(\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{dx}\mathbf{+}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

So, the found equation is exact. Then,

$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\right)\mathbf{dy}$

Integrate with respect to y.

$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\mathbf{+}\frac{\mathbf{2}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{2}\left({\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\mathbf{y}\mathbf{-}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\right)}{\mathbf{x}}\mathbf{+}\mathbf{h}\left(\mathbf{x}\right)$

Differentiate F with respect to x.

$\begin{array}{c}\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left(\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{+}\mathbf{h}\mathbf{\text{'}}\left(\mathbf{x}\right)\\ \mathbf{=}\mathbf{M}\end{array}$$\begin{array}{c}\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left(\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\right)\mathbf{+}\mathbf{h}\mathbf{\text{'}}\left(\mathbf{x}\right)\\ \mathbf{=}\mathbf{M}\end{array}$

Then equalize theMvalues.

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

Integrate on both sides with respect tox.

$\begin{array}{c}\int \mathbf{h}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{=}\int \mathbf{0}\mathbf{dx}\\ \mathbf{h}\left(\mathbf{x}\right)\mathbf{=}\mathbf{C}\end{array}$

Substitute inF.

$\begin{array}{c}\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\mathbf{+}\frac{\mathbf{2}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{2}\left({\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\mathbf{y}\mathbf{-}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\right)}{\mathbf{x}}\mathbf{+}\mathbf{C}\\ \mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{-}\mathbf{2}}{\mathbf{x}}\right)\mathbf{+}\mathbf{C}\\ \mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{2}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{-}\mathbf{2}}{\mathbf{x}}\right)\mathbf{+}\mathbf{C}\\ \mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}{\mathbf{x}}\right)\mathbf{+}\mathbf{C}\\ \mathbf{=}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\right)\mathbf{+}\mathbf{C}\end{array}$

Therefore, the required solution is ${\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}\mathbf{y}}\left({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\right)\mathbf{=}\mathbf{C}$.