Use the method discussed under “Equations with Linear Coefficients” to solve Problems 29–32.

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

Compared to the general form of “Equation with Linear Coefficients”,

$\mathbf{\left(}\mathbf{\right(}{\mathbf{a}}_{\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{y}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{(}{\mathbf{a}}_{\mathbf{2}}\mathbf{x}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}$

where${\mathbf{a}\mathbf{\text{'}}}_{\mathbf{i}}\mathbf{s}\mathbf{,}{\mathbf{b}\mathbf{\text{'}}}_{\mathbf{i}}\mathbf{s}\mathbf{and}{\mathbf{c}\mathbf{\text{'}}}_{\mathbf{i}}\mathbf{s}$,are constants.

The values are,${\mathbf{a}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{b}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{,}{\mathbf{a}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{b}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}$

In this case,

${\mathbf{b}}_{\mathbf{1}}\ne {\mathbf{a}}_{\mathbf{2}}{,}_{}{\mathbf{a}}_{\mathbf{1}}{\mathbf{b}}_2\ne {\mathbf{a}}_{\mathbf{2}}{\mathbf{b}}_{\mathbf{1}}{\mathbf{andc}}_{\mathbf{1}}\ne 0\mathbf{\&}{\mathbf{c}}_{\mathbf{2}}\ne 0.$

The translation of axes $\mathbf{x}\mathbf{=}\mathbf{u}\mathbf{+}\mathbf{h}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{k}$, where h and k satisfy the system

$\begin{array}{c}\mathbf{2}\mathbf{h}\mathbf{-}\mathbf{k}\mathbf{=}\mathbf{-}\mathbf{42}\\ \mathbf{h}\mathbf{-}\mathbf{2}\mathbf{k}\mathbf{=}\mathbf{22}\mathbf{.}\end{array}$

Solving the system gives

$\mathbf{1}\mathbf{h}\mathbf{=}\mathbf{-}\frac{\mathbf{106}}{\mathbf{23}}\mathbf{,}\mathbf{k}\mathbf{=}\mathbf{-}\frac{\mathbf{306}}{\mathbf{23}}$

Next, one substitute $\mathbf{x}\mathbf{=}\mathbf{u}\mathbf{-}\frac{\mathbf{106}}{\mathbf{23}}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{v}\mathbf{-}\frac{\mathbf{306}}{\mathbf{23}}\mathbf{}$into the original equation observing that data-custom-editor="chemistry" $\mathbf{dx}\mathbf{=}\mathbf{du}\mathrm{and}\mathbf{dy}\mathbf{=}\mathbf{dv}$, and we get

$\begin{array}{l}\mathbf{(}\mathbf{12}\mathbf{u}\mathbf{-}\mathbf{v}\mathbf{)}\mathbf{du}\mathbf{+}\mathbf{(}\mathbf{u}\mathbf{-}\mathbf{2}\mathbf{v}\mathbf{)}\mathbf{dv}\mathbf{=}\mathbf{0}\\ \frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\mathbf{-}\left(\mathbf{12}\mathbf{u}\mathbf{-}\mathbf{v}\right)\left(\mathbf{u}\mathbf{-}\mathbf{2}\mathbf{v}\right)\\ \frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\left(\mathbf{v}\mathbf{-}\mathbf{12}\mathbf{u}\right)\left(\mathbf{u}\mathbf{-}\mathbf{2}\mathbf{v}\right)\mathbf{}\end{array}$

The last equation is homogeneous, so one let z = v/u.

Then, $\frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\mathbf{z}\mathbf{+}\mathbf{u}\left(\frac{\mathbf{dz}}{\mathbf{du}}\right)$and substituting for v/u yields

$\mathbf{z}\mathbf{+}\mathbf{u}\frac{\mathbf{dz}}{\mathbf{du}}\mathbf{=}\frac{\mathbf{z}\mathbf{-}\mathbf{12}}{\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{z}}$

Separating variables gives

$\frac{\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{z}}{\mathbf{2}\mathbf{(}{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{)}}\mathbf{dz}\mathbf{=}\int \frac{\mathbf{1}}{\mathbf{u}}\mathbf{du}$

From which, after utilizing a partial fraction expansion of the first integrand, one finds

$\begin{array}{l}\frac{\sqrt{\mathbf{6}}}{\mathbf{24}}\mathbf{ln}\left|\frac{\mathbf{z}\mathbf{-}\sqrt{\mathbf{6}}}{\mathbf{z}\mathbf{+}\sqrt{\mathbf{6}}}\right|\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{ln}\left|{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\right|\mathbf{=}\mathbf{ln}\left|\mathbf{u}\right|\mathbf{+}\mathbf{C}\\ \sqrt{\mathbf{6}}\mathbf{ln}\left|\frac{\mathbf{z}\mathbf{-}\sqrt{\mathbf{6}}}{\mathbf{z}\mathbf{+}\sqrt{\mathbf{6}}}\right|\mathbf{-}\mathbf{12}\mathbf{ln}\left|{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\right|\mathbf{=}\mathbf{24}\mathbf{ln}\left|\mathbf{u}\right|\mathbf{+}\mathbf{C}\\ \mathbf{ln}{\left(\left|\frac{\mathbf{z}\mathbf{-}\sqrt{\mathbf{6}}}{\mathbf{z}\mathbf{+}\sqrt{\mathbf{6}}}\right|\right)}^{\sqrt{\mathbf{6}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{(}{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{)}}^{\mathbf{12}}{\mathbf{u}}^{\mathbf{24}}}\mathbf{=}\mathbf{C}\\ \int \mathrm{l}\mathbf{n}{\left(\frac{\mathbf{v}\mathbf{-}\sqrt{\mathbf{6}}\mathbf{u}}{\mathbf{v}\mathbf{+}\sqrt{\mathbf{6}}\mathbf{u}}\right)}^{\sqrt{\mathbf{6}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{(}{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{6}{\mathbf{u}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{12}}}\mathbf{=}\mathbf{C}\\ \sqrt{\mathbf{6}}\mathbf{ln}\mathbf{(}\mathbf{v}\mathbf{-}\sqrt{\mathbf{6}}\mathbf{u}\mathbf{)}\mathbf{-}\sqrt{\mathbf{6}}\mathbf{ln}\mathbf{(}\mathbf{v}\mathbf{+}\sqrt{\mathbf{6}}\mathbf{u}\mathbf{)}\mathbf{-}\mathbf{12}\mathbf{ln}\mathbf{(}\mathbf{v}\mathbf{-}\sqrt{\mathbf{6}}\mathbf{u}\mathbf{)}\mathbf{-}\mathbf{12}\mathbf{ln}\mathbf{(}\mathbf{v}\mathbf{+}\sqrt{\mathbf{6}}\mathbf{u}\mathbf{)}\mathbf{=}\mathbf{C}\\ \mathbf{(}\sqrt{\mathbf{6}}\mathbf{-}\mathbf{12}\mathbf{\left)}\mathbf{ln}\mathbf{\right(}\mathbf{23}\mathbf{-}\mathbf{23}\sqrt{\mathbf{6}}\mathbf{x}\mathbf{+}\mathbf{306}\mathbf{-}\mathbf{106}\sqrt{\mathbf{6}}\mathbf{)}\mathbf{-}\mathbf{(}\sqrt{\mathbf{6}}\mathbf{+}\mathbf{12}\mathbf{\left)}\mathbf{ln}\mathbf{\right(}\mathbf{23}\mathbf{y}\mathbf{+}\mathbf{23}\sqrt{\mathbf{6}}\mathbf{x}\mathbf{+}\mathbf{306}\mathbf{+}\mathbf{106}\sqrt{\mathbf{6}}\mathbf{=}\mathbf{C}\end{array}$

Therefore, the last equation gives an implicit solution to the original equation$\mathbf{(}\sqrt{\mathbf{6}}\mathbf{-}\mathbf{12}\mathbf{\left)}\mathbf{ln}\mathbf{\right(}\mathbf{23}\mathbf{-}\mathbf{23}\sqrt{\mathbf{6}}\mathbf{x}\mathbf{+}\mathbf{306}\mathbf{-}\mathbf{106}\sqrt{\mathbf{6}}\mathbf{)}\mathbf{-}\mathbf{(}\sqrt{\mathbf{6}}\mathbf{+}\mathbf{12}\mathbf{\left)}\mathbf{ln}\mathbf{\right(}\mathbf{23}\mathbf{y}\mathbf{+}\mathbf{23}\sqrt{\mathbf{6}}\mathbf{x}\mathbf{+}\mathbf{306}\mathbf{+}\mathbf{106}\sqrt{\mathbf{6}}\mathbf{=}\mathbf{C}$