Question

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

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

Short answer

$\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{3}{\mathbf{)}}^{\mathbf{3}}\mathbf{=}\mathbf{C}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{2}}$

Step-by-step solution

Find the constants.

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.

We get ${\mathbf{a}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{b}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{a}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{,}{\mathbf{b}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{32}$,

Find the value of  and

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

$\begin{array}{c}\mathbf{12}\mathbf{hk}\mathbf{=}\mathbf{0}\\ \mathbf{14}\mathbf{h}\mathbf{+}\mathbf{k}\mathbf{=}\mathbf{32}\mathbf{.}\end{array}$

Solving the system gives

$\mathbf{h}\mathbf{=}\frac{\mathbf{16}}{\mathbf{13}}\mathrm{and}\mathrm{k}=0$

Find the value of dv/du

Next, one substitute$\mathbf{x}\mathbf{=}\mathbf{u}\mathbf{+}\frac{\mathbf{16}}{\mathbf{13}},\mathbf{y}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{0}$, into the original equation observing that $\mathbf{dx}\mathbf{=}\mathbf{du}\mathrm{and}\mathbf{dy}\mathbf{=}\mathbf{dv}$, and we get

$\begin{array}{l}\mathbf{(}\mathbf{12}\mathbf{u}\mathbf{-}\mathbf{v}\mathbf{)}\mathbf{du}\mathbf{+}\mathbf{(}\mathbf{14}\mathbf{u}\mathbf{+}\mathbf{v}\mathbf{)}\mathbf{dv}\mathbf{=}\mathbf{0}\\ \frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\mathbf{-}\left(\mathbf{12}\mathbf{u}\mathbf{-}\mathbf{v}\right)\left(\mathbf{14}\mathbf{u}\mathbf{+}\mathbf{v}\right)\\ \frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\left(\mathbf{v}\mathbf{-}\mathbf{12}\mathbf{u}\right)\left(\mathbf{14}\mathbf{u}\mathbf{+}\mathbf{v}\right)\end{array}$

Find the implicit equation

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

$\mathbf{z}\mathbf{+}\mathbf{u}\frac{\mathbf{dz}}{\mathbf{du}}\mathbf{=}\frac{\mathbf{z}\mathbf{-}\mathbf{12}}{\mathbf{z}\mathbf{+}\mathbf{14}}$

Separating variables gives

$\int \frac{\mathbf{z}\mathbf{+}\mathbf{14}}{\mathbf{(}\mathbf{-}{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{13}\mathbf{z}\mathbf{-}\mathbf{12}\mathbf{)}}\mathbf{dz}\mathbf{=}\int \frac{\mathbf{1}}{\mathbf{u}}\mathbf{du}$

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

$\begin{array}{l}\frac{\mathbf{-}\mathbf{13}}{11}\mathbf{ln}\left|\mathbf{z}\mathbf{+}\mathbf{1}\right|\mathbf{-}\frac{\mathbf{2}}{\mathbf{11}}\mathbf{ln}\left|\mathbf{z}\mathbf{+}\mathbf{12}\right|\mathbf{=}\mathbf{ln}\left|\mathbf{u}\right|\mathbf{+}\mathbf{C}\\ \mathbf{-}\mathbf{13}\mathbf{ln}\left|\mathbf{z}\mathbf{+}\mathbf{1}\right|\mathbf{-}\mathbf{2}\mathbf{ln}\left|\mathbf{z}\mathbf{+}\mathbf{12}\right|\mathbf{=}\mathbf{11}\mathbf{ln}\left|\mathbf{u}\right|\mathbf{+}\mathbf{C}\\ \mathbf{ln}\frac{{\mathbf{(}\mathbf{z}\mathbf{+}\mathbf{12}\mathbf{)}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{z}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{13}}{\mathbf{u}}^{\mathbf{11}}}\mathbf{=}\mathbf{C}\\ \mathbf{ln}\frac{{\mathbf{(}\mathbf{v}\mathbf{+}\mathbf{12}\mathbf{u}\mathbf{)}}^{\mathbf{2}}}{{\mathbf{(}\mathbf{v}\mathbf{+}\mathbf{u}\mathbf{)}}^{\mathbf{13}}}\mathbf{=}\mathbf{C}\\ \frac{{\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{3}\mathbf{)}}^{\mathbf{3}}}{{\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{C}\mathbf{\text{'}}\end{array}$

Therefore the last equation gives an implicit solution${\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{3}\mathbf{)}}^{\mathbf{3}}\mathbf{=}\mathbf{C}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{2}}$