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

$\left(\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{1}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{3}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Equation of Linear coefficients

We have used various substitutions for y to transform the original equation into a new equation that we could solve. In some cases, we must transform both x and y into new variables, say u and v. This is the situation for equations with linear coefficients-that is, equations of the form

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

Given, $\left(\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{1}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{3}\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(1\right)$

Compare the given equation with the general form of linear coefficients. Since,

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

Then,

$\begin{array}{l}{\mathbf{a}}_{\mathbf{1}}{\mathbf{b}}_{\mathbf{2}}\mathbf{=}\left(\mathbf{-}\mathbf{4}\right)\left(\mathbf{1}\right)\mathbf{=}\mathbf{-}\mathbf{4}\\ {\mathbf{a}}_{\mathbf{2}}{\mathbf{b}}_{\mathbf{1}}\mathbf{=}\left(\mathbf{1}\right)\left(\mathbf{-}\mathbf{1}\right)\mathbf{=}\mathbf{-}\mathbf{1}\end{array}$

So, ${\mathbf{a}}_{\mathbf{1}}{\mathbf{b}}_{\mathbf{2}}\ne {\mathbf{a}}_{\mathbf{2}}{\mathbf{b}}_{\mathbf{1}}$.

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

$\begin{array}{l}\mathbf{-}\mathbf{4}\mathbf{h}\mathbf{-}\mathbf{k}\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{0}······\left(2\right)\\ \mathbf{h}\mathbf{+}\mathbf{k}\mathbf{+}3\mathbf{=}\mathbf{0}······\left(3\right)\end{array}$

Add equations (2) and (3)

$\mathbf{h}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}$

Substitutedata-custom-editor="chemistry" $\mathbf{h}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}$ in equation (3).

$\begin{array}{l}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{+}\mathbf{k}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}\\ \mathbf{2}\mathbf{+}\mathbf{3}\mathbf{k}\mathbf{+}\mathbf{9}\mathbf{=}\mathbf{0}\\ \mathbf{3}\mathbf{k}\mathbf{=}\mathbf{-}\mathbf{11}\\ \mathbf{k}\mathbf{=}\mathbf{-}\frac{\mathbf{11}}{\mathbf{3}}\end{array}$

Substitute the value of h and k in translation axes.

$\begin{array}{c}\mathbf{x}\mathbf{=}\mathbf{u}+\frac{2}{3}\to \mathbf{dx}\mathbf{=}\mathbf{du}\\ \mathbf{y}\mathbf{=}\mathbf{v}-\frac{11}{3}\to \mathbf{dy}\mathbf{=}\mathbf{dv}\end{array}$

Substitute the values of x, y, and dx on the equation (1)

$\begin{array}{c}\left(\mathbf{-}\mathbf{4}\left(\mathbf{u}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\right)\mathbf{-}\left(\mathbf{v}\mathbf{-}\frac{\mathbf{11}}{\mathbf{3}}\right)\mathbf{-}\mathbf{1}\right)\mathbf{du}\mathbf{+}\left(\left(\mathbf{u}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\right)\mathbf{+}\left(\mathbf{v}\mathbf{-}\frac{\mathbf{11}}{\mathbf{3}}\right)\mathbf{+}\mathbf{3}\right)\mathbf{dv}\mathbf{=}\mathbf{0}\\ \left(\mathbf{-}\mathbf{4}\mathbf{u}\mathbf{-}\mathbf{v}\right)\mathbf{du}\mathbf{+}\left(\mathbf{u}\mathbf{+}\mathbf{v}\right)\mathbf{dv}\mathbf{=}\mathbf{0}\\ \frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\frac{\mathbf{4}\mathbf{u}\mathbf{+}\mathbf{v}}{\mathbf{u}\mathbf{+}\mathbf{v}}\\ \frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\frac{\mathbf{4}\mathbf{+}\frac{\mathbf{v}}{\mathbf{u}}}{\mathbf{1}\mathbf{+}\frac{\mathbf{v}}{\mathbf{u}}}\end{array}$

So, this is a homogeneous equation, we can solve it by the next move.

Let us assume,

$\mathbf{z}\mathbf{=}\frac{\mathbf{v}}{\mathbf{u}}\to \mathbf{v}\mathbf{=}\mathbf{zu}$

Now differentiate with respect to u.

$\frac{\mathbf{dv}}{\mathbf{du}}\mathbf{=}\mathbf{z}\mathbf{+}\mathbf{u}\frac{\mathbf{dz}}{\mathbf{du}}$

Where last found one becomes,

$\begin{array}{c}\mathbf{z}\mathbf{+}\mathbf{u}\frac{\mathbf{dz}}{\mathbf{du}}\mathbf{=}\frac{4\mathbf{+}\mathbf{z}}{\mathbf{1}+\mathbf{z}}\\ \mathbf{u}\frac{\mathbf{dz}}{\mathbf{du}}\mathbf{=}\frac{4\mathbf{+}\mathbf{z}}{\mathbf{1}+\mathbf{z}}\mathbf{-}\mathbf{z}\\ \mathbf{u}\frac{\mathbf{dz}}{\mathbf{du}}\mathbf{=}\frac{4-{\mathbf{z}}^{\mathbf{2}}}{\mathbf{1}+\mathbf{z}}\\ \frac{\mathbf{1}+\mathbf{z}}{4-{\mathbf{z}}^{\mathbf{2}}}\mathbf{dz}\mathbf{=}\frac{\mathbf{1}}{\mathbf{u}}\mathbf{du}\end{array}$

Now integrate both sides.

$\begin{array}{l}\int \frac{\mathbf{1}\mathbf{+}\mathbf{z}}{\mathbf{4}\mathbf{-}{\mathbf{z}}^{\mathbf{2}}}\mathbf{dz}\mathbf{=}\int \frac{\mathbf{1}}{\mathbf{u}}\mathbf{du}\\ \mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{ln}\left|\mathbf{2}\mathbf{-}\mathbf{z}\right|\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{ln}\left|\mathbf{2}\mathbf{+}\mathbf{z}\right|\mathbf{=}\mathbf{ln}\left|\mathbf{u}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{3}\mathbf{ln}\left|\mathbf{2}\mathbf{-}\mathbf{z}\right|\mathbf{+}\mathbf{ln}\left|\mathbf{2}\mathbf{+}\mathbf{z}\right|\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{ln}\left|\mathbf{u}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\\ \mathbf{ln}\left|{\left(\mathbf{2}\mathbf{-}\mathbf{z}\right)}^{\mathbf{3}}\left(\mathbf{2}\mathbf{+}\mathbf{z}\right)\right|\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{ln}\left|\mathbf{u}\right|\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\\ {\left(\mathbf{2}\mathbf{-}\mathbf{z}\right)}^{\mathbf{3}}\left(\mathbf{2}\mathbf{+}\mathbf{z}\right)\mathbf{=}{\mathbf{Cu}}^{\mathbf{-}\mathbf{4}}\end{array}$

Substitute $\mathbf{z}\mathbf{=}\frac{\mathbf{v}}{\mathbf{u}}$.

$\begin{array}{c}{\left(\mathbf{2}\mathbf{-}\frac{\mathbf{v}}{\mathbf{u}}\right)}^{\mathbf{3}}\left(\mathbf{2}\mathbf{+}\frac{\mathbf{v}}{\mathbf{u}}\right)\mathbf{=}\mathbf{C}\frac{\mathbf{1}}{{\mathbf{u}}^{\mathbf{4}}}\\ \frac{{\left(\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{v}\right)}^{\mathbf{3}}}{{\mathbf{u}}^{\mathbf{3}}}\mathbf{\times }\frac{\left(\mathbf{2}\mathbf{u}\mathbf{+}\mathbf{v}\right)}{\mathbf{u}}\mathbf{=}\mathbf{C}\frac{\mathbf{1}}{{\mathbf{u}}^{\mathbf{4}}}\\ {\left(\mathbf{2}\mathbf{u}\mathbf{-}\mathbf{v}\right)}^{\mathbf{3}}\left(\mathbf{2}\mathbf{u}\mathbf{+}\mathbf{v}\right)\mathbf{=}\mathbf{C}\end{array}$

Substitute$\mathbf{u}\mathbf{=}\mathbf{x}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}\mathbf{v}\mathbf{=}\mathbf{y}\mathbf{+}\frac{\mathbf{11}}{\mathbf{3}}$

$\begin{array}{l}\mathbf{C}\mathbf{=}{\left(\mathbf{2}\mathbf{x}\mathbf{-}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{-}\mathbf{y}\mathbf{-}\frac{\mathbf{11}}{\mathbf{3}}\right)}^{\mathbf{3}}\left(\mathbf{2}\mathbf{x}\mathbf{-}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{+}\mathbf{y}\mathbf{+}\frac{\mathbf{11}}{\mathbf{3}}\right)\\ \mathbf{C}\mathbf{=}{\left(\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{15}\right)}^{\mathbf{3}}\left(\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{+}\mathbf{7}\right)\end{array}$

Hence the solution is$\mathbf{C}\mathbf{=}{\left(\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{15}\right)}^{\mathbf{3}}\left(\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{+}\mathbf{7}\right)$