Question

Question: In Problems 1-30, solve the equation.

$\left(y-2x-1\right)\mathbf{dx}\mathbf{+}\left(x+y-4\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

The solution of the given equation is $\mathbf{xy}\mathbf{-}{\mathbf{x}}^2\mathbf{-}\mathbf{x}\mathbf{+}\frac{{\mathbf{y}}^2}{2}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{C}$.

Step-by-step solution

Given information and simplification

Given that, $\left(y-2x-1\right)\mathbf{dx}\mathbf{+}\left(x+y-4\right)\mathbf{dy}\mathbf{=}\mathbf{0}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Since the given equation is linear coefficients. So, we can take

$\begin{array}{l}\mathbf{x}\mathbf{=}\mathbf{u}\mathbf{+}\mathbf{h}\to \mathbf{dx}\mathbf{=}\mathbf{du}\\ \mathbf{y}\mathbf{=}\mathbf{v}\mathbf{+}\mathbf{k}\to \mathbf{dy}\mathbf{=}\mathbf{dv}\end{array}$

Evaluate the equation (1).

$\begin{array}{c}\left(y-2x-1\right)\mathbf{dx}\mathbf{+}\left(x+y-4\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{-}\frac{y-2x-1}{x+y-4}\\ \frac{\mathrm{dv}}{\mathrm{du}}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\left(v+k\right)\mathbf{+}\mathbf{2}\left(u+h\right)}{u+h+v+k-4}\\ \mathbf{=}\frac{2u-v+2h-k+1}{u+v+h+k-4}\end{array}$

Using the condition, we get,

$\begin{array}{c}2h-k+1=0\\ h+k-4=0\end{array}$

Solve the above equations to find the values of h and k.

$\begin{array}{c}3h-3=0\\ 3h=3\\ h=1\end{array}$

Substitute h = 1.

$\begin{array}{c}1+k-4=0\\ k=3\end{array}$

Then,

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

So, the founded equation is homogeneous.

Let us take $\mathbf{s}\mathbf{=}\frac{v}{u}\to \frac{\mathrm{dv}}{\mathrm{du}}\mathbf{=}\mathbf{s}\mathbf{+}\mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}$

Then,

$\begin{array}{c}\mathbf{s}\mathbf{+}\mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}\mathbf{=}\frac{2-s}{1+s}\\ \mathbf{u}\frac{\mathrm{ds}}{\mathrm{du}}\mathbf{=}\frac{2-s}{1+s}\mathbf{-}\mathbf{s}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{-}\mathbf{s}\mathbf{-}\mathbf{s}\mathbf{-}{\mathbf{s}}^2}{1+s}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{-}\mathbf{2}\mathbf{s}\mathbf{-}{\mathbf{s}}^2}{1+s}\\ \frac{1+s}{{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}}\mathbf{ds}\mathbf{=}\mathbf{-}\frac{\mathrm{du}}{u}\end{array}$

Evaluation method

Now integrate the equation (2) on both sides.

$\int \frac{1+s}{{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}}\mathbf{ds}\mathbf{=}\mathbf{-}\int \frac{\mathrm{du}}{u}$

Find the value of $\int \frac{1+s}{{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}}\mathbf{ds}$ separately.

Let $\mathbf{t}\mathbf{=}{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}\to \mathbf{dt}\mathbf{=}\mathbf{2}\left(s+1\right)\mathbf{ds}$.

Then,

$\begin{array}{c}\int \frac{1+s}{{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}}\mathbf{ds}\mathbf{=}\int \frac{1}{t}\frac{\mathrm{dt}}{2}\\ \mathbf{=}\frac{1}{2}\int \frac{1}{t}\mathbf{dt}\\ \mathbf{=}\frac{1}{2}\mathbf{ln}\left|t\right|\\ \mathbf{=}\frac{1}{2}\mathbf{ln}\left|{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}\right|\end{array}$

Substitute the value here.

$\begin{array}{c}\int \frac{1+s}{{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}}\mathbf{ds}\mathbf{=}\mathbf{-}\int \frac{\mathrm{du}}{u}\\ \frac{1}{2}\mathbf{ln}\left|{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}\right|\mathbf{=}\mathbf{-}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_1\\ \mathbf{ln}\left|{\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}\right|\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{ln}\left|u\right|\mathbf{+}{\mathbf{C}}_1\\ \mathbf{=}\mathbf{-}\mathbf{ln}\left|{\mathbf{u}}^2\right|\mathbf{+}{\mathbf{C}}_1\\ \left({\mathbf{s}}^2\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{2}\right){\mathbf{u}}^2\mathbf{=}{\mathbf{C}}_2\end{array}$

Again, substitute the value of s, u and v.

$\begin{array}{c}\left({\left(\frac{v}{u}\right)}^2\mathbf{+}\mathbf{2}\left(\frac{v}{u}\right)\mathbf{-}\mathbf{2}\right){\mathbf{u}}^2\mathbf{=}{\mathbf{C}}_2\\ \left(\frac{{\mathbf{v}}^2}{{\mathbf{u}}^2}\mathbf{+}\mathbf{2}\frac{v}{u}\mathbf{-}\mathbf{2}\right){\mathbf{u}}^2\mathbf{=}{\mathbf{C}}_2\\ \frac{{\mathbf{v}}^2\mathbf{+}\mathbf{2}\mathbf{uv}\mathbf{-}\mathbf{2}{\mathbf{u}}^2}{{\mathbf{u}}^2}{\mathbf{u}}^2\mathbf{=}{\mathbf{C}}_2\\ {\mathbf{v}}^2\mathbf{+}\mathbf{2}\mathbf{uv}\mathbf{-}\mathbf{2}{\mathbf{u}}^2\mathbf{=}{\mathbf{C}}_2\\ {\left(y-3\right)}^2\mathbf{+}\mathbf{2}\left(x-1\right)\left(y-3\right)\mathbf{-}\mathbf{2}{\left(x-1\right)}^2\mathbf{=}{\mathbf{C}}_2\\ {\mathbf{y}}^2\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{+}\mathbf{9}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{6}\mathbf{-}\mathbf{2}\left({\mathbf{x}}^2\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}\right)\mathbf{=}{\mathbf{C}}_2\\ {\mathbf{y}}^2\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{-}\mathbf{2}{\mathbf{x}}^2\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{=}{\mathbf{C}}_3\\ \mathbf{xy}\mathbf{-}{\mathbf{x}}^2\mathbf{-}\mathbf{x}\mathbf{+}\frac{{\mathbf{y}}^2}{2}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{C}\end{array}$

So, the solution is $\mathbf{xy}\mathbf{-}{\mathbf{x}}^2\mathbf{-}\mathbf{x}\mathbf{+}\frac{{\mathbf{y}}^2}{2}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{C}$