Answer to Question #108029 in Differential Equations for Peter Engalla Garcia

Question

Find a set of solutions for:
2(x-y)dx+(3x-y-1)dy=0

(2x+y-4)dx+(x-3y+12)dy=0

dy/dx=tanycotx - secycosx

Answer 1

1)

$2(x-y)dx+(3x-y-1)dy=0\\ y'=\frac{2(y-x)}{3x-y-1}$

Solving the system

$\left \{ \begin{matrix} y_0-x_0=0 \\ 3x_0-y_0-1=0 \end{matrix}\right.\\ \left\{\begin{matrix} x_0=\frac{1}{2} \\ y_0=\frac{1}{2} \end{matrix}\right.$

We shall replace

$\left\{\begin{matrix} x=t+\frac{1}{2} \\ y=z+\frac{1}{2} \end{matrix}\right.\\ y'=z'$

where $t$ is a new independent variable, $z$ is a new function.

Substitute into equation

$z'=\frac{2(z+\frac{1}{2}-t-\frac{1}{2})}{3t+\frac{3}{2}-z-\frac{1}{2}-1}\\ z'=\frac{2(z-t)}{3t-z}$

We shall replace

$\frac{z}{t}=u(t), z=t\cdot u, z'= tu'+u$

then

$tu'+u=\frac{2(tu-t)}{3t-tu}\\ tu'+u=\frac{2u-2}{3-u}\\ tu'=\frac{2u-2}{3-u}-u\\ tu'=\frac{2u-2-3u+u^2}{3-u}\\ tu'=\frac{u^2-u-2}{3-u}\\ \frac{3-u}{(u+1)(u-2)}du=\frac{dt}{t}\\ \int(-\frac{4}{3}\frac{1}{u+1}+\frac{1}{3}\frac{1}{u-2})dy=\int\frac{dt}{t}\\ -\frac{4}{3}ln|u+1|+\frac{1}{3}ln|u-2|=ln|t|+\frac{1}{3}ln|c|\\ \frac{u-2}{(u+1)^4}=ct^3, u+1=0$

where $u=\frac{y-\frac{1}{2}}{x-\frac{1}{2}}$

Then solutions of the equation are

$\frac{\frac{y-\frac{1}{2}}{x-\frac{1}{2}}-2}{(\frac{y-\frac{1}{2}}{x-\frac{1}{2}}+1)^4}=c(x-\frac{1}{2})^3,\\ \frac{y-\frac{1}{2}}{x-\frac{1}{2}}+1=0$

2)

$(2x+y-4)dx+(x-3y+12)dy=0\\ y'=-\frac{2x+y-4}{x-3y+12}$

Solving the system

$\left\{\begin{matrix} 2x_0+y_0-4=0 \\ x_0-3y_0+12=0 \end{matrix}\right.\\ \left\{\begin{matrix} x_0=0 \\ y_0=4 \end{matrix}\right.$

We shall replace

$\left\{\begin{matrix} x=t \\ y=z+4 \end{matrix}\right.\\ y'=z'\\ z'=\frac{2t+z+4-4}{t-3z-12+12}\\ z'=-\frac{2t+z}{t-3z}$

We shall replace

$\frac{z}{t}=u(t), z=t\cdot u, z'= tu'+u$

then

$tu'+u=\frac{2t+tu}{3tu-t}\\ tu'=\frac{2+u}{3u-1}-u\\ tu'=\frac{2+u-3u^2+u}{3u-1}\\ tu'=\frac{-3u^2+2u+2}{3u-1}\\ -\frac{3u-1}{3u^2-2u-2}du=\frac{dt}{t}\\ -\frac{1}{2}\int\frac{d(3u^2-2u-2)}{3u^2-2u-2}du=\frac{dt}{t}\\ -\frac{1}{2}ln|3u^2-2u-2|=ln|t|+\frac{1}{2}ln|c|\\ \frac{1}{3u^2-2u-2}=ct^2\\ 3u^2-2u-2=0$

where $u=\frac{y-4}{x}$

Then solutions of the equation are

$\frac{1}{3(\frac{y-4}{x})^2-2\cdot\frac{y-4}{x}-2}=cx^2\\ 3(\frac{y-4}{x})^2-2\cdot \frac{y-4}{x}-2=0$

3)

$\frac{dy}{dx}=\tan y\cot x-\sec y\cos x\\ \frac{dy}{dx}=\frac{\sin y}{\cos y}\frac{\cos x}{\sin x}-\frac{1}{\cos y}\cos x\\ \frac{dy}{dx}=\frac{\sin y\cos x-\sin x \cos x}{\sin x\cos y}\\ (\sin x-\sin y)\cos x dx+\sin x \cos y dy=0$

Put $u=\sin y, du=\cos ydy$

Substituting, we get

$(\sin x-u)\cos xdx+\sin x du=0\\ \frac{du}{dx}-u\cdot\frac{\cos x}{\sin x}=-\cos x$

The equation is first - order linear in $u$

The integrating factor is

$I=e^{-\int\frac{\cos x}{\sin x}dx}=e^{-\ln\sin x}=\frac{1}{\sin x}$

Hence

$u\frac{1}{\sin x}=-\int\frac{\cos x}{\sin x}dx=-ln|\sin x|+c$

solve for $u$

$u=-\sin x\ln|\sin x|+c\sin x$

Put $y$ back

$\sin y=-\sin x\ln|\sin x|+c\sin x$