Question

dy/dx=x+3y-5/x-y-1

Answer 1

$\displaystyle\frac{dy}{dx}=\frac{x+3y-5}{x-y-1}$

Solution:

Introduce replacements: $x=x_0+t$ and $y=y_0+z$ , where $x_0$ and $y_0$ are solutions of the system of equations:

$\begin{cases} x_0+3y_0-5=0 \\ x_0-y_0-1=0 \end{cases}$

$x_0=2$ , $y_0=1$ .

Substitute $x=2+t$ and $y=1+z$ into the initial equation:

$\displaystyle\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dz}{dt}=\frac{2+t+3(1+z)-5}{2+t-(1+z)-1}$

$\displaystyle\frac{dz}{dt}=\frac{t+3z}{t-z}$

Now introduce another replacement: $z=ut$ . Then $\frac{dz}{dt}=\frac{du}{dt}t+u$ .

$\displaystyle\frac{du}{dt}t+u=\frac{t+3ut}{t-ut}$

$\displaystyle\frac{du}{dt}t=\frac{1+3u}{1-u}-u=\frac{1+3u-u+u^2}{1-u}=\frac{(1+u)^2}{1-u}$

$\displaystyle\frac{1-u}{(1+u)^2}du=\frac{dt}{t}$

$\displaystyle-\frac{(u+1-2)}{(1+u)^2}du=\frac{dt}{t}$

$\displaystyle(-\frac{1}{1+u}+\frac{2}{(1+u)^2})du=\frac{dt}{t}$

$\displaystyle-\ln|1+u|-\frac{2}{1+u}=\ln{|t|}+\ln{C}$

$\displaystyle(1+u)\cdot t\cdot C=e^{-\frac{2}{1+u}}$ .

Returning to replacements: $z=y-1$ , $t=x-2$ and $\displaystyle u=\frac{z}{t}=\frac{y-1}{x-2}$

$\displaystyle(1+\frac{y-1}{x-2})(x-2)\cdot C=e^{-\frac{2}{1+\frac{y-1}{x-2}}}$

$\displaystyle(x+y-3)\cdot C=e^{-\frac{2(x-2)}{x+y-3}}$

Answer: $\displaystyle(x+y-3)\cdot C=e^{-\frac{2(x-2)}{x+y-3}}$ .