Question

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

Answer 1

Given the differential equation,

$\frac{dy}{dx} =\frac{x-y+1}{x+y-3}$. Make the substitution,

$x=X+h,y=Y+k$, then

$\frac{dy}{dx} =\frac{dY}{dY} =\frac{X+h-Y-k+1}{X+h+Y+k-3}\\ \frac{dY}{dY} =\frac{X-Y+h-k+1}{X+Y+h+k-3}$

Now set

$h-k+1=0,h+k-3=0$. Solve the above 2 equations,

$h=1,k=2$

The differential equation becomes,

$\frac{dY}{dX} =\frac{X-Y}{X+Y}$. This is a homogenous differential equation,

Let $Y=vX$, then $\frac{dY}{dX} =v+X \frac{dv}{dX}$

$\frac{X-vX}{X+vX} =\frac{1-v}{1+v}$

The differential equation is,

$v+X \frac{dv}{dX} =\frac{1-v}{1+v} \\ X \frac{dv}{dX} =\frac{1-v}{1+v} -v\\ X \frac{dv}{dX} =\frac{1-2v-v^2}{1+v} \\ \frac{(1+v)dv}{1-2v-v^2} =\frac{dX}{X} \\ \frac{(1+v)dv}{2-(1+v)^2} =\frac{dX}{X}$

Integrating both sides,

$\int \frac{(1+v)dv}{2-(1+v)^2} =\int \frac{dX}{X} \\ -\frac{1}{2}\ln |2-(1+v)^2|=\n X+C\\ \ln |2-(1+v)^2|=-2(\n X+C)\\ \ln |1-2v-v^2|=-2(\n X+C)\\ \ln |1-2\frac{Y}{X}-(\frac{Y}{X})^2|=-2(\n X+C)$

$\ln |1-2\frac{y-2}{x-1}-(\frac{y-2}{x-1})^2|=-2(\n |x-1|+C)\\ \ln |(x-1)^2-2(x-1)(y-2)-(y-2)^2|-2\n |x-1|=-2(\n |x-1|+C)\\ \ln |(x-1)^2-2(x-1)(y-2)-(y-2)^2|=C\\ (x-1)^2-2(x-1)(y-2)-(y-2)^2=C\\ (x-1)^2-2(x-1)(y-2)-(y-2)^2=C$

Answer 2

Answer:

Step-by-step explanation:

I discoverd an error in final step of simplification. The correction is as follows.

ln | (x-1)^2 - 2(x-1)(y-2) -(y-2)^2| - 2ln|(x-1)| = -2(|(x-1) +c)|

Final answer is

(x-1)^2 -2(x-1)(y-2) - (y-2)^2 = 2ln|(x-1)| -2(x-1) + K