Question

If the general solution for differential equation $\frac{dv}{dx}=\frac{x+2y-3}{2x+y-3}is|x+y-2|=c|(x-y{)}^n|$ c > 0 then n = ___

Answer 1

$\frac{x+2y-3}{2x+y-3}$ would have been a homogeneous function of degree zero if the 3 in numerator and denominator were not there. Our aim is to make constant terms zero. Then it will be a homogeneous function of degree zero and hence the equation will be homogeneous differential equation.
Put x = X + h ….(1)
y = Y + k ….(2)
$So,\frac{dy}{dx}=\frac{dY}{dx}So,\frac{dY}{dx}=\frac{X+2Y+(h+2k-3)}{2X+Y(2h+k-3)}$
Now for constant terms to be zero h + 2k - 3 = 0 and 2h + k - 3 should be equal to 0
Solving these 2 equations simultaneously we get h = k =1.
So for h = k = 1.
$\frac{dY}{dX}=\frac{X+2Y}{2X+Y}$ which is a homogeneous differential equation.
$PutY=vX\text{so}\frac{dY}{dX}=v+\frac{Xdv}{dX}\Rightarrow v+\frac{Xdv}{dx}=\frac{1+2v}{2+v}\Rightarrow \left(\frac{2+v}{1-v^2}\right)dv=\frac{dx}{x}$
which can be solved by variable separable method.
$\Rightarrow [\frac{3}{2(1-v)}+\frac{1}{2(1+v)}]dv=\frac{dx}{x}\Rightarrow \frac{-3}{2}\log |1-v|+\frac{1}{2}\log |1+v|=\log |X|+\frac{1}{2}\log c.\Rightarrow \log \left|\frac{1+v}{(1-v{)}^3}\right|=\log x^2+\log c$
$\Rightarrow \left|\frac{1+v}{\left(2\right(1-v{)}^3)}\right|=c{X}^{2}c>0\Rightarrow |Y+X|=c|(Y-X{)}^3|\Rightarrow |x+y-2|=c|(x-y{)}^3.|$, c > 0 (Putting values of x and y from (1) and (2).
So $n=3$.