If the general solution for differential equation dvdx=x+2y−32x+y−3 is |x+y−2|=c|(x−y)n| c > 0 then n = ___
x+2y−32x+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)
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.
dYdX=X+2Y2X+Y which is a homogeneous differential equation.
Put Y=vX so dYdX=v+XdvdX⇒v+Xdvdx=1+2v2+v⇒(2+v1−v2)dv=dxx
which can be solved by variable separable method.
⇒[32(1−v)+12(1+v)]dv=dxx⇒−32log|1−v|+12log|1+v|=log|X|+12logc.⇒log∣∣∣1+v(1−v)3∣∣∣=log x2+log c
⇒∣∣∣1+v(2(1−v)3)∣∣∣=cX2c>0⇒|Y+X|=c|(Y−X)3|⇒|x+y−2|=c|(x−y)3.|, c > 0 (Putting values of x and y from (1) and (2).