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Question

If the general solution for differential equation dvdx=x+2y32x+y3 is |x+y2|=c|(xy)n| c > 0 then n = ___


Solution

x+2y32x+y3 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,dydx=dYdxSo,dYdx=X+2Y+(h+2k3)2X+Y(2h+k3)
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+XdvdXv+Xdvdx=1+2v2+v(2+v1v2)dv=dxx 
which can be solved by variable separable method.
[32(1v)+12(1+v)]dv=dxx32log|1v|+12log|1+v|=log|X|+12logc.log1+v(1v)3=log x2+log c 
1+v(2(1v)3)=cX2c>0|Y+X|=c|(YX)3||x+y2|=c|(xy)3.|, c > 0 (Putting values of x and y from (1) and (2). 
So n=3.
 

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