Answer to Question #164071 in Differential Equations for mohit upadhyay

Question #164071

dx/1+y = dy/1+x = dz/z


1
Expert's answer
2021-02-19T15:32:06-0500

dxdy(1+y)(1+x)=d(xy)xy\dfrac{dx-dy}{(1+y)-(1+x)} =- \dfrac{d(x-y)}{x-y}

Also, since d2=0,dx+dy1+y+1+x=dx+dy+d2x+y+2=d(x+y+2)x+y+2d2=0, \dfrac{dx+dy}{1+y+1+x} = \dfrac{dx+dy+d2}{x+y+2} = \dfrac{d(x+y+2)}{x+y+2}

Each of the fraction of the given DE is equal to

d(xy)xy-\dfrac{d(x-y)}{x-y} and d(x+y+2)x+y+2\dfrac{d(x+y+2)}{x+y+2}

Solving

d(xy)xy=dzz-\dfrac{d(x-y)}{x-y} = \dfrac{dz}{z}

ln(xy)1=lnz+c1c1=1z(xy)ln(x-y)^{-1} = lnz + c_1\\ c_1 = \frac{1}{z(x-y)}

Also, solving

d(x+y+2)x+y+2=dzz\dfrac{d(x+y+2)}{x+y+2} = \dfrac{dz}{z}

ln(x+y+2)=lnz+c2c2=x+y+2zln(x+y+2) = lnz + c_2\\ c_2 = \frac{x+y+2}{z}

So the solution is F(1z(xy),x+y+2z)=0.F(\frac{1}{z(x-y)}, \frac{x+y+2}{z})=0.


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