Question

x+y+z=d
x+(1+d)y+z=2d
x+y+(1+d)z=0​

Answer 1

Step-by-step explanation:

yzdx−2xzdy+(xy−y3z)dz=0.(1)

(1)yzdx−2xzdy+(xy−y3z)dz=0.

P=yzP=yz, Q=−2xzQ=−2xz and R=xy−y3zR=xy−y3z

Now

P(∂Q∂z−∂R∂y)+Q(∂R∂x−∂P∂z)+R(∂P∂y−∂Q∂z)=P(−2x−x+3y2z)+Q(y−y)+R(z+2z)=yz(−3x+3y2z)+(−2xz)(0)+(xy−y3z)(z+2z)=−3xzy+3y3z2+0+3xyz−3y3z2=0

P(∂Q∂z−∂R∂y)+Q(∂R∂x−∂P∂z)+R(∂P∂y−∂Q∂z)=P(−2x−x+3y2z)+Q(y−y)+R(z+2z)=yz(−3x+3y2z)+(−2xz)(0)+(xy−y3z)(z+2z)=−3xzy+3y3z2+0+3xyz−3y3z2=0

Hence equation (1)(1) is integrable.

Let zz be a constant in (1)(1), then dz=0dz=0. Integrating both sides

∫12xzdx−∫12yzdy12zlnx−1zlnyx1/2y=0=const=Φ(z)(2)

∫12xzdx−∫12yzdy=012zln⁡x−1zln⁡y=const(2)x1/2y=Φ(z)

Differentiating w.r.t xx, yy, zz

x−1/22ydx+(−x1/2y2)dy+(−Φ′(z))dz=0(3)

(3)x−1/22ydx+(−x1/2y2)dy+(−Φ′(z))dz=0

From equations (1)(1) and (3)(3)

yz1/(2x1/2y)=−2xz−x1/2/y2=xy−y3z−Φ′(z).

yz1/(2x1/2y)=−2xz−x1/2/y2=xy−y3z−Φ′(z).