Answer to Question #316062 in Differential Equations for sandip

Question #316062

Solve the PDE whose auxiliary equations as follows: 𝑑π‘₯/ 2𝑦(𝑧 βˆ’ 3) = 𝑑𝑦 /𝑦(2π‘₯ βˆ’ 𝑧) = 𝑑𝑧 /𝑦(2π‘₯ βˆ’ 3)


1
Expert's answer
2022-03-29T11:44:37-0400

The auxiliary equations is:

dx2y(zβˆ’3)=dyy(2xβˆ’z)=dzy(2xβˆ’3)\frac{dx}{2y(z-3)}=\frac{dy}{y(2x-z)}=\frac{dz}{y(2x-3)}

A first characteristic equation comes from

dx2y(zβˆ’3)=dzy(2xβˆ’3)\frac{dx}{2y(z-3)}=\frac{dz}{y(2x-3)}

(2xβˆ’3)dx=2(zβˆ’3)dz(2x-3)dx=2(z-3)dz

x2βˆ’3x+94=z2βˆ’6z+9+C1x^2-3x+\frac94=z^2-6z+9+C_1

(xβˆ’32)2=(zβˆ’3)2+C1(x-\frac32)^2=(z-3)^2+C_1

C1=(xβˆ’32)2βˆ’(zβˆ’3)2C_1=(x-\frac32)^2-(z-3)^2

A second characteristic equation comes from

dzβˆ’dyy(2xβˆ’3βˆ’2x+z)=dx2y(zβˆ’3)\frac{dz-dy}{y(2x-3-2x+z)}=\frac{dx}{2y(z-3)}

2d(zβˆ’y)=dx2d(z-y)=dx

2(zβˆ’y)=x+C22(z-y)=x+C_2

C2=2zβˆ’2yβˆ’xC_2=2z-2y-x.

General solution of the PDE on the form of implicit equation:

Ξ¦(C1,C2)=0\Phi(C_1,C_2)=0


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