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:

"\\frac{dx}{2y(z-3)}=\\frac{dy}{y(2x-z)}=\\frac{dz}{y(2x-3)}"

A first characteristic equation comes from

"\\frac{dx}{2y(z-3)}=\\frac{dz}{y(2x-3)}"

"(2x-3)dx=2(z-3)dz"

"x^2-3x+\\frac94=z^2-6z+9+C_1"

"(x-\\frac32)^2=(z-3)^2+C_1"

"C_1=(x-\\frac32)^2-(z-3)^2"

A second characteristic equation comes from

"\\frac{dz-dy}{y(2x-3-2x+z)}=\\frac{dx}{2y(z-3)}"

"2d(z-y)=dx"

"2(z-y)=x+C_2"

"C_2=2z-2y-x".

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

"\\Phi(C_1,C_2)=0"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS