Answer to Question #279219 in Differential Equations for hak5

Question

Find the equation of integral surface to the differential equation

2y(z-3)p+(2x-z)q=y(2x-3) which passes through the circle z=0 , (x)^2 + (y)^2 = 2x

Answer 1

Given equation is 2y(z-3)p+(2x-z)q=y(2x-3), circle is x²+y²=2x,z=0

Langrange's auxiliary equations for equation are dx/(2y(z-3))=dy/(2x-z)=dz/(y(2x-3))

Taking the first and third fractions of it, (2x-3)dz-2(z-3)dz=0

Integrating, x²-3x-z²+6z=c₁.c₁ being an arbitrary constant.

Choosing 1/2,y,-1 as multipliers, each fraction of dx/(2y(z-3))=dy/(2x-z)=dz/(y(2x-3))=((1/2)dx+ydy-dz)/(y(z-3)+y(2x-3)-y(2x-3))=((1/2)dz+ydy-dz)/0

Hence (1/2)dx+ydy-dz=0 or dx+2ydy-2dz=0.

Integrating,x+y²-2z=c₂,c₂ being an arbitrary constant.

Now, the parametric equations of given circle are x=t, y=(2t-t²)^1/2,z=0

substituting this values we have

t²-3t=c₁ and 3t-t²=c₂

Eliminating t from the above equations we have c₁+c₂=0

Substituting the values of c₁ and c₂ the desired integral surface is

x²-3x-z²+6z+x+y²-2z=0 or x²+y²-z²-2x+4z=0