Answer to Question #237125 in Differential Equations for Anuj

Question #237125

Find the integral surface of (x-y)(p-q) = z(q+1) which contains the circle z=4; x2+y2=42.


1
Expert's answer
2021-09-16T08:03:56-0400

(x-y)(p-q)=z(q+1)

xp-qx-yp+yq=z(q+1)

xp-qx-yp+yq-zq=z

(x-y)p+(-x+y-z)q=z

Corresponding symmetric equation of the given system is;

"\\frac{dx}{x-y}=\\frac{dy}{-x+y-z}=\\frac{dz}{z}"

We need two independent first integral

Adding numerator and denominator we get,

"\\frac{dx+dy+dz}{x-y-x+y-z+z}=\\frac{d(x+y+z)}{0}"

Which means d(x+y+z)=0 so,

1st independent integral is f1=x+y+z

Next:

"\\frac{dx-dy+dz}{x-y+x-y+z+z}=\\frac{dz}{z}"

"\\frac{d(x-y+z)}{x-y+z}=\\frac{2dz}{z}"

"d(\\frac{ln|x-y+z|}{z^{2}})=0"

and we get 2nd independent integral

f2="\\frac{x-y+z}{z^{2}}"

Taking t as a parameter and the given equation of x2+y2=42,z=4. Can be put in parametric form as;

x=t

y=4+/-t

z=4

We then rewrite f1 and f2 as:

f1=t+4+t+4(taking positive value of t in y)

f1=2t+8

t="\\frac{f_1-8}{2}"

f2="\\frac{t-(4+t)+4}{4^{2}}"

f2="\\frac{2t}{4^{2}}"

Replace t in f2

f2="\\frac{f_1-8}{16}"

f1-8-16f2=0

Putting the values of f1 and f2 the desired integral surface is;

x+y+z-16("\\frac{x-y+z}{z^{2}})" -8=0


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