b) Solve the PDE
dz/dx.+dz/dy =z^2
Answers
Answer:
please write the class and this object which this question belongs.
Answer:
What is the solution of dx/x^2(x-y) = dy/y^2(x-y) = dz/z (x^2+y^2)?
Raida Innab
Answered 1 year ago
dx/(x^2 (x-y) = dy/y^ 2x-y) = dz/z (x^2 + y ^2)
indefinite integrals:
integrate each side
and
and
then:
{[x ln (x-y) - x ln (x) +y ]/ x(y^2) } + constant
{[y ln (y-x) + x -y ln (y)]/(x^2 (y))} + constant
{(ln z (x^2 + y^2))/(x^2 + y^2)} + constant
((x ln (x - y) - x ln(x) + y)/(x y^2)) + c = ((y ln(y - x) + x - y ln(y))/(x^2 y) )+ c
here log is base 10,
{(x log(x - y) - x log(x) + y)/(x y^2)} + c =
{(y log(y - x) + x - (y log(y))/((x^2) y))} + c=
series expansion at x=0 ==>Laurent expansion
series expansion at x= infinity ==> Piuseux series
{(ln z (x^2 + y^2))/(x^2 + y^2)} + c=
+c
log (x) is the natural logarithm ln
The above relate to 2D Euclidean Quantum Field Theory:
The Laurent series for a complex function f(z) about a point c is given by, used to express complex functions when the Taylor series cannot be applied:
In mathematics, Puiseux series are a generalization of power series that allow for negative and fractional exponents of the indeterminate T. They were first introduced by Isaac Newton in 167 and rediscovered by Victor Puiseux in 1850.
For example:
is a Puiseux series in T.
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Ragu Rajagopalan
Answered 1 year ago
dxx2(x−y)=dyy2(x−y)=dzz(x2+y2)
Considering the first 2 terms:
dxx2(x−y)=dyy2(x−y)⟹dxx2=dyy2
Integrating: −1y=K−1x=Kx−1x
⟹y=x1−Kx
Considering the first and last terms:
dxx2(x−y)=dzz(x2+y2)
⟹dxx2(x−x1−Kx)=dzz(x2+(x1−Kx)2)
⟹dzz=(1−Kx)x2(−Kx2)(1−Kx)2(x2(1−Kx)2+x2)⋅dx
=−((1−Kx)2+1)⋅dxKx2(1−Kx)=−(1−Kx)⋅dxKx2+dxKx2(1−Kx)
=−dxKx2+dxx+1K⋅[dxx2(1−Kx)]
Applying partial fractions:
1x2(1−Kx)=Px+Qx2+R1−Kx
⟹1=(Px+Q)(1−Kx)+Rx2=(R−PK)x2+(P−QK)x+Q
⟹Q=1;P−QK=0⟹P=K;
R=PK=K2
∴1x2(1−Kx)=Kx+1x2+K21−Kx
∴dzz=−dxKx2+dxx+dxK⋅[Kx+1x2+K21−Kx]
⟹dzz=2⋅dxx+K1−Kx
Integrating:
ln|z|=2⋅ln|x|−ln|1−Kx|+lnK2
⟹ln(z(1−Kx)x2)=lnK2
⟹z=K2x21−Kx=K2⋅x⋅⎛⎝⎜⎜⎜x1−Kx=y⎞⎠⎟⎟⎟=K2xy
Ans:
y=x1−Kx;z=K2x21−Kx=K2xy
Vipin Pandey
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Aldino Piva
Answered November 4, 2020
How do you solve -dx/x(x+y) = dy/y (x + y) = dz/ (x -y) (2x+2y+z)?
The equation -dx/x(x+y) = dy/y(x+y) = dZ/(x-y)(2x+2y +z) seems as the
characteristic system of the following PDE
-x(x+y)z_x + y(x+y)z_y = (x-y)( 2x +2y +z) in the unknown z(x,y).
The first two equations give
dy/dx = - y/x from which obtain y = C/x. From the second and third
equations follows
dz/dy - (x-y)z/y(x+y) = 2(x-y)/y . This equation becomes
dz/dy - (C- y^2)z/y( C+y^2) = 2(C- y^2)/y^2 . Thi is a first order DE in
normal form . The integrating factor is IF = (C+y^2)/y and the solution
is given by
z = 1/IF[ Integral of ( 2( C- y^2)IF/y^2)dy + K ] . Follows
z = Ky/(C + y^2) - (C^2 + y^4)/y(C+y^2 ).