Question #323276

45) Find the integral surface of the equation x(y ^ 2 + z) * p - y(x ^ 2 + z) * q = (x ^ 2 - y ^ 2) * z whic the straight line x + y = 0, z = 1


1
Expert's answer
2022-04-05T13:03:13-0400

x(y2+z)zdxy(x2+z)zdy=(x2y2)zx(y ^ 2 + z)\frac{\partial z}{dx}- y(x ^ 2 + z)\frac{\partial z}{dy}=(x ^ 2 - y ^ 2) z

x+y=0,z=1x + y = 0, z = 1

The auxiliary equations is:

dxx(y2+z)=dyy(x2+z)=dz(x2y2)z\frac{dx}{x(y ^ 2 + z)}=\frac{dy}{- y(x ^ 2 + z)}=\frac{dz}{(x ^ 2 - y ^ 2) z}

A first characteristic equation comes from

xdx+ydyx2(y2+z)y2(x2+z)=dz(x2y2)z\frac{xdx+ydy}{x^2(y ^ 2 + z)- y^2(x ^ 2 + z)}=\frac{dz}{(x ^ 2 - y ^ 2) z}

xdx+ydy(x2y2)z=dz(x2y2)z\frac{xdx+ydy}{(x ^ 2 - y ^ 2) z}=\frac{dz}{(x ^ 2 - y ^ 2) z}

xdx+ydy=dzxdx+ydy=dz

x2+y22=z+C12\frac{x^2+y^2}{2}=z+\frac{C_1}{2}

C1=x2+y22zC_1=x^2+y^2-2z

A second characteristic equation comes from

ydx+xdyxy(y2+z)xy(x2+z)=dz(x2y2)z\frac{ydx+xdy}{xy(y ^ 2 + z)- xy(x ^ 2 + z)}=\frac{dz}{(x ^ 2 - y ^ 2) z}

ydx+xdyxy(x2y2)=dz(x2y2)z\frac{ydx+xdy}{-xy(x ^ 2 - y ^ 2) }=\frac{dz}{(x ^ 2 - y ^ 2) z}

d(xy)xy=dzz\frac{d(xy)}{-xy}=\frac{dz}{ z}

lnxy+lnC2=lnz-\ln{|xy|}+\ln{C_2}=\ln {|z|}

C2xy=z\frac{C_2}{xy}=z

C2=xyzC_2=xyz

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

Φ(C1,C2)=0\Phi(C_1,C_2)=0.

C1=x2+y22z=x2+y2+2xy2xy2z=C_1=x^2+y^2-2z=x^2+y^2+2xy-2xy-2z=(x+y)22xy2z(x+y)^2-2xy-2z (1)

C2=xyzC_2=xyz (2)

Substitute (2) into (1):

C1=(x+y)22C2z2zC_1=(x+y)^2-2\frac{C_2}{z}-2z

Use x+y=0,z=1x + y = 0, z = 1 to exclude C1C_1 and C2C_2 :

C1=(0)22C2121=2C22C_1=(0)^2-2\frac{C_2}{1}-2\cdot1=-2C_2-2

2C22=x2+y22z-2C_2-2=x^2+y^2-2z

2xyz2=x2+y22z-2xyz-2=x^2+y^2-2z

z=x2+y22+xyz+1z=\frac{x^2+y^2}{2}+xyz+1

Answer: z=x2+y22+xyz+1z=\frac{x^2+y^2}{2}+xyz+1 .


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