Find the integral surface of the pde pq=z containing the curve x-0, z-y2
"F(z,p,q)=pq-z=0"
Trivial solution:
"z=f(x+ay)"
Let "x+ay=u"
then:
"z=f(u), \\frac{\\partial u}{\\partial x}=1, \\frac{\\partial u}{\\partial y}=a"
"p=\\frac{\\partial z}{\\partial x}=\\frac{dz}{du}\\frac{\\partial u}{\\partial x}=\\frac{dz}{du}"
"q=\\frac{\\partial z}{\\partial y}=\\frac{dz}{du}\\frac{\\partial u}{\\partial y}=a\\frac{dz}{du}"
Substituing values of "p" and "q" in the initial equation, we get:
"a(\\frac{dz}{du})^2=z"
"\\frac{dz}{du}=\\sqrt{z\/a}"
"du=dz\/\\sqrt{z\/a}"
"u=2\\sqrt{z\/a}+b"
"x+ay=2\\sqrt{z\/a}+b"
If "x=0, z=y^2" , then the solution is:
"ay=2y\/\\sqrt{a}+b"
"y=\\frac{b}{a-2\/\\sqrt{a}}, z=(\\frac{b}{a-2\/\\sqrt{a}})^2"
Also:
"\\frac{dz}{du}=-\\sqrt{z\/a}"
"x+ay=-2\\sqrt{z\/a}+b"
"ay=-2y\/\\sqrt{a}+b"
"y=\\frac{b}{a+2\/\\sqrt{a}}, z=(\\frac{b}{a+2\/\\sqrt{a}})^2"
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