Question

TI find the equation of the integral surface of the partial differential equation xp+yq=z, which passes through the curve x+y=1,yz=1
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The equation of integral surface for the given partial differential equation is (x+y)²=yz.

Step-by-step explanation:

The given partial differential equation is:

                $xp+yq=z$

The Lagrange's differential equation:

              $\frac{dx}{x} =\frac{dy}{y}=\frac{dz}{z}$

From the first two parts:

on integration,  $\frac{dx}{x}=\frac{dy}{y} \\ \\ \int\limits \frac{1}{x} \, dx =\int\limits \frac{1}{y} \, dy$

                     $lnx=lny+lnc_{1} \\ lnc_{1}=lnx-lny\\ c_{1} =\frac{x}{y}$                        ...................(1)

From the next two parts:

                 $\frac{dy}{y}=\frac{dz}{z} \\ \\ \int\limits \frac{1}{y} \, dy =\int\limits \frac{1}{z} \, dz$

               $lny=lnz+lnc_{2} \\ lnc_{2}=lny-lnz\\ c_{2} =\frac{y}{z}$                        .................(2)

The given curve:  $x+y=1, yz=1$

Consider that  $x=t$   therefore   $y=1-x=1-t$

Put the value of y in eq. $yz=1$  we get, $z=\frac{1}{y}=\frac{1}{1-t}$  

Put the value of x and y in eq. (1)

we get        $c_{1}=\frac{t}{(1-t)}$

Put the value of y and z in eq. (2)

we get        $c_{2}=(1-t)^{2}$

We need eq. only in terms of coefficients

Consider that       $c_{2} (c_{1} +1)^{2} =(1-t)^{2}(\frac{t}{1-t} +1)^{2}$    

                            $c_{2} (c_{1} +1)^{2} =(1-t)^{2}(\frac{t+1-t}{1-t} )^{2}=(1-t)^{2}\frac{1}{(1-t)^{2}} \\ \\ c_{2} (c_{1} +1)^{2}=1$..........(3)

Put the values of c₁ and c₂ in eq. (3),

                    $(\frac{y}{z} )(1+\frac{x}{y} )^{2}=1 \\\\ (\frac{y}{z} )(\frac{x+y}{y} )^{2}=1\\(x+y)^{2}=yz$

Therefore the integral surface equation for PDE is $(x+y)^{2}=yz$  which passed through a curve.