Question

Find the characteristics of the equation pq=z and determine the integral surface which passes
through the parabola x=0, y2=z.

Answer 1

Answer:

The characteristic of the given differential equation is $x=2E(e^{t} -1)$, $y=\frac{E}{2}(e^{t} +1)$ and $z=E^{2} e^{2t}$.

The equation of the integral surface is $16z=(x+4y)^{2}$

Step-by-step explanation:

Given:

The equation of the parabola $pq=z$

The given curve of the parabola is $x=0,y^{2} =z$

To find: The characteristic of the given equation and to determine its integral surface.

Formula used: The  equation of a parabola is: y = a(x-h)2 + k

Here $F(x,y,z,p,q)=pq-z=0$

Therefore the characteristic equation are,

$x'(t)=F_{p}=q$,

$y'(t)=F_{q} =p,$

$z'(t)=pF_{p} +qF_{q}=2pq$,

$p'(t)=-(F_{x} +pF_{z} )=p,$

$q'(t)=-(F_{y} +qF_{z} )=q$

The equation of the given curve is $x=0,y^{2} =z$

we can take the initial values as $x_{0} =0,$ $y_{0} =E,Z_{0} =E^{2}$

From the equation ,$z'_{0}=p_{0} x'_{0}+q_{0} y'_{0},$

Then we get, $q_{0} =2E$

Therefore the given equation $pq=z$ provides $p_{0} =\frac{E}{2}$

Now the equation,

$x'(t)=q,$ and $q'(t)=q$ on integration, which give $x=q+c_{1}$

$y'(t)=p$ and $p'(t)=p$ on integration, which give $y=p+c_{2}$ where $c_{1} , c_{2}$ are constant.

Using the given initial condition, we get

$0=2E+c_{1} ,$

$E=\frac{E}{2} +c_{2}$

From the above equation we get,

$c_{1} =-2E$

$c_{2} =\frac{E}{2}$

Hence the value of x and y is

$x=q-2E$

$y=p+\frac{E}{2}$

Again integrating the equation, $p'(t)=p$ and $q'(t)=q$ then we get

$p=c_{3} e^{t}$

$q=c_{4} e^{t}$  where $c_{3}, c_{4}$  are constant.

Using the initial condition we can get ,

$c_{3} =\frac{E}{2}$

$c_{4} =2E$

Therefore the values of p and q are,

$p=\frac{E}{2} e^{t}$

$q=2Ee^{t}$

Hence, $x=2E(e^{t} -1)$,

$y=\frac{E}{2} (e^{t} +1)$ ⇒$e^{t} =\frac{4y+x}{4y-x}$

$E=\frac{1}{4} (4y-x)$

Integrating the characteristic equation $z'(t)=2pq=2E^{2} e^{2t}$ we get $z=E^{2} e^{2t}$ we have used the initial condition $Z_{0} =E^{2}$  at $t=0$

Finally,

$z=\frac{(4y-x)^{2} }{16} .\frac{(4y+x)^{2} }{4y-x)^{2} }$

$16z=(x+4y)^{2}$

Final Answer:

Hence the characteristic of the differential equation are

$x=2E(e^{t} -1)$

$y=\frac{E}{2}(e^{t} +1)$

$z=E^{2} e^{2t}$

And the equation of the required integral surface is $16z=(x+4y)^{2}$

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