Question

find the complete integral of 2zx-px2-2qxy+pq=0​

Answer 1

Answer:

Step-by-step explanation:

Applying Charpit auxiliary equation we get

dq / 0 = dp/2z - 2qy = dz/px^2 - pq + 2xyq - pq = dy/2xy - p = dx/x^2 - q = df/0

Consider the equation

2zx - px^2 - 2qxy + pq = 0

Now dq = 0 or q = a

2xz - px^2 - 2axy + pa = 0

p(x^2 - a) = 2x(z - ay)

p = 2x(z - ay) / x^2 - a

dz = p dx + a dy

dz = 2x / x^2 - a dx(z - ay) + ady

2x / x^2 - a dx = dz - a dy / z - ay

dz - a dy / z - ay = 2x / x^2 - a dx

After integrating we get

log (z - ay) = log(x^2 - a) + log b

z - ay = b(x^2 - a)

z = b(x^2 - a) + ay

Answer 2

Answer:

Step-by-step explanation:

Let,

$f(x,y,z.p,q)=2zx-px^2-2qxy+pq=0..................\text{equation-1}$

Perform derivation w.r.t p,q,x,y,z then write charpits relation

$\frac{dx}{-f{p}}=\frac{dy}{-f{q}}=\frac{dz}{-p\times f{p}-q\times f{q}}=\frac{dp}{fx+x\times fz}=\frac{dq}{fy+y\times fz}$

Therefore,

$\frac{dx}{x^{2}-q}=\frac{dy}{2xy-p}}=\frac{dz}{px^{2}-2pq+2qxy}=\frac{dp}{2z-2qy}=\frac{dq}{0}$

As ${dq}=0$ or q=c

As

$\frac{dx}{x^{2}-q}=\frac{dp}{2z-2qy}\\\\p=\frac{2x(z-qy)}{x^{2}-q}\\\\p=\frac{2x(z-cy)}{x^{2}-c}\text{ [ put the value of q=c we get ]}$

Substituting for $p$ and $q$ in $dz$ we get

$dz=p\times dx+q\times dy\\\\\Rightarrow{dz}=\frac{2x(z-cy)}{x^{2}-c}\times dx+c\times dy\\\\\Rightarrow{dz}-c\times dy=\frac{2x(z-cy)}{x^{2}-c}\times dx\\\\\Rightarrow\frac{{dz}-c\times dy}{(z-cy)}=\frac{2x}{x^{2}-c}\times dx$

Integrating we get

$\log (z-cy)=\log (x^{2}-c)+\log k\\\\\Rightarrow\frac{z-cy}{x^{2}-c}=k\\\\\Rightarrow(z-cy)=(x^{2}-c)\times k$

Hence the solution is

$(z-cy)=(x^{2}-c)\times k$