Integral surface of PDE x2p + y2q = —z2 which passes through hyperbola xy = x + y, z = 1 isgiven bya)xy + 2y2 + x2 = 3xy2b)yz + 2xy + xz = 3xyzc)xz + 2xz + xy = 3xyzd)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Integral surface of PDE x2p + y2q = &mdash;z2 which passes through hyperbola xy = x + y, z = 1 isgiven bya)xy + 2y2 + x2 = 3xy2b)yz + 2xy + xz = 3xyzc)xz + 2xz + xy = 3xyzd)None of theseCorrect answer is option 'B'. Can you explain this answer?

Answer

Answer

To find the integral surface of the PDE x^2p + y^2q = 0, we can use the method of characteristics.

Let's define a new variable u = xy. Then, we have du = xdy + ydx.

Now, let's rewrite the PDE in terms of u and p:

(u/y^2)p + y^2q = 0

Multiply through by y^2:

up + y^4q = 0

Now, we have a PDE in terms of u and p. We can separate the variables by dividing through by u and q:

p/u + y^4 = 0

This gives us two equations:

p/u = -y^4 (1)

q = 0 (2)

From equation (2), we see that q is constant.

Now, let's solve equation (1) for p:

p = -u/y^4

Substitute this expression for p into equation (2):

q = 0

Therefore, q is constant and equal to 0.

So, the integral surface of the PDE x^2p + y^2q = 0 is given by the equation:

p = -u/y^4

In terms of the original variables x and y, the integral surface is:

p = -xy/(y^4) = -x/y^3

Answer

Solve the pfaffian differential equation by first solving it's intergrability.
(y²+yz)dx+(xz+z²) dy+(y²-xy)dz=0

Answer

Solve the lagrange's auxiliary equations and eliminate the arbitrary constants by solving general solution with the given hyperbola which I think you give wrong equation

Answer

B

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