Partial Differential Equations Questions and Answers

Answer: A

Partial derivatives exist only when there is more than one independent variable to differentiate with respect to, so the distinction is the number of independent variables and nothing else. The power of the unknown decides the degree, arbitrary constants arise in both kinds, and both demand the derivatives that appear to exist

Answer: D

Differentiating partially gives \( p = \partial z/\partial x = a \) and \( q = \partial z/\partial y = b \), so substituting back into the original relation removes both constants:

\[ z \;=\; px + qy. \]

Two constants are removed by two differentiations, and the result is of the first order

Answer: C

Each arbitrary function eliminated costs one differentiation, so one function gives a first-order equation and two give a second-order one. The wave equation, for instance, comes of eliminating the two functions in \( y = f(x - ct) + g(x + ct) \). Arbitrary functions rather than constants are what make the general solution of a partial equation so much larger than an ordinary one's

Answer: B

With \( u = x^{2} + y^{2} \) the chain rule gives \( p = 2x f'(u) \) and \( q = 2y f'(u) \), so dividing removes the unknown derivative of \( f \):

\[ \frac{p}{q} \;=\; \frac{x}{y} \quad \Longrightarrow \quad yp - xq \;=\; 0. \]

The relation says the surface is unchanged along circles about the axis, which is exactly what dependence on \( x^{2} + y^{2} \) means

Answer: A

The complete integral \( F(x,y,z,a,b) = 0 \) gives the singular solution when \( a \) and \( b \) are eliminated between it and \( \partial F/\partial a = 0 \), \( \partial F/\partial b = 0 \). Geometrically it is the envelope of the family. Setting the constants to zero merely picks one member, and replacing them by functions leads towards the general solution

Answer: D

Fixing the constants selects one surface out of the two-parameter family, and that is what a particular solution is. The envelope of the family is the singular solution, a different object that need not belong to the family at all, and no meaning attaches to adding two solutions of a nonlinear equation

Answer: C

The equation says the normal to the surface is perpendicular to the vector \( (P, Q, R) \), so that vector lies in the surface and its direction gives the characteristic curves:

\[ \frac{dx}{P} \;=\; \frac{dy}{Q} \;=\; \frac{dz}{R}. \]

Two independent integrals \( u = a \) and \( v = b \) of these give the general solution \( \phi(u,v) = 0 \). Each denominator must sit under the differential of its own variable

Answer: B

The subsidiary equations are \( dx/1 = dy/1 = dz/0 \). The last gives \( z = b \), and the first pair gives \( x - y = a \), so

\[ z \;=\; f(x - y). \]

Checking, \( p = f' \) and \( q = -f' \), which indeed sum to zero. The combination \( x + y \) would satisfy \( p - q = 0 \) instead

Answer: A

The subsidiary equations \( dx/y = dy/(-x) = dz/0 \) give \( z \) constant and \( x\,dx + y\,dy = 0 \), whose integral is

\[ x^{2} + y^{2} \;=\; \text{constant}. \]

So the solution is any function of \( x^{2} + y^{2} \), a surface of revolution about the \( z \)-axis. This is the converse of forming the equation by eliminating that same function

Answer: D

Clairaut's form is \( z = px + qy + f(p,q) \), whose complete integral is obtained simply by replacing \( p \) and \( q \) by constants \( a \) and \( b \):

\[ z \;=\; ax + by + f(a,b). \]

The bare \( f(p,q) = 0 \) and the separable \( f(x,p) = g(y,q) \) are two of the other standard types, each with its own recipe