Partial Differential Equations Questions and Answers

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

Answer: C

The equation is in Clairaut's form, so \( p \) and \( q \) are replaced by constants directly:

\[ z \;=\; ax + by + ab. \]

Substituting back, \( p = a \) and \( q = b \), and the right-hand side reads \( ax + by + ab \) as required. Dropping the product term would satisfy a different equation, \( z = px + qy \)

Answer: B

Charpit's method is the general procedure: it adjoins an auxiliary equation in \( dp \), \( dq \), \( dx \), \( dy \) and \( dz \), from which a second relation between \( p \) and \( q \) is found and integrated. The standard types are shortcuts worth using where they apply, and a linear equation is Lagrange's case

Answer: A

Putting \( D' = mD \) reduces the operator polynomial to one in \( m \), whose roots \( m_{1}, m_{2}, \dots \) give the solution as a sum of arbitrary functions:

\[ z \;=\; f_{1}(y + m_{1}x) + f_{2}(y + m_{2}x) + \cdots \]

Each root contributes one arbitrary function, in place of the arbitrary constant an ordinary equation would contribute

Answer: D

With \( D' = mD \) the auxiliary equation is

\[ 1 - m^{2} \;=\; 0 \quad \Longrightarrow \quad m \;=\; \pm 1, \]

so the two roots give \( z = f(y + x) + g(y - x) \). Imaginary roots \( m = \pm i \) would come from the *sum* of the two second derivatives, which is Laplace's equation

Answer: C

The discriminant \( B^{2} - 4AC \) decides the type, exactly as it does for a conic: negative gives elliptic, zero parabolic and positive hyperbolic. The classification governs what kind of data the problem needs and how disturbances travel, so it is settled before any method of solution is chosen