Partial Differential Equations Questions and Answers

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

Answer: B

Here \( A = 1 \), \( B = 0 \) and \( C = 1 \), so

\[ B^{2} - 4AC \;=\; 0 - 4 \;=\; -4 \;<\; 0, \]

which is the elliptic case. Elliptic problems carry data on the whole boundary of a closed region and have smooth solutions throughout, with no direction of propagation

Answer: A

Written in the standard form the only second derivative is \( z_{xx} \), so \( A = \alpha \) with \( B = C = 0 \) and

\[ B^{2} - 4AC \;=\; 0, \]

the parabolic case. Parabolic problems march forward in time from an initial state, and their solutions smooth out as they go, which is why heat conduction has no reverse

Answer: D

Rearranged as \( z_{tt} - c^{2}z_{xx} = 0 \) we have \( A = 1 \), \( B = 0 \), \( C = -c^{2} \), so

\[ B^{2} - 4AC \;=\; 4c^{2} \;>\; 0 \]

for any non-zero \( c \), the hyperbolic case. Hyperbolic equations carry disturbances along characteristics at finite speed, which is what makes a wave a wave

Answer: C

The two arbitrary functions represent disturbances travelling in opposite directions without change of shape:

\[ y(x,t) \;=\; f(x - ct) + g(x + ct). \]

A single function would describe a wave running one way only, which cannot meet two initial conditions, and a product would not satisfy the equation at all

Answer: B

The shape \( f \) is unchanged if \( x \) and \( t \) increase together so as to keep \( x - ct \) fixed, which requires \( dx/dt = c \). So the profile moves bodily at speed \( c \). Amplitude belongs to \( f \) itself, and frequency and wavelength appear only once a particular periodic \( f \) is chosen