First Order Partial Differential Equations Questions and Answers

Q: The integral surface satisfying the partial differential equation \(\frac{\partial z}{\partial x} + z^{2}\frac{\partial z}{\partial y} = 0\) and passing through the straight line \(x = 1, y = z\) is

The integral surface is \((x - 1)z^{2} + z = y\).This is a quasi-linear equation \(Pp + Qq = R\) with \(P = 1\), \(Q = z^{2}\), \(R = 0\). Lagrange's auxiliary equations are \(\frac{dx}{1} = \frac{dy}{z^{2}} = \frac{dz}{0}\), giving the characteristics:\[ \begin{aligned} z &= c_{1} \\ y - z^{2}x &= c_{2} \end{aligned} \]On the initial line write \(x = 1\), \(y = z = t\). Then \(c_{1} = t\) and \(c_{2} = t - t^{2}\), so \(c_{2} = c_{1} - c_{1}^{2}\). Substituting back:\[ y - z^{2}x = z - z^{2} \;\Rightarrow\; (x - 1)z^{2} + z = y. \]Check: at \(x = 1\) it reduces to \(z = y\). All four surfaces happen to contain that line, so the equation decides. On the other three, \(z\) is not constant along the characteristics; for example on \(x^{2} + y^{2} - z^{2} = 1\), \(z_{x} = x/z\) and \(z_{y} = y/z\), so \(z_{x} + z^{2}z_{y} = x/z + yz \ne 0\).

Q: Let \(u(x,y) = 2f(y)\cos(x - 2y),\ (x,y) \in \mathbb{R}^{2}\), be a solution of the initial value problem\[ \begin{aligned} 2u_{x} + u_{y} &= u \\ u(x, 0) &= \cos(x). \end{aligned} \]Then \(f(1)\) is equal to

\(f(1)\) is equal to \(\frac{e}{2}\).Substitute \(u = 2f(y)\cos(x - 2y)\). Writing \(c = \cos(x - 2y)\) and \(s = \sin(x - 2y)\):\[ \begin{aligned} u_{x} &= -2f s, \quad u_{y} = 2f' c + 4f s \\ 2u_{x} + u_{y} &= 2f' c \end{aligned} \]Setting this equal to \(u = 2f c\) gives \(f' = f\), so \(f(y) = Ae^{y}\). The initial condition \(u(x, 0) = 2A\cos x = \cos x\) gives \(A = \frac{1}{2}\).So \(f(y) = \frac{1}{2}e^{y}\) and \(f(1) = \frac{e}{2}\). The value \(\frac{1}{2}\) is \(f(0)\), and \(e\) forgets the factor 2 in the given form of \(u\).

Q: The Cauchy problem \(uu_{x} + yu_{y} = x\) with \(u(x, 1) = 2x\), when solved using its characteristic equations with an independent variable \(t\), is found to admit of a solution in the form\[ \begin{aligned} & x = \frac{3}{2}se^{t} - \frac{1}{2}se^{-t}, \\ & y = e^{t}, \quad u = f(s, t). \end{aligned} \]Then \(f(s, t) =\)

\(f(s, t) = \frac{3}{2}se^{t} + \frac{1}{2}se^{-t}\).The characteristic equations of \(uu_{x} + yu_{y} = x\) are \(\frac{dx}{dt} = u\), \(\frac{dy}{dt} = y\), \(\frac{du}{dt} = x\), started from the initial curve \(x = s\), \(y = 1\), \(u = 2s\) at \(t = 0\).Since \(\frac{dx}{dt} = u\), the solution is simply the derivative of the given \(x\):\[ \begin{aligned} u &= \frac{dx}{dt} = \frac{d}{dt}\left(\frac{3}{2}se^{t} - \frac{1}{2}se^{-t}\right) \\ &= \frac{3}{2}se^{t} + \frac{1}{2}se^{-t} \end{aligned} \]Check: at \(t = 0\), \(u = \frac{3}{2}s + \frac{1}{2}s = 2s\), matching \(u(x, 1) = 2x\), and \(\frac{du}{dt} = \frac{3}{2}se^{t} - \frac{1}{2}se^{-t} = x\). The option equal to \(x\) itself gives \(u = s\) at \(t = 0\), and the other two fail the initial condition.

Q: Consider the Cauchy problem\[ \begin{aligned} x\frac{\partial u}{\partial x} &+ y\frac{\partial u}{\partial y} \\ &= u; \end{aligned} \]u = f(t) on the initial curve \(\Gamma = (t,t)\); \(t \gt 0\).Consider the following statements:P: If f(t) = 2t + 1, then there exists a unique solution to the Cauchy problem in a neighbourhood of \(\Gamma\).Q: If f(t) = 2t - 1, then there exist infinitely many solutions to the Cauchy problem in a neighbourhood of \(\Gamma\).Then

Both P and Q are FALSE.The characteristic equations are \(\frac{dx}{x} = \frac{dy}{y} = \frac{du}{u}\). The characteristic curves are rays through the origin, y/x constant, along which u grows linearly with the distance: every solution is homogeneous of degree 1, \(u = x\,g(y/x)\).The initial curve \(\Gamma\), the line y = x, is itself a characteristic, so the Cauchy problem is characteristic. Along it, any solution must satisfy u(t, t) = t g(1) = ct for a constant c. A solution exists only if the data f(t) has this form, and then there are infinitely many (g is free away from the ray).Neither 2t + 1 nor 2t - 1 is a constant multiple of t, so in both cases there is no solution at all: no unique solution for P and no infinitely many for Q.

Q: The Pfaffian differential equation :\[ \begin{aligned} \bar{X} \cdot d\bar{r} &= P(x, y, z)\, dx + Q(x, y, z)\, dy \\ &\quad + R(x, y, z)\, dz = 0 \end{aligned} \]is integrable if :

The Pfaffian equation is integrable if \(\bar{X} \cdot \text{curl}\, \bar{X} = 0\).The equation \(P\,dx + Q\,dy + R\,dz = 0\) is integrable, meaning it has a family of solution surfaces \(u(x, y, z) = c\), exactly when there is an integrating factor \(\mu\) with \(\mu\bar{X} = \nabla u\). Taking the curl of \(\mu\bar{X}\) and dotting with \(\bar{X}\) gives the necessary condition\[ \bar{X} \cdot \text{curl}\, \bar{X} = 0 \]and the converse is also true (it is the standard integrability theorem). In components it reads \(P(Q_{z} - R_{y}) + Q(R_{x} - P_{z}) + R(P_{y} - Q_{x}) = 0\).\(\bar{X} \cdot \text{curl}\, \bar{X} \ne 0\) is the condition for non-integrability. \(\bar{X} \cdot \bar{X} = |\bar{X}|^{2}\) only measures the length of the field and says nothing about integrability.

Q: The general solution of the partial differential equation :\[ \begin{aligned} &xzp - yzq = y^{2} - x^{2}, \\ &\left(\frac{\partial z}{\partial x} = p, \frac{\partial z}{\partial y} = q\right) \end{aligned} \]is :

The general solution is \(\phi(xy, x^{2} + y^{2} + z^{2}) = 0\).This is Lagrange's linear equation \(Pp + Qq = R\) with \(P = xz\), \(Q = -yz\), \(R = y^{2} - x^{2}\). The auxiliary equations are \(\dfrac{dx}{xz} = \dfrac{dy}{-yz} = \dfrac{dz}{y^{2} - x^{2}}\).From the first two, \(dx/x = -dy/y\), so \(xy = c_{1}\). Using multipliers \(x, y, z\): \(xP + yQ + zR = x^{2}z - y^{2}z + z(y^{2} - x^{2}) = 0\), so \(x\,dx + y\,dy + z\,dz = 0\) and \(x^{2} + y^{2} + z^{2} = c_{2}\).The general solution is \(\phi(c_{1}, c_{2}) = 0\). The other options pair \(x^{2} + y^{2} + z^{2}\) or \(xyz\) with expressions that are not constant along the characteristics.

Q: The following partial differential equation :\[ \begin{aligned} &(x^{2} + z^{2})\frac{\partial z}{\partial x} - xy\frac{\partial z}{\partial y} \\ &= z^{3}x + y^{2} \end{aligned} \]is :

The equation is quasi-linear.A first-order equation \(a\,z_{x} + b\,z_{y} = c\) is quasi-linear when it is linear in the derivatives \(z_{x}, z_{y}\) but the coefficients \(a, b\) may depend on \(z\) itself. Here the coefficient of \(z_{x}\) is \(x^{2} + z^{2}\), which contains \(z\), and the derivatives appear only to the first power.It is not linear or semi-linear, because those need \(a\) and \(b\) to depend on \(x, y\) only. It is not fully non-linear, since there are no products or powers of the derivatives. The \(z^{3}\) on the right is allowed in a quasi-linear equation.

Q: A set of all surfaces of revolution with \(z\)-axis as the axis of revolution is characterized by the partial differential equation :

The equation is \(y\,\partial z/\partial x - x\,\partial z/\partial y = 0\).A surface of revolution about the \(z\)-axis has the form \(z = f(x^{2} + y^{2})\). Differentiating,\[ \begin{aligned} z_{x} &= 2x f'(x^{2} + y^{2}) \\ z_{y} &= 2y f'(x^{2} + y^{2}) \end{aligned} \]Eliminating \(f'\): \(y z_{x} = 2xy f' = x z_{y}\), so \(y z_{x} - x z_{y} = 0\).The other combinations do not cancel: \(x z_{x} + y z_{y} = 2(x^{2} + y^{2})f'\), and \(x z_{x} - y z_{y}\) and \(y z_{x} + x z_{y}\) leave terms in \(4xyf'\) or \(2(x^{2} - y^{2})f'\).

Q: The partial differential equation\(u_{t} + u_{x} = u^{2}\)can be classified as :

The equation is semilinear.A first-order PDE is semilinear when the derivatives appear linearly with coefficients depending only on the independent variables, while the remaining term may depend non-linearly on \(u\). In \(u_{t} + u_{x} = u^{2}\) the derivative part \(u_{t} + u_{x}\) is linear with constant coefficients, and the non-linearity is only in \(u^{2}\).It is not linear, because of \(u^{2}\). It is non-linear in the broad sense, but the standard classification (linear, semilinear, quasilinear, fully non-linear) places it in the more specific class of semilinear equations.

Q: The complete integral of the partial differential equation\(f(p, q) = 0\)is :

The complete integral is \(z = \phi(a)x + ay + b\).For an equation in \(p = z_{x}\) and \(q = z_{y}\) alone, try \(z = \alpha x + \beta y + b\). Then \(p = \alpha\) and \(q = \beta\), so the equation holds whenever \(f(\alpha, \beta) = 0\). Taking \(q = a\) and solving for \(p = \phi(a)\) gives a two-parameter family \(z = \phi(a)x + ay + b\), which is the complete integral.Option A fixes \(q = 1\), leaving only one effective parameter. Options with \(x^{2}\) give \(p = 2\phi(a)x\) or \(p = 2x\), which depend on \(x\) and do not satisfy a relation between constants \(p\) and \(q\).

The integral surface satisfying the partial differential equation \(\frac{\partial z}{\partial x} + z^{2}\frac{\partial z}{\partial y} = 0\) and passing through the straight line \(x = 1, y = z\) is

A.\((x - 1)z + z^{2} = y^{2}\)
B.\(x^{2} + y^{2} - z^{2} = 1\)
C.\((y - z)x + x^{2} = 1\)
D.\((x - 1)z^{2} + z = y\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The integral surface is \((x - 1)z^{2} + z = y\).

This is a quasi-linear equation \(Pp + Qq = R\) with \(P = 1\), \(Q = z^{2}\), \(R = 0\). Lagrange's auxiliary equations are \(\frac{dx}{1} = \frac{dy}{z^{2}} = \frac{dz}{0}\), giving the characteristics:

\[ \begin{aligned} z &= c_{1} \\ y - z^{2}x &= c_{2} \end{aligned} \]

On the initial line write \(x = 1\), \(y = z = t\). Then \(c_{1} = t\) and \(c_{2} = t - t^{2}\), so \(c_{2} = c_{1} - c_{1}^{2}\). Substituting back:

\[ y - z^{2}x = z - z^{2} \;\Rightarrow\; (x - 1)z^{2} + z = y. \]

Check: at \(x = 1\) it reduces to \(z = y\). All four surfaces happen to contain that line, so the equation decides. On the other three, \(z\) is not constant along the characteristics; for example on \(x^{2} + y^{2} - z^{2} = 1\), \(z_{x} = x/z\) and \(z_{y} = y/z\), so \(z_{x} + z^{2}z_{y} = x/z + yz \ne 0\).

Discuss in Forum

Let \(u(x,y) = 2f(y)\cos(x - 2y),\ (x,y) \in \mathbb{R}^{2}\), be a solution of the initial value problem

\[ \begin{aligned} 2u_{x} + u_{y} &= u \\ u(x, 0) &= \cos(x). \end{aligned} \]

Then \(f(1)\) is equal to

A.\(\frac{1}{2}\)
B.\(\frac{e}{2}\)
C.\(e\)
D.\(\frac{3e}{2}\)

Exam Year(s)

GATE Mathematics - 2015,

Answer: B

\(f(1)\) is equal to \(\frac{e}{2}\).

Substitute \(u = 2f(y)\cos(x - 2y)\). Writing \(c = \cos(x - 2y)\) and \(s = \sin(x - 2y)\):

\[ \begin{aligned} u_{x} &= -2f s, \quad u_{y} = 2f' c + 4f s \\ 2u_{x} + u_{y} &= 2f' c \end{aligned} \]

Setting this equal to \(u = 2f c\) gives \(f' = f\), so \(f(y) = Ae^{y}\). The initial condition \(u(x, 0) = 2A\cos x = \cos x\) gives \(A = \frac{1}{2}\).

So \(f(y) = \frac{1}{2}e^{y}\) and \(f(1) = \frac{e}{2}\). The value \(\frac{1}{2}\) is \(f(0)\), and \(e\) forgets the factor 2 in the given form of \(u\).

Discuss in Forum

The Cauchy problem \(uu_{x} + yu_{y} = x\) with \(u(x, 1) = 2x\), when solved using its characteristic equations with an independent variable \(t\), is found to admit of a solution in the form

\[ \begin{aligned} & x = \frac{3}{2}se^{t} - \frac{1}{2}se^{-t}, \\ & y = e^{t}, \quad u = f(s, t). \end{aligned} \]

Then \(f(s, t) =\)

A.\(\frac{3}{2}se^{t} + \frac{1}{2}se^{-t}\)
B.\(\frac{1}{2}se^{t} + \frac{3}{2}se^{-t}\)
C.\(\frac{1}{2}se^{t} - \frac{3}{2}se^{-t}\)
D.\(\frac{3}{2}se^{t} - \frac{1}{2}se^{-t}\)

Exam Year(s)

GATE Mathematics - 2018,

Answer: A

\(f(s, t) = \frac{3}{2}se^{t} + \frac{1}{2}se^{-t}\).

The characteristic equations of \(uu_{x} + yu_{y} = x\) are \(\frac{dx}{dt} = u\), \(\frac{dy}{dt} = y\), \(\frac{du}{dt} = x\), started from the initial curve \(x = s\), \(y = 1\), \(u = 2s\) at \(t = 0\).

Since \(\frac{dx}{dt} = u\), the solution is simply the derivative of the given \(x\):

\[ \begin{aligned} u &= \frac{dx}{dt} = \frac{d}{dt}\left(\frac{3}{2}se^{t} - \frac{1}{2}se^{-t}\right) \\ &= \frac{3}{2}se^{t} + \frac{1}{2}se^{-t} \end{aligned} \]

Check: at \(t = 0\), \(u = \frac{3}{2}s + \frac{1}{2}s = 2s\), matching \(u(x, 1) = 2x\), and \(\frac{du}{dt} = \frac{3}{2}se^{t} - \frac{1}{2}se^{-t} = x\). The option equal to \(x\) itself gives \(u = s\) at \(t = 0\), and the other two fail the initial condition.

Discuss in Forum

Consider the Cauchy problem

\[ \begin{aligned} x\frac{\partial u}{\partial x} &+ y\frac{\partial u}{\partial y} \\ &= u; \end{aligned} \]

u = f(t) on the initial curve \(\Gamma = (t,t)\); \(t \gt 0\).

Consider the following statements:

P: If f(t) = 2t + 1, then there exists a unique solution to the Cauchy problem in a neighbourhood of \(\Gamma\).
Q: If f(t) = 2t - 1, then there exist infinitely many solutions to the Cauchy problem in a neighbourhood of \(\Gamma\).

Then

A.both P and Q are TRUE
B.P is FALSE and Q is TRUE
C.P is TRUE and Q is FALSE
D.both P and Q are FALSE

Exam Year(s)

GATE Mathematics - 2023,

Answer: D

Both P and Q are FALSE.

The characteristic equations are \(\frac{dx}{x} = \frac{dy}{y} = \frac{du}{u}\). The characteristic curves are rays through the origin, y/x constant, along which u grows linearly with the distance: every solution is homogeneous of degree 1, \(u = x\,g(y/x)\).

The initial curve \(\Gamma\), the line y = x, is itself a characteristic, so the Cauchy problem is characteristic. Along it, any solution must satisfy u(t, t) = t g(1) = ct for a constant c. A solution exists only if the data f(t) has this form, and then there are infinitely many (g is free away from the ray).

Neither 2t + 1 nor 2t - 1 is a constant multiple of t, so in both cases there is no solution at all: no unique solution for P and no infinitely many for Q.

Discuss in Forum

The Pfaffian differential equation :

\[ \begin{aligned} \bar{X} \cdot d\bar{r} &= P(x, y, z)\, dx + Q(x, y, z)\, dy \\ &\quad + R(x, y, z)\, dz = 0 \end{aligned} \]

is integrable if :

A.\(\bar{X} \cdot \bar{X} \ne 0\)
B.\(\bar{X} \cdot \text{curl}\, \bar{X} = 0\)
C.\(\bar{X} \cdot \text{curl}\, \bar{X} \ne 0\)
D.\(\bar{X} \cdot \bar{X} = 0\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The Pfaffian equation is integrable if \(\bar{X} \cdot \text{curl}\, \bar{X} = 0\).

The equation \(P\,dx + Q\,dy + R\,dz = 0\) is integrable, meaning it has a family of solution surfaces \(u(x, y, z) = c\), exactly when there is an integrating factor \(\mu\) with \(\mu\bar{X} = \nabla u\). Taking the curl of \(\mu\bar{X}\) and dotting with \(\bar{X}\) gives the necessary condition

\[ \bar{X} \cdot \text{curl}\, \bar{X} = 0 \]

and the converse is also true (it is the standard integrability theorem). In components it reads \(P(Q_{z} - R_{y}) + Q(R_{x} - P_{z}) + R(P_{y} - Q_{x}) = 0\).

\(\bar{X} \cdot \text{curl}\, \bar{X} \ne 0\) is the condition for non-integrability. \(\bar{X} \cdot \bar{X} = |\bar{X}|^{2}\) only measures the length of the field and says nothing about integrability.

Discuss in Forum

The general solution of the partial differential equation :

\[ \begin{aligned} &xzp - yzq = y^{2} - x^{2}, \\ &\left(\frac{\partial z}{\partial x} = p, \frac{\partial z}{\partial y} = q\right) \end{aligned} \]

is :

A.\(\phi(xy, x^{2} + y^{2} + z^{2}) = 0\)
B.\(\phi(x^{2} - y^{2}, x^{2} + y^{2} + z^{2}) = 0\)
C.\(\phi(xyz, x^{2} - y^{2}) = 0\)
D.\(\phi(xyz, x^{2} + y^{2}) = 0\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The general solution is \(\phi(xy, x^{2} + y^{2} + z^{2}) = 0\).

This is Lagrange's linear equation \(Pp + Qq = R\) with \(P = xz\), \(Q = -yz\), \(R = y^{2} - x^{2}\). The auxiliary equations are \(\dfrac{dx}{xz} = \dfrac{dy}{-yz} = \dfrac{dz}{y^{2} - x^{2}}\).

From the first two, \(dx/x = -dy/y\), so \(xy = c_{1}\). Using multipliers \(x, y, z\): \(xP + yQ + zR = x^{2}z - y^{2}z + z(y^{2} - x^{2}) = 0\), so \(x\,dx + y\,dy + z\,dz = 0\) and \(x^{2} + y^{2} + z^{2} = c_{2}\).

The general solution is \(\phi(c_{1}, c_{2}) = 0\). The other options pair \(x^{2} + y^{2} + z^{2}\) or \(xyz\) with expressions that are not constant along the characteristics.

Discuss in Forum

The following partial differential equation :

\[ \begin{aligned} &(x^{2} + z^{2})\frac{\partial z}{\partial x} - xy\frac{\partial z}{\partial y} \\ &= z^{3}x + y^{2} \end{aligned} \]

is :

A.linear
B.non-linear
C.semi-linear
D.Quasi-linear

Exam Year(s)

MH SET - 2018, MH SET - 2021, SET Exam - 2018, SET Exam - 2021,

Answer: D

The equation is quasi-linear.

A first-order equation \(a\,z_{x} + b\,z_{y} = c\) is quasi-linear when it is linear in the derivatives \(z_{x}, z_{y}\) but the coefficients \(a, b\) may depend on \(z\) itself. Here the coefficient of \(z_{x}\) is \(x^{2} + z^{2}\), which contains \(z\), and the derivatives appear only to the first power.

It is not linear or semi-linear, because those need \(a\) and \(b\) to depend on \(x, y\) only. It is not fully non-linear, since there are no products or powers of the derivatives. The \(z^{3}\) on the right is allowed in a quasi-linear equation.

Discuss in Forum

A set of all surfaces of revolution with \(z\)-axis as the axis of revolution is characterized by the partial differential equation :

A.\(x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} = 0\)
B.\(x\dfrac{\partial z}{\partial x} - y\dfrac{\partial z}{\partial y} = 0\)
C.\(y\dfrac{\partial z}{\partial x} - x\dfrac{\partial z}{\partial y} = 0\)
D.\(y\dfrac{\partial z}{\partial x} + x\dfrac{\partial z}{\partial y} = 0\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The equation is \(y\,\partial z/\partial x - x\,\partial z/\partial y = 0\).

A surface of revolution about the \(z\)-axis has the form \(z = f(x^{2} + y^{2})\). Differentiating,

\[ \begin{aligned} z_{x} &= 2x f'(x^{2} + y^{2}) \\ z_{y} &= 2y f'(x^{2} + y^{2}) \end{aligned} \]

Eliminating \(f'\): \(y z_{x} = 2xy f' = x z_{y}\), so \(y z_{x} - x z_{y} = 0\).

The other combinations do not cancel: \(x z_{x} + y z_{y} = 2(x^{2} + y^{2})f'\), and \(x z_{x} - y z_{y}\) and \(y z_{x} + x z_{y}\) leave terms in \(4xyf'\) or \(2(x^{2} - y^{2})f'\).

Discuss in Forum

The partial differential equation

\(u_{t} + u_{x} = u^{2}\)

can be classified as :

A.Linear
B.Non-linear
C.Semilinear
D.None of the above

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The equation is semilinear.

A first-order PDE is semilinear when the derivatives appear linearly with coefficients depending only on the independent variables, while the remaining term may depend non-linearly on \(u\). In \(u_{t} + u_{x} = u^{2}\) the derivative part \(u_{t} + u_{x}\) is linear with constant coefficients, and the non-linearity is only in \(u^{2}\).

It is not linear, because of \(u^{2}\). It is non-linear in the broad sense, but the standard classification (linear, semilinear, quasilinear, fully non-linear) places it in the more specific class of semilinear equations.

Discuss in Forum

10.

The complete integral of the partial differential equation

\(f(p, q) = 0\)

is :

A.\(z = \phi(a) x + y + b\)
B.\(z = \phi(a) x^{2} + ay + b\)
C.\(z = \phi(a) x + ay + b\)
D.\(z = x^{2} + ay + b\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The complete integral is \(z = \phi(a)x + ay + b\).

For an equation in \(p = z_{x}\) and \(q = z_{y}\) alone, try \(z = \alpha x + \beta y + b\). Then \(p = \alpha\) and \(q = \beta\), so the equation holds whenever \(f(\alpha, \beta) = 0\). Taking \(q = a\) and solving for \(p = \phi(a)\) gives a two-parameter family \(z = \phi(a)x + ay + b\), which is the complete integral.

Option A fixes \(q = 1\), leaving only one effective parameter. Options with \(x^{2}\) give \(p = 2\phi(a)x\) or \(p = 2x\), which depend on \(x\) and do not satisfy a relation between constants \(p\) and \(q\).

First Order Partial Differential Equations Questions and Answers

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