First Order Partial Differential Equations Questions and Answers

Q: A partial differential equation which represents a surface of revolution obtained by revolving a plane curve \(f(x, y) = 0\) about x-axis is :

The equation is y + zq = 0.Revolving a curve about the x-axis gives a surface whose cross-sections perpendicular to the axis are circles, so it has the form\[ y^{2} + z^{2} = \phi(x) \]Treating z as a function of x and y, differentiate with respect to y: \(2y + 2zq = 0\), where \(q = \partial z/\partial y\). So \(y + zq = 0\), and the arbitrary function \(\phi\) has dropped out.Differentiating with respect to x instead gives \(2zp = \phi'(x)\), which still contains \(\phi\). The options \(yp - xq = 0\) and \(xp - yq = 0\) describe other families (for example surfaces of revolution about the z-axis satisfy \(yp - xq = 0\)), and \(x + zp = 0\) is the analogue for an axis along y.

Q: Consider the following two first order partial differential equations\(xp - yq = x\) ................. (1)and \(x^{2}p + q = xz\) ................ (2)Then :

The two equations have common solutions.Solve the pair for p and q. From (2), \(q = xz - x^{2}p\); substituting in (1):\[ \begin{aligned} p &= \frac{1 + yz}{1 + xy} \\ q &= \frac{x(z - x)}{1 + xy} \end{aligned} \]The equations are compatible if \(dz = p\,dx + q\,dy\) is integrable, that is, if \(p_{y} + qp_{z} = q_{x} + pq_{z}\). Working it out:\[ \begin{aligned} p_{y} + qp_{z} &= \frac{(z - x)(1 + xy)}{(1 + xy)^{2}} \\ q_{x} + pq_{z} &= \frac{z - x - x^{2}y + xyz}{(1 + xy)^{2}} \end{aligned} \]and the second numerator factorises as \((z - x)(1 + xy)\), so the two sides agree. The equations are therefore compatible and share a one-parameter family of solutions. Neither solution set contains the other, since each single equation also has solutions that fail the other.

Q: The characteristic curve and the envelope of the one parameter family of surfaces\[ (x - a)^{2} + (y - 2a)^{2} + z^{2} = 1 \]are :

The characteristics are great circles of the spheres and the envelope is (y − 2x)2 + 5z2 = 5.The family is spheres of radius 1 centred at \((a, 2a, 0)\). Differentiate \(F = (x - a)^{2} + (y - 2a)^{2} + z^{2} - 1\) with respect to a:\[ \begin{aligned} -2(x - a) - 4(y - 2a) &= 0 \\ a &= \frac{x + 2y}{5} \end{aligned} \]This plane \(x + 2y = 5a\) passes through the centre \((a, 2a, 0)\), so it cuts each sphere in a great circle: that is the characteristic curve. Substituting a back, \(x - a = \frac{2(2x - y)}{5}\) and \(y - 2a = \frac{y - 2x}{5}\), so\[ \begin{aligned} \frac{4(y - 2x)^{2} + (y - 2x)^{2}}{25} + z^{2} &= 1 \\ (y - 2x)^{2} + 5z^{2} &= 5 \end{aligned} \]The envelope is this cylinder, not \((y - 2x)^{2} + 2z^{2} = 2\) and not a pair of planes.

Q: The solution of the first order partial differential equation \(f(x, y, z, p, q) = 0\) of the form \(F(u, v) = 0\), where u and v are functions of x, y, z is called :

The solution \(F(u, v) = 0\) is the general integral.For a first order equation such as Lagrange's linear equation \(Pp + Qq = R\), if \(u = c_{1}\) and \(v = c_{2}\) are two independent integrals of the auxiliary equations, then \(F(u, v) = 0\) with F an arbitrary function is the general integral (general solution): it contains an arbitrary function, not just arbitrary constants.A complete integral contains two arbitrary constants, \(\phi(x, y, z, a, b) = 0\). A particular integral is obtained from it by fixing the constants, and a singular integral is the envelope of the complete integral, containing no arbitrary element.

Q: The partial differential equations\[ f(x, y, p, q) = 0 \text{ and } g(x, y, p, q) = 0 \]are compatible if :

The equations are compatible if ∂(f, g)/∂(x, p) + ∂(f, g)/∂(y, q) = 0.Two first order equations \(f(x, y, z, p, q) = 0\) and \(g = 0\) are compatible when\[ \begin{aligned} [f, g] &= \frac{\partial(f, g)}{\partial(x, p)} + p\frac{\partial(f, g)}{\partial(z, p)} \\ &+ \frac{\partial(f, g)}{\partial(y, q)} + q\frac{\partial(f, g)}{\partial(z, q)} = 0 \end{aligned} \]Here f and g do not involve z, so the Jacobians with respect to z vanish and the condition reduces to the sum in option C.The options with Jacobians in (x, z) and (y, z) are zero identically for these f and g and test nothing, and the difference of the (x, p) and (y, q) Jacobians has the wrong sign.

Q: The partial differential equation\[ yp - xq = 0 \]has :

The equation has infinitely many surfaces containing the circle.The Lagrange auxiliary equations are \(\frac{dx}{y} = \frac{dy}{-x} = \frac{dz}{0}\), giving the characteristics \(x^{2} + y^{2} = c_{1}\), \(z = c_{2}\). The general integral is \(z = F(x^{2} + y^{2})\) for an arbitrary function F.The circle \(x^{2} + y^{2} = 4\), \(z = 2\) is itself a characteristic curve. Any F with \(F(4) = 2\) gives an integral surface through it, for example \(z = 2\), \(z = \frac{x^{2} + y^{2}}{2}\) or \(z = x^{2} + y^{2} - 2\). There are infinitely many such F.This is the characteristic case of the Cauchy problem, where uniqueness fails; uniqueness holds only when the initial curve is not a characteristic. So 'unique', 'finitely many' and 'none' are all wrong.

Q: Consider the following two statements :(I) The characteristic curve of a one parameter family of surfaces \(f(x, y, z, a) = 0\) is a locus of the common points of intersection of :\[ \begin{aligned} &f(x, y, z, a) = 0 \;\text{ and} \\ &\frac{\partial f(x, y, z, a)}{\partial a} = 0. \end{aligned} \](II) The envelope of the one parameter family of surfaces \(f(x, y, z, a) = 0\) is obtained by eliminating the parameter 'a' from the equations :\[ \begin{aligned} &f(x, y, z, a) = 0 \;\text{ and} \\ &\frac{\partial f(x, y, z, a)}{\partial a} = 0. \end{aligned} \]Then :

(I) and (II) both are true.For a one-parameter family \(f(x, y, z, a) = 0\), two neighbouring members with parameters a and \(a + \delta a\) meet in a curve. As \(\delta a \to 0\), that curve tends to the set satisfying both \(f = 0\) and \(\partial f / \partial a = 0\). This limiting curve is the characteristic curve of the family, so (I) is the definition.The envelope is the surface traced out by all the characteristic curves as a varies. It is found by eliminating a between the same two equations, which is statement (II). The two statements describe the same pair of equations, one before and one after the parameter is eliminated.

Q: Let P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz = 0 be a Pfaffian differential equation.Consider the following three statements.(I) Every Pfaffian differential equation is integrable.(II) A Pfaffian differential equation is integrable if and only if \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 0\), where \(\overline{X} = (P, Q, R)\).(III) A Pfaffian differential equation is integrable if and only if :\[ \begin{aligned} &P\left(\frac{\partial Q}{\partial z} - \frac{\partial R}{\partial y}\right) + Q\left(\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z}\right) + \\ &R\left(\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}\right) = 0 \end{aligned} \]Then :

(II) and (III) are true.A Pfaffian equation \(\overline{X} \cdot d\overline{r} = 0\) is integrable (has a family of integral surfaces) exactly when \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 0\). Expanding,\[ \begin{aligned} \overline{X} \cdot \mathrm{curl}\, \overline{X} = &P(R_{y} - Q_{z}) + Q(P_{z} - R_{x}) \\ &+ R(Q_{x} - P_{y}) \end{aligned} \]The expression in (III) is the negative of this, so setting it to zero is the same condition; (II) and (III) are equivalent forms of one test.(I) is false, because many Pfaffian equations fail this test. For instance \(-y\,dx + x\,dy + dz = 0\) has \(\overline{X} = (-y, x, 1)\), \(\mathrm{curl}\, \overline{X} = (0, 0, 2)\) and \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 2 \neq 0\), so it is not integrable.

Q: The envelope of the one parameter family of planes through (0, 0, 0) to the integral surface of the partial differential equation \(p^{2} + q^{2} = 1\) is :

The envelope is the cone \(x^{2} + y^{2} = z^{2}\).A complete integral of \(p^{2} + q^{2} = 1\) is \(z = ax + by + c\) with \(a^{2} + b^{2} = 1\). Planes through the origin have \(c = 0\); write \(a = \cos\alpha\), \(b = \sin\alpha\):\[ \begin{aligned} z &= x\cos\alpha + y\sin\alpha \\ 0 &= -x\sin\alpha + y\cos\alpha \end{aligned} \]The second line is the derivative with respect to \(\alpha\). Squaring and adding the two lines eliminates \(\alpha\) and gives \(z^{2} = x^{2} + y^{2}\).\(x^{2} + y^{2} = 4\) and \(x^{2} - y^{2} = 4\) do not pass through the origin, and \(x^{2} - z^{2} = y^{2}\) has the sign of \(z^{2}\) reversed.

21.

A partial differential equation which represents a surface of revolution obtained by revolving a plane curve \(f(x, y) = 0\) about x-axis is :

A.\(yp - xq = 0\)
B.\(xp - yq = 0\)
C.\(y + zq = 0\)
D.\(x + zp = 0\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The equation is y + zq = 0.

Revolving a curve about the x-axis gives a surface whose cross-sections perpendicular to the axis are circles, so it has the form

\[ y^{2} + z^{2} = \phi(x) \]

Treating z as a function of x and y, differentiate with respect to y: \(2y + 2zq = 0\), where \(q = \partial z/\partial y\). So \(y + zq = 0\), and the arbitrary function \(\phi\) has dropped out.

Differentiating with respect to x instead gives \(2zp = \phi'(x)\), which still contains \(\phi\). The options \(yp - xq = 0\) and \(xp - yq = 0\) describe other families (for example surfaces of revolution about the z-axis satisfy \(yp - xq = 0\)), and \(x + zp = 0\) is the analogue for an axis along y.

Discuss in Forum

22.

Consider the following two first order partial differential equations

\(xp - yq = x\) ................. (1)
and \(x^{2}p + q = xz\) ................ (2)

Then :

A.they have common solutions
B.every solution of (1) is also the solution of (2)
C.every solution of (2) is also the solution of (1)
D.they do not have any common solutions

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The two equations have common solutions.

Solve the pair for p and q. From (2), \(q = xz - x^{2}p\); substituting in (1):

\[ \begin{aligned} p &= \frac{1 + yz}{1 + xy} \\ q &= \frac{x(z - x)}{1 + xy} \end{aligned} \]

The equations are compatible if \(dz = p\,dx + q\,dy\) is integrable, that is, if \(p_{y} + qp_{z} = q_{x} + pq_{z}\). Working it out:

\[ \begin{aligned} p_{y} + qp_{z} &= \frac{(z - x)(1 + xy)}{(1 + xy)^{2}} \\ q_{x} + pq_{z} &= \frac{z - x - x^{2}y + xyz}{(1 + xy)^{2}} \end{aligned} \]

and the second numerator factorises as \((z - x)(1 + xy)\), so the two sides agree. The equations are therefore compatible and share a one-parameter family of solutions. Neither solution set contains the other, since each single equation also has solutions that fail the other.

Discuss in Forum

23.

The characteristic curve and the envelope of the one parameter family of surfaces

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

are :

A.the great circle of the sphere and \((y - 2x)^{2} + 5z^{2} = 5\) respectively
B.any circle on the sphere and \((y - 2x)^{2} + 2z^{2} = 2\) respectively
C.exactly the same great circles
D.a circle on the sphere and \(x^{2} + y^{2} = 1,\ z = 2a\) respectively

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The characteristics are great circles of the spheres and the envelope is (y − 2x)² + 5z² = 5.

The family is spheres of radius 1 centred at \((a, 2a, 0)\). Differentiate \(F = (x - a)^{2} + (y - 2a)^{2} + z^{2} - 1\) with respect to a:

\[ \begin{aligned} -2(x - a) - 4(y - 2a) &= 0 \\ a &= \frac{x + 2y}{5} \end{aligned} \]

This plane \(x + 2y = 5a\) passes through the centre \((a, 2a, 0)\), so it cuts each sphere in a great circle: that is the characteristic curve. Substituting a back, \(x - a = \frac{2(2x - y)}{5}\) and \(y - 2a = \frac{y - 2x}{5}\), so

\[ \begin{aligned} \frac{4(y - 2x)^{2} + (y - 2x)^{2}}{25} + z^{2} &= 1 \\ (y - 2x)^{2} + 5z^{2} &= 5 \end{aligned} \]

The envelope is this cylinder, not \((y - 2x)^{2} + 2z^{2} = 2\) and not a pair of planes.

Discuss in Forum

24.

The solution of the first order partial differential equation \(f(x, y, z, p, q) = 0\) of the form \(F(u, v) = 0\), where u and v are functions of x, y, z is called :

A.complete integral
B.general integral
C.particular integral
D.singular integral

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The solution \(F(u, v) = 0\) is the general integral.

For a first order equation such as Lagrange's linear equation \(Pp + Qq = R\), if \(u = c_{1}\) and \(v = c_{2}\) are two independent integrals of the auxiliary equations, then \(F(u, v) = 0\) with F an arbitrary function is the general integral (general solution): it contains an arbitrary function, not just arbitrary constants.

A complete integral contains two arbitrary constants, \(\phi(x, y, z, a, b) = 0\). A particular integral is obtained from it by fixing the constants, and a singular integral is the envelope of the complete integral, containing no arbitrary element.

Discuss in Forum

25.

The partial differential equations

\[ f(x, y, p, q) = 0 \text{ and } g(x, y, p, q) = 0 \]

are compatible if :

A.\(\dfrac{\partial(f,g)}{\partial(x,z)}+\dfrac{\partial(f,g)}{\partial(y,z)}=0\)
B.\(\dfrac{\partial(f,g)}{\partial(x,z)}-\dfrac{\partial(f,g)}{\partial(y,z)}=0\)
C.\(\dfrac{\partial(f,g)}{\partial(x,p)}+\dfrac{\partial(f,g)}{\partial(y,q)}=0\)
D.\(\dfrac{\partial(f,g)}{\partial(x,p)}-\dfrac{\partial(f,g)}{\partial(y,q)}=0\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The equations are compatible if ∂(f, g)/∂(x, p) + ∂(f, g)/∂(y, q) = 0.

Two first order equations \(f(x, y, z, p, q) = 0\) and \(g = 0\) are compatible when

\[ \begin{aligned} [f, g] &= \frac{\partial(f, g)}{\partial(x, p)} + p\frac{\partial(f, g)}{\partial(z, p)} \\ &+ \frac{\partial(f, g)}{\partial(y, q)} + q\frac{\partial(f, g)}{\partial(z, q)} = 0 \end{aligned} \]

Here f and g do not involve z, so the Jacobians with respect to z vanish and the condition reduces to the sum in option C.

The options with Jacobians in (x, z) and (y, z) are zero identically for these f and g and test nothing, and the difference of the (x, p) and (y, q) Jacobians has the wrong sign.

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26.

The partial differential equation

\[ yp - xq = 0 \]

has :

A.unique integral surface containing the circle \(x^{2} + y^{2} = 4\), \(z = 2\)
B.finitely many integral surfaces containing the circle \(x^{2} + y^{2} = 4\), \(z = 2\)
C.infinitely many surfaces containing the circle \(x^{2} + y^{2} = 4\), \(z = 2\)
D.no integral surface containing the circle \(x^{2} + y^{2} = 4\), \(z = 2\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The equation has infinitely many surfaces containing the circle.

The Lagrange auxiliary equations are \(\frac{dx}{y} = \frac{dy}{-x} = \frac{dz}{0}\), giving the characteristics \(x^{2} + y^{2} = c_{1}\), \(z = c_{2}\). The general integral is \(z = F(x^{2} + y^{2})\) for an arbitrary function F.

The circle \(x^{2} + y^{2} = 4\), \(z = 2\) is itself a characteristic curve. Any F with \(F(4) = 2\) gives an integral surface through it, for example \(z = 2\), \(z = \frac{x^{2} + y^{2}}{2}\) or \(z = x^{2} + y^{2} - 2\). There are infinitely many such F.

This is the characteristic case of the Cauchy problem, where uniqueness fails; uniqueness holds only when the initial curve is not a characteristic. So 'unique', 'finitely many' and 'none' are all wrong.

Discuss in Forum

27.

Consider the following two statements :

(I) The characteristic curve of a one parameter family of surfaces \(f(x, y, z, a) = 0\) is a locus of the common points of intersection of :

\[ \begin{aligned} &f(x, y, z, a) = 0 \;\text{ and} \\ &\frac{\partial f(x, y, z, a)}{\partial a} = 0. \end{aligned} \]

(II) The envelope of the one parameter family of surfaces \(f(x, y, z, a) = 0\) is obtained by eliminating the parameter 'a' from the equations :

\[ \begin{aligned} &f(x, y, z, a) = 0 \;\text{ and} \\ &\frac{\partial f(x, y, z, a)}{\partial a} = 0. \end{aligned} \]

Then :

A.(I) is true and (II) is false
B.(I) is false and (II) is true
C.(I) and (II) both are false
D.(I) and (II) both are true

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

(I) and (II) both are true.

For a one-parameter family \(f(x, y, z, a) = 0\), two neighbouring members with parameters a and \(a + \delta a\) meet in a curve. As \(\delta a \to 0\), that curve tends to the set satisfying both \(f = 0\) and \(\partial f / \partial a = 0\). This limiting curve is the characteristic curve of the family, so (I) is the definition.

The envelope is the surface traced out by all the characteristic curves as a varies. It is found by eliminating a between the same two equations, which is statement (II). The two statements describe the same pair of equations, one before and one after the parameter is eliminated.

Discuss in Forum

28.

Let P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz = 0 be a Pfaffian differential equation.

Consider the following three statements.

(I) Every Pfaffian differential equation is integrable.
(II) A Pfaffian differential equation is integrable if and only if \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 0\), where \(\overline{X} = (P, Q, R)\).
(III) A Pfaffian differential equation is integrable if and only if :

\[ \begin{aligned} &P\left(\frac{\partial Q}{\partial z} - \frac{\partial R}{\partial y}\right) + Q\left(\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z}\right) + \\ &R\left(\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}\right) = 0 \end{aligned} \]

Then :

A.(I) and (II) are true
B.(I) and (III) are true
C.(II) and (III) are true
D.(I), (II) and (III) are true

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

(II) and (III) are true.

A Pfaffian equation \(\overline{X} \cdot d\overline{r} = 0\) is integrable (has a family of integral surfaces) exactly when \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 0\). Expanding,

\[ \begin{aligned} \overline{X} \cdot \mathrm{curl}\, \overline{X} = &P(R_{y} - Q_{z}) + Q(P_{z} - R_{x}) \\ &+ R(Q_{x} - P_{y}) \end{aligned} \]

The expression in (III) is the negative of this, so setting it to zero is the same condition; (II) and (III) are equivalent forms of one test.

(I) is false, because many Pfaffian equations fail this test. For instance \(-y\,dx + x\,dy + dz = 0\) has \(\overline{X} = (-y, x, 1)\), \(\mathrm{curl}\, \overline{X} = (0, 0, 2)\) and \(\overline{X} \cdot \mathrm{curl}\, \overline{X} = 2 \neq 0\), so it is not integrable.

Discuss in Forum

29.

The envelope of the one parameter family of planes through (0, 0, 0) to the integral surface of the partial differential equation \(p^{2} + q^{2} = 1\) is :

A.\(x^{2} - y^{2} = 4\)
B.\(x^{2} + y^{2} = 4\)
C.\(x^{2} + y^{2} = z^{2}\)
D.\(x^{2} - z^{2} = y^{2}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The envelope is the cone \(x^{2} + y^{2} = z^{2}\).

A complete integral of \(p^{2} + q^{2} = 1\) is \(z = ax + by + c\) with \(a^{2} + b^{2} = 1\). Planes through the origin have \(c = 0\); write \(a = \cos\alpha\), \(b = \sin\alpha\):

\[ \begin{aligned} z &= x\cos\alpha + y\sin\alpha \\ 0 &= -x\sin\alpha + y\cos\alpha \end{aligned} \]

The second line is the derivative with respect to \(\alpha\). Squaring and adding the two lines eliminates \(\alpha\) and gives \(z^{2} = x^{2} + y^{2}\).

\(x^{2} + y^{2} = 4\) and \(x^{2} - y^{2} = 4\) do not pass through the origin, and \(x^{2} - z^{2} = y^{2}\) has the sign of \(z^{2}\) reversed.

First Order Partial Differential Equations Questions and Answers

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