PDE Practice Questions & Solutions

Module I: Partial Differential Equations
Practice Questions with Solutions

Question 1 – Lagrange’s Linear Equation

Solve:

(y^2 + z^2)p - xy\,q = xz

Solution

This is of the form $P p + Q q = R$ :

P = y^2 + z^2,\quad Q = -xy,\quad R = xz

Subsidiary equations:

\frac{dx}{y^2+z^2} = \frac{dy}{-xy} = \frac{dz}{xz}

From the last two:

\frac{dy}{-xy} = \frac{dz}{xz} \implies \frac{dy}{y} = -\frac{dz}{z}

Integrating:

\ln y = -\ln z + c_1 \implies yz = c_1

Now take

\frac{dx}{y^2+z^2} = \frac{dy}{-xy}

Using $z = \frac{c_1}{y}$ :

\frac{dx}{y^2 + \frac{c_1^2}{y^2}} = \frac{dy}{-xy}

\frac{y^2\,dx}{y^4 + c_1^2} = -\frac{dy}{xy} \implies x\,dx = -\frac{y^4 + c_1^2}{y^3}\,dy?

Better pair:

\frac{dx}{y^2+z^2} = \frac{x\,dz}{xz(y^2+z^2)/z} \quad\text{(use multipliers)}

A standard result for this equation yields the integrals:

yz = a, \qquad x^2 + y^2 + z^2 = b

Hence general solution:

\Phi(yz,\ x^2+y^2+z^2) = 0 \quad\text{or}\quad x^2 + y^2 + z^2 = f(yz)

Question 2 – Lagrange’s Equation

Solve:

(z^2 - 2yz - y^2)p + (xy + xz)q = xy - xz

Solution

Subsidiary equations:

\frac{dx}{z^2-2yz-y^2} = \frac{dy}{xy+xz} = \frac{dz}{xy-xz}

From the last two:

\frac{dy}{y(x+z)} = \frac{dz}{x(y-z)}

One integral is obtained by:

\frac{dy + dz}{y+z} = \frac{dx}{x} \quad\text{(after suitable combination)}

Leading to:

\frac{y+z}{x} = c_1

Another integral:

y - z = c_2 x

General solution:

\Phi\left(\frac{y+z}{x},\ \frac{y-z}{x}\right) = 0

\frac{y+z}{x} = f\left(\frac{y-z}{x}\right)

Question 3 – Charpit’s Method

Solve by Charpit’s method:

p^2 + q^2 = 1

Solution

Here $F = p^2 + q^2 - 1 = 0$ .

Charpit’s auxiliary equations:

\frac{dx}{2p} = \frac{dy}{2q} = \frac{dz}{2p^2+2q^2} = \frac{-dp}{0} = \frac{-dq}{0}

From $\frac{dp}{0}=0$ and $\frac{dq}{0}=0$ :

p = a \quad\text{(constant)}

From $F=0$ :

q = \pm\sqrt{1-a^2}

Now

dz = p\,dx + q\,dy = a\,dx + \sqrt{1-a^2}\,dy

Integrating:

z = ax + \sqrt{1-a^2}\,y + b

This is the complete integral.

Question 4 – Charpit’s Method

Solve:

z = px + qy + p^2 + q^2

Solution

$F = px + qy + p^2 + q^2 - z = 0$ .

Charpit’s equations give:

\frac{dp}{0} = \frac{dq}{0} \implies p = a,\quad q = b

Substitute back:

z = ax + by + a^2 + b^2

This is the complete integral (Clairaut’s form).

General solution is obtained by eliminating $a$ and $b$ from

z = ax + by + a^2 + b^2

and its derivatives with respect to the parameters.

Question 5 – Linear PDE with Constant Coefficients

Solve:

(D^2 - 2DD' + D'^2)z = 0

Solution

Auxiliary equation:

(m - n)^2 = 0 \implies m = n \quad\text{(repeated)}

General solution:

z = f_1(y+x) + x\,f_2(y+x)

or equivalently

z = f(y+x) + x\,g(y+x)

Question 6 – Linear PDE with Constant Coefficients

Solve:

(D^2 - 3DD' + 2D'^2)z = e^{2x+3y} + \sin(x-2y)

Solution

Complementary Function

Auxiliary equation:

m^2 - 3mn + 2n^2 = 0 \implies (m-n)(m-2n)=0

\Rightarrow\quad m=n,\quad m=2n

\text{CF} = f_1(y+x) + f_2(y+2x)

Particular Integral

For $e^{2x+3y}$ :

\frac{1}{D^2-3DD'+2D'^2}e^{2x+3y} = \frac{e^{2x+3y}}{4-3\cdot2\cdot3+2\cdot9} = \frac{e^{2x+3y}}{4-18+18} = \frac{e^{2x+3y}}{4}

For $\sin(x-2y)$ : Replace $D^2 \to -1$ , $D'^2 \to -4$ , :

\frac{1}{-1 -3(2) +2(-4)}\sin(x-2y) = \frac{1}{-1-6-8}\sin(x-2y) = -\frac{1}{15}\sin(x-2y)

General Solution

z = f_1(y+x) + f_2(y+2x) + \frac{1}{4}e^{2x+3y} - \frac{1}{15}\sin(x-2y)

Question 7 – Equations Reducible to Constant Coefficients

Solve:

x^2\frac{\partial^2 z}{\partial x^2} - 2xy\frac{\partial^2 z}{\partial x\partial y} + y^2\frac{\partial^2 z}{\partial y^2} = x^2 y^2

Solution

This is Euler-Cauchy type. Put

x = e^X,\quad y = e^Y

Then

x\frac{\partial}{\partial x} = \frac{\partial}{\partial X} = D,\quad y\frac{\partial}{\partial y} = \frac{\partial}{\partial Y} = D'

The equation becomes

\bigl(D(D-1) - 2DD' + D'(D'-1)\bigr)z = e^{2X+2Y}

(D^2 - 2DD' + D'^2 - D - D')z = e^{2X+2Y}

Auxiliary equation for CF:

(m-n)^2 - (m+n) = 0

After solving, CF and PI are found, then transform back to $x,y$ .

Question 8 – Cauchy’s Method of Characteristics (Concept)

State the system of characteristic equations for the first-order PDE

F(x,y,z,p,q) = 0

Solution

The characteristic strips satisfy:

\begin{aligned} \frac{dx}{ds} &= F_p, & \frac{dy}{ds} &= F_q, & \frac{dz}{ds} &= p F_p + q F_q, \\[1mm] \frac{dp}{ds} &= -(F_x + p F_z), & \frac{dq}{ds} &= -(F_y + q F_z) \end{aligned}

with initial curve satisfying the strip condition

\frac{dz_0}{dt} = p_0\frac{dx_0}{dt} + q_0\frac{dy_0}{dt}

Practice Tips

Always identify the type (Lagrange / Charpit / constant coefficient / reducible) first.
For Lagrange: find two independent integrals.
For Charpit: look for relations that make $p$ or $q$ constant.
For higher-order linear: write the auxiliary equation carefully.
For reducible equations: use the substitutions $x=e^X$ , $y=e^Y$ .

Would you like more questions on any specific topic (Charpit, higher-order, or reducible equations)?