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Differential equations

A differential equation relates an unknown function to its derivatives. In a GATE question you usually have to classify the equation, choose the method that fits it, and fix the constants from initial or boundary conditions. These notes cover first-order methods (separable, homogeneous, exact, linear, Bernoulli, Clairaut), linear equations with constant coefficients (complementary function, particular integral, variation of parameters, reduction of order), Cauchy-Euler equations, IVPs and BVPs with their existence and uniqueness conditions, and separation-of-variables solutions of first-order PDEs and of the heat, wave and Laplace equations.

19 previous-year questions · 19 solved · 6 practice questions

Classifying an ODE: order, degree, linearity

Conventions used throughout. $y^{'} = \frac{d y}{d x}$ , $D \equiv \frac{d}{d x}$ , $j = - 1$ is the imaginary unit, and $ln$ is the natural logarithm. An ordinary differential equation (ODE) has one independent variable. A partial differential equation (PDE) has two or more. Wherever $ln x$ appears, take $x > 0$ unless stated otherwise (for $x < 0$ , replace $ln x$ by $ln ∣ x ∣$ ). All coefficients are real unless stated otherwise.

The order of an ODE is the order of the highest derivative in it. The degree is the power of that highest-order derivative after the equation has been cleared of radicals and fractions in the derivatives, so that it is a polynomial in $y^{'}, y^{''}, \dots$ . If the equation cannot be made a polynomial in the derivatives (for example $y^{''} + sin (y^{'}) = 0$ ), the degree is not defined.

LINEAR ODE OF ORDER N

a_{0} (x) y^{(n)} + a_{1} (x) y^{(n - 1)} + \dots + a_{n - 1} (x) y^{'} + a_{n} (x) y = r (x)

Linear means $y$ and every derivative appear only to the first power, are never multiplied together, and never sit inside another function. The coefficients may depend on $x$ only, and the equation has order $n$ wherever $a_{0} (x) \neq = 0$ . The equation is homogeneous when $r (x) = 0$ and non-homogeneous otherwise.

Equation	Order	Degree	Linear?
$y^{'''} + x^{2} y^{'} - e^{x} y = cos x$	3	1	Yes (variable coefficients are allowed)
$y^{''} + sin y = 0$	2	1	No: $sin y$
$y y^{''} + (y^{'})^{2} = 0$	2	1	No: the product $y y^{''}$
$(y^{'})^{2} + y = x$	1	2	No: $(y^{'})^{2}$
$y = x y^{'} + \frac{1}{y ^{'}}$	1	2 (multiply by $y^{'}$ : $x (y^{'})^{2} - y y^{'} + 1 = 0$ )	No
$[1 + (y^{'})^{2}]^{3/2} = y^{''}$	2	2 (square both sides)	No
$y^{''} + sin (y^{'}) = 0$	2	Not defined	No

The general solution of an $n$ th-order ODE contains exactly $n$ independent (essential) arbitrary constants. A particular solution fixes them from given conditions. Conversely, eliminating $n$ essential constants from a family gives an $n$ th-order ODE. For example, $y = c_{1} e^{2 x} + c_{2} e^{- 3 x}$ gives $y^{''} + y^{'} - 6 y = 0$ .
A singular solution satisfies the ODE but no choice of constants produces it from the general solution. For a linear ODE whose coefficients and $r (x)$ are continuous and whose leading coefficient is non-zero on an interval, the general solution contains every solution on that interval, so singular solutions are met with nonlinear equations (for example Clairaut's equation, in the next section).
Superposition: if $y_{1}$ and $y_{2}$ solve a linear homogeneous ODE, so does $c_{1} y_{1} + c_{2} y_{2}$ . It fails for non-homogeneous and nonlinear equations.
For a linear non-homogeneous ODE, the general solution is the complementary function (the general solution of the homogeneous equation) plus any one particular integral: $y = y_{c} + y_{p}$ .

EXAM TIP

Two traps. (1) A nonlinear equation can still have degree 1. Degree looks only at the highest derivative, while linearity looks at every term: $y^{''} + sin y = 0$ has degree 1 and is nonlinear. (2) 'Homogeneous' has two meanings. A first-order equation $\frac{d y}{d x} = F (\frac{y}{x})$ is homogeneous in the sense of homogeneous functions. A linear equation is homogeneous when $r (x) = 0$ . Check which meaning the question uses.

First-order equations

A first-order equation is written $\frac{d y}{d x} = f (x, y)$ or $M (x, y) d x + N (x, y) d y = 0$ . Test the standard forms roughly in this order: separable, homogeneous, exact, linear, Bernoulli. Use the first one that fits. The general solution has one arbitrary constant, which one initial condition fixes.

VARIABLES SEPARABLE

\frac{d y}{d x} = f (x) g (y) \Rightarrow \int \frac{d y}{g ( y )} = \int f (x) d x + C

Valid where $g (y) \neq = 0$ . Each root $y = y_{0}$ of $g (y) = 0$ is a separate constant solution, and dividing by $g (y)$ can lose it (sometimes it reappears for a particular value of the constant). For $\frac{d y}{d x} = f (a x + b y + c)$ , substitute $z = a x + b y + c$ to get the separable equation $\frac{d z}{d x} = a + b f (z)$ .

HOMOGENEOUS FIRST-ORDER EQUATION

\frac{d y}{d x} = F (\frac{y}{x}), y = v x \Rightarrow \int \frac{d v}{F ( v ) - v} = ln ∣ x ∣ + C

Applies when $f (x, y)$ is unchanged by $x \to k x, y \to k y$ , that is, when $M$ and $N$ are homogeneous functions of the same degree. It uses $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Valid where $F (v) \neq = v$ ; each root $v_{0}$ of $F (v) = v$ gives the straight-line solution $y = v_{0} x$ .

EQUATIONS REDUCIBLE TO HOMOGENEOUS FORM

\frac{d y}{d x} = \frac{a _{1} x + b _{1} y + c _{1}}{a _{2} x + b _{2} y + c _{2}}

If $a_{1} b_{2} - a_{2} b_{1} \neq = 0$ , put $x = X + h$ , $y = Y + k$ , where $(h, k)$ solves $a_{1} h + b_{1} k + c_{1} = 0$ and $a_{2} h + b_{2} k + c_{2} = 0$ . The equation becomes $\frac{d Y}{d X} = \frac{a _{1} X + b _{1} Y}{a _{2} X + b _{2} Y}$ , which is homogeneous. If $a_{1} b_{2} - a_{2} b_{1} = 0$ , the two linear expressions are proportional apart from their constants, so the right-hand side is a function of one combination $a x + b y$ . Substitute $z = a x + b y$ as in the separable case above.

EXACT EQUATION

M d x + N d y = 0 is exact ⟺ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}; \int_{y const} M d x + \int (terms of N free of x) d y = C

The test holds on a simply connected region where $M$ , $N$ and their first partial derivatives are continuous. The solution is $ϕ (x, y) = C$ , where $ϕ_{x} = M$ and $ϕ_{y} = N$ . The shortcut on the right needs $\int M d x$ written without hidden $x$ -free parts (write $ln x$ , not $ln (x y)$ ). The safe route is $ϕ = \int M d x + h (y)$ with $h^{'} (y) = N - \frac{\partial}{\partial y} \int M d x$ .

INTEGRATING FACTORS FOR A NON-EXACT EQUATION

\frac{M _{y} - N _{x}}{N} = f (x) \Rightarrow μ = e^{\int f (x) d x}; \frac{N _{x} - M _{y}}{M} = g (y) \Rightarrow μ = e^{\int g (y) d y}

Multiply the whole equation by $μ$ to make it exact. Two more rules: if $M$ and $N$ are homogeneous of the same degree and $M x + N y \neq = 0$ , then $μ = \frac{1}{M x + N y}$ ; if $M = y f_{1} (x y)$ and $N = x f_{2} (x y)$ with $M x - N y \neq = 0$ , then $μ = \frac{1}{M x - N y}$ .

LINEAR FIRST-ORDER EQUATION

\frac{d y}{d x} + P (x) y = Q (x) : IF = e^{\int P d x}, y \cdot IF = \int Q \cdot IF d x + C

First divide by the coefficient of $\frac{d y}{d x}$ so that it becomes 1. $P$ keeps its sign. If the equation is linear in $x$ instead, $\frac{d x}{d y} + P (y) x = Q (y)$ , use IF $= e^{\int P d y}$ and $x \cdot IF = \int Q \cdot IF d y + C$ .

BERNOULLI EQUATION

\frac{d y}{d x} + P (x) y = Q (x) y^{n} (n \neq = 0, 1) : z = y^{1 - n} \Rightarrow \frac{d z}{d x} + (1 - n) P z = (1 - n) Q

Divide by $y^{n}$ first. The substitution turns the equation into a linear one in $z$ . When $n > 0$ , $y = 0$ is also a solution, which the division by $y^{n}$ loses.

CLAIRAUT'S EQUATION

y = x p + f (p), p = \frac{d y}{d x} : y = c x + f (c)

Differentiating gives $\frac{d p}{d x} [x + f^{'} (p)] = 0$ . The factor $\frac{d p}{d x} = 0$ gives the general solution, the family of straight lines $y = c x + f (c)$ (replace $p$ by $c$ ). The factor $x + f^{'} (p) = 0$ , with $p$ eliminated between it and the equation, gives the singular solution, which is the envelope of those lines. Example: $y = x p + \frac{1}{p}$ has general solution $y = c x + \frac{1}{c}$ ; $x - \frac{1}{p ^{2}} = 0$ gives the singular solution $y^{2} = 4 x$ .

ORTHOGONAL TRAJECTORIES

Cartesian: \frac{d y}{d x} \to - \frac{d x}{d y}; polar: \frac{d r}{d θ} \to - r^{2} \frac{d θ}{d r}

Form the family's ODE with the parameter eliminated, make the replacement, then solve. Example: $y = c x^{2}$ gives $\frac{d y}{d x} = \frac{2 y}{x}$ . The trajectories satisfy $- \frac{d x}{d y} = \frac{2 y}{x}$ , which gives the ellipses $x^{2} + 2 y^{2} = k$ .

WORKED EXAMPLE

Solve $(x^{2} + y^{2} + x) d x + x y d y = 0$ for $x > 0$ .

Solution

$M = x^{2} + y^{2} + x$ and $N = x y$ , so $M_{y} = 2 y$ and $N_{x} = y$ . The equation is not exact.
$\frac{M _{y} - N _{x}}{N} = \frac{2 y - y}{x y} = \frac{1}{x}$ depends on $x$ alone, so $μ = e^{\int d x / x} = x$ .
Multiplying by $x$ gives $(x^{3} + x y^{2} + x^{2}) d x + x^{2} y d y = 0$ . Now $M_{y} = 2 x y = N_{x}$ , so it is exact.
$\int M d x$ (with $y$ constant) $= \frac{x ^{4}}{4} + \frac{x ^{2} y ^{2}}{2} + \frac{x ^{3}}{3}$ . $N = x^{2} y$ has no terms free of $x$ .
So $\frac{x ^{4}}{4} + \frac{x ^{2} y ^{2}}{2} + \frac{x ^{3}}{3} = C$ . Multiply by 12 and rename the constant.

Answer: $3 x^{4} + 6 x^{2} y^{2} + 4 x^{3} = C$ , where $C$ is an arbitrary constant.

WORKED EXAMPLE

Solve $\frac{d y}{d x} + \frac{y}{x} = x^{2} y^{2}$ , $x > 0$ , with $y (1) = 1$ , and find $y (1.5)$ .

Solution

This is Bernoulli with $n = 2$ , $P = \frac{1}{x}$ and $Q = x^{2}$ . Put $z = y^{1 - 2} = \frac{1}{y}$ .
The linear equation is $\frac{d z}{d x} + (1 - 2) \frac{1}{x} z = (1 - 2) x^{2}$ , that is, $\frac{d z}{d x} - \frac{z}{x} = - x^{2}$ .
IF $= e^{- \int d x / x} = \frac{1}{x}$ , so $\frac{d}{d x} (\frac{z}{x}) = - x$ and $\frac{z}{x} = - \frac{x ^{2}}{2} + C$ .
$z (1) = 1$ gives $1 = - \frac{1}{2} + C$ , so $C = \frac{3}{2}$ . Hence $z = \frac{3 x - x ^{3}}{2}$ and $y = \frac{2}{3 x - x ^{3}}$ .
$y (1.5) = \frac{2}{4.5 - 3.375} = \frac{2}{1.125} = \frac{16}{9}$ .

Answer: $y = \frac{2}{3 x - x ^{3}}$ (valid for $0 < x < 3$ ), so $y (1.5) = 16/9 \approx 1.778$ .

EXAM TIP

Three checks. (1) Put a linear equation in standard form before reading off $P$ . For $x y^{'} + 2 y = x^{3}$ , $P = 2/ x$ and IF $= x^{2}$ , not $e^{2 x}$ . (2) For $x > 0$ , $e^{l n x} = x$ and $e^{- l n x} = 1/ x$ . (3) If you divide by $g (y)$ to separate variables, check whether the given initial condition needs one of the constant solutions $g (y) = 0$ .

Linear ODEs with constant coefficients: complementary function and Wronskian

Write the equation as $f (D) y = r (x)$ with real constant coefficients $a_{i}$ . The complementary function $y_{c}$ solves $f (D) y = 0$ . Trying $y = e^{m x}$ turns the ODE into the auxiliary (characteristic) equation $f (m) = 0$ . An $n$ th-order equation has $n$ roots, counted with multiplicity, and $y_{c}$ has $n$ arbitrary constants. Because the coefficients are real, complex roots occur in conjugate pairs $α \pm j β$ .

STRUCTURE OF THE SOLUTION

f (D) y = r (x), f (D) = a_{0} D^{n} + a_{1} D^{n - 1} + \dots + a_{n}; y = y_{c} + y_{p}

$y_{c}$ comes from the roots of $f (m) = 0$ and is the general solution of $f (D) y = 0$ . $y_{p}$ is any one solution of the full equation.

Roots of the auxiliary equation	Contribution to $y_{c}$
Real and distinct: $m_{1}, m_{2}, \dots$	$c_{1} e^{m_{1} x} + c_{2} e^{m_{2} x} + \dots$
Real $m$ , repeated $k$ times	$(c_{1} + c_{2} x + \dots + c_{k} x^{k - 1}) e^{m x}$
Complex pair $α \pm j β$	$e^{α x} (c_{1} cos β x + c_{2} sin β x)$
Complex pair $α \pm j β$ , repeated twice	$e^{α x} [(c_{1} + c_{2} x) cos β x + (c_{3} + c_{4} x) sin β x]$

WRONSKIAN

W (y_{1}, y_{2}) = y_{1} y_{2}^{'} - y_{2} y_{1}^{'} = det [y_{1} y_{1}^{'} y_{2} y_{2}^{'}]

For $n$ functions, use the $n \times n$ determinant whose rows are the functions and their derivatives up to order $n - 1$ . If $W \neq = 0$ at some point of an interval, the functions are linearly independent on that interval.

ABEL'S FORMULA

y^{''} + p (x) y^{'} + q (x) y = 0 \Rightarrow W (x) = K e^{- \int p (x) d x}

Divide by the coefficient of $y^{''}$ first so that it is 1. Holds for two solutions of a linear homogeneous ODE with $p, q$ continuous on an interval. There $W$ is either identically zero (the solutions are dependent) or never zero (they are independent and form a fundamental set).

REDUCTION OF ORDER

y^{''} + p (x) y^{'} + q (x) y = 0, y_{1} known : y_{2} = y_{1} \int \frac{e ^{- \int p d x}}{y _{1}^{2}} d x

Gives a second, independent solution on an interval where $y_{1} \neq = 0$ . It follows from Abel's formula, because $(\frac{y _{2}}{y _{1}})^{'} = \frac{W}{y _{1}^{2}}$ . Example: $x^{2} y^{''} - 3 x y^{'} + 4 y = 0$ ( $x > 0$ ) has $y_{1} = x^{2}$ . Dividing by $x^{2}$ gives $p = - \frac{3}{x}$ , so $e^{- \int p d x} = x^{3}$ and $y_{2} = x^{2} \int \frac{x ^{3}}{x ^{4}} d x = x^{2} ln x$ .

Pair of functions	Wronskian $W (y_{1}, y_{2})$
$e^{m_{1} x}, e^{m_{2} x}$	$(m_{2} - m_{1}) e^{(m_{1} + m_{2}) x}$
$e^{m x}, x e^{m x}$	$e^{2 m x}$
$cos β x, sin β x$	$β$
$e^{α x} cos β x, e^{α x} sin β x$	$β e^{2 α x}$
$x^{m_{1}}, x^{m_{2}}$	$(m_{2} - m_{1}) x^{m_{1} + m_{2} - 1}$

WORKED EXAMPLE

A linear homogeneous ODE with constant coefficients has the general solution $y = c_{1} + e^{- x} (c_{2} cos 2 x + c_{3} sin 2 x)$ . Find the ODE.

Solution

There are three constants, so the order is 3. The constant $c_{1} = c_{1} e^{0 x}$ gives the root $m = 0$ , and $e^{- x} cos 2 x$ , $e^{- x} sin 2 x$ give $m = - 1 \pm 2 j$ .
The quadratic with roots $- 1 \pm 2 j$ is $(m + 1)^{2} + 4 = m^{2} + 2 m + 5$ .
The auxiliary equation is $m (m^{2} + 2 m + 5) = m^{3} + 2 m^{2} + 5 m = 0$ .
Replace each $m^{k}$ by the $k$ th derivative of $y$ .

Answer: $y^{'''} + 2 y^{''} + 5 y^{'} = 0$ .

EXAM TIP

(1) Every solution tends to 0 as $x \to \infty$ exactly when every root has a negative real part. A simple root with zero real part gives a bounded solution (a constant or a pure oscillation). A repeated root with zero real part gives an unbounded one, such as $x$ or $x cos β x$ , and any root with a positive real part gives an unbounded one. (2) For arbitrary functions the Wronskian test works in one direction only. $W \neq = 0$ somewhere proves independence, but $W \equiv 0$ proves dependence only for solutions of one linear ODE. For example, $x^{2}$ and $x ∣ x ∣$ have $W \equiv 0$ yet are independent on $(- 1, 1)$ . (3) For $m^{2} + bm + c = 0$ , the roots add to $- b$ and multiply to $c$ , which quickly checks a factorisation.

Particular integral: operator method, undetermined coefficients, variation of parameters

The particular integral is $y_{p} = \frac{1}{f ( D )} r (x)$ , where $\frac{1}{f ( D )}$ is the inverse of $f (D)$ . $\frac{1}{D}$ means integrate (with no constant), and $\frac{1}{D - a} r = e^{a x} \int e^{- a x} r d x$ . The shortcuts below cover the standard right-hand sides. Use variation of parameters for any other $r (x)$ (such as $tan x$ , $sec x$ , $ln x$ or $1/ x$ ) and for variable coefficients once two independent solutions are known.

PI FOR AN EXPONENTIAL

\frac{1}{f ( D )} e^{a x} = \frac{e ^{a x}}{f ( a )} (f (a) \neq = 0)

Replace $D$ by $a$ . This also covers a constant ( $a = 0$ ) and $cosh a x$ or $sinh a x$ written as exponentials.

PI FOR AN EXPONENTIAL WHEN A IS A ROOT

\frac{1}{f ( D )} e^{a x} = \frac{x ^{k} e ^{a x}}{f ^{(k)} ( a )} if a is a root of f (m) = 0 of multiplicity k

Differentiate $f$ with respect to $D$ $k$ times. For a simple root this gives $\frac{x e ^{a x}}{f ^{'} ( a )}$ , and for a double root $\frac{x ^{2} e ^{a x}}{f ^{''} ( a )}$ .

PI FOR SIN AX AND COS AX

\frac{1}{ϕ ( D ^{2} )} sin a x = \frac{sin a x}{ϕ ( - a ^{2} )}, \frac{1}{ϕ ( D ^{2} )} cos a x = \frac{cos a x}{ϕ ( - a ^{2} )} (ϕ (- a^{2}) \neq = 0)

Replace $D^{2}$ by $- a^{2}$ . Never replace $D$ itself. If odd powers of $D$ remain, as in $\frac{1}{p D + q}$ , multiply numerator and denominator by $p D - q$ so that $D^{2}$ appears. Replace it by $- a^{2}$ again, then apply the $D$ now in the numerator as a derivative.

PI FOR SIN AX AND COS AX AT RESONANCE (D^2 + A^2 = 0)

\frac{1}{D ^{2} + a ^{2}} sin a x = - \frac{x}{2 a} cos a x, \frac{1}{D ^{2} + a ^{2}} cos a x = \frac{x}{2 a} sin a x

Used when $\pm j a$ are roots of the auxiliary equation, so that $ϕ (- a^{2}) = 0$ . Note the minus sign in the sine case.

PI FOR A POLYNOMIAL

\frac{1}{f ( D )} x^{m} = [f (D)]^{- 1} x^{m}, \frac{1}{1 + t} = 1 - t + t^{2} - \dots, \frac{1}{1 - t} = 1 + t + t^{2} + \dots

Take out any factor $D^{k}$ first. Then write the rest as $c [1 \pm ϕ (D)]$ , expand in ascending powers of $D$ , and stop at $D^{m}$ , because higher derivatives of $x^{m}$ vanish. Example: $\frac{1}{1 + D} x^{2} = (1 - D + D^{2}) x^{2} = x^{2} - 2 x + 2$ .

EXPONENTIAL SHIFT

\frac{1}{f ( D )} [e^{a x} V (x)] = e^{a x} \frac{1}{f ( D + a )} V (x)

Move $e^{a x}$ to the left and replace $D$ by $D + a$ . Then handle $V$ (a polynomial, sine or cosine) with the rules above. This covers $x^{n} e^{a x}$ and $e^{a x} sin b x$ .

PI FOR X TIMES V

\frac{1}{f ( D )} [x V (x)] = x \frac{1}{f ( D )} V - \frac{f ^{'} ( D )}{[ f ( D ) ] ^{2}} V

Use for $x sin a x$ or $x cos a x$ . Alternatively, write $sin a x$ as the imaginary part of $e^{j a x}$ and use the exponential shift.

Term in $r (x)$	Trial $y_{p}$ (before the modification rule)
$k e^{a x}$	$A e^{a x}$
A polynomial of degree $n$	$A_{n} x^{n} + A_{n - 1} x^{n - 1} + \dots + A_{0}$
$k cos β x$ or $k sin β x$	$K_{1} cos β x + K_{2} sin β x$
$k e^{α x} cos β x$ or $k e^{α x} sin β x$	$e^{α x} (K_{1} cos β x + K_{2} sin β x)$
$x^{n} e^{a x}$	$e^{a x} (A_{n} x^{n} + \dots + A_{0})$
$x^{n} cos β x$ or $x^{n} sin β x$	$(A_{n} x^{n} + \dots + A_{0}) cos β x + (B_{n} x^{n} + \dots + B_{0}) sin β x$

Undetermined coefficients, sum rule: for $r = r_{1} + r_{2}$ , find a particular integral for each term and add them.
Modification rule: if any term of the trial form already appears in $y_{c}$ (it solves the homogeneous equation), multiply the trial form by $x$ . Use $x^{2}$ if the corresponding root is double, and $x^{k}$ for multiplicity $k$ .
Substitute the trial form into the ODE and equate the coefficients of like terms to find the constants.
The method needs constant coefficients and an $r (x)$ built only from the forms in the table.

VARIATION OF PARAMETERS

y_{p} = - y_{1} \int \frac{y _{2} r ( x )}{W} d x + y_{2} \int \frac{y _{1} r ( x )}{W} d x, W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}

For $y^{''} + p (x) y^{'} + q (x) y = r (x)$ with the coefficient of $y^{''}$ equal to 1, so divide through first. $y_{1}$ and $y_{2}$ are independent solutions of the homogeneous equation. It works for any $r (x)$ continuous on an interval where $p$ and $q$ are continuous.

WORKED EXAMPLE

Find the general solution of $y^{''} - 2 y^{'} + y = 6 x e^{x} + 2 sin x$ .

Solution

The auxiliary equation is $m^{2} - 2 m + 1 = (m - 1)^{2} = 0$ , with double root $m = 1$ , so $y_{c} = (c_{1} + c_{2} x) e^{x}$ .
For $6 x e^{x}$ , use the exponential shift ( $D \to D + 1$ ): $\frac{1}{( D - 1 ) ^{2}} [e^{x} \cdot 6 x] = e^{x} \frac{1}{D ^{2}} (6 x) = e^{x} x^{3}$ .
For $2 sin x$ , put $D^{2} = - 1$ in $D^{2} - 2 D + 1$ to get $- 2 D$ . Then $\frac{1}{- 2 D} (2 sin x) = - \int sin x d x = cos x$ .
Check: for $y = cos x$ , $y^{''} - 2 y^{'} + y = - cos x + 2 sin x + cos x = 2 sin x$ .

Answer: $y = (c_{1} + c_{2} x) e^{x} + x^{3} e^{x} + cos x$ .

WORKED EXAMPLE

Use variation of parameters to find a particular integral of $y^{''} + 4 y = tan 2 x$ on $- \frac{π}{4} < x < \frac{π}{4}$ .

Solution

$y_{1} = cos 2 x$ , $y_{2} = sin 2 x$ and $W = 2$ . The coefficient of $y^{''}$ is already 1, so $r = tan 2 x$ , which is continuous on the interval because $cos 2 x \neq = 0$ there.
First term: $- y_{1} \int \frac{y _{2} r}{W} d x = - \frac{c o s 2 x}{2} \int \frac{s i n ^{2} 2 x}{c o s 2 x} d x = - \frac{c o s 2 x}{2} \int (sec 2 x - cos 2 x) d x$ .
$\int (sec 2 x - cos 2 x) d x = \frac{1}{2} ln ∣ sec 2 x + tan 2 x ∣ - \frac{1}{2} sin 2 x$ , so the first term is $- \frac{1}{4} cos 2 x ln ∣ sec 2 x + tan 2 x ∣ + \frac{1}{4} sin 2 x cos 2 x$ .
Second term: $y_{2} \int \frac{y _{1} r}{W} d x = \frac{s i n 2 x}{2} \int sin 2 x d x = - \frac{1}{4} sin 2 x cos 2 x$ .
Adding, the $sin 2 x cos 2 x$ terms cancel.

Answer: $y_{p} = - \frac{1}{4} cos 2 x ln ∣ sec 2 x + tan 2 x ∣$ , so $y = c_{1} cos 2 x + c_{2} sin 2 x - \frac{1}{4} cos 2 x ln ∣ sec 2 x + tan 2 x ∣$ on any interval where $cos 2 x \neq = 0$ .

EXAM TIP

(1) A particular integral is not unique. Two answers that differ by a term of $y_{c}$ are both correct, so remove any $y_{c}$ terms before comparing with the options. (2) Before variation of parameters, make the coefficient of $y^{''}$ equal to 1. For $x^{2} y^{''} + \dots = r (x)$ , the $r$ in the formula is $r (x) / x^{2}$ . (3) Check for resonance first. If the $e^{a x}$ or $sin a x$ on the right already appears in $y_{c}$ , the ordinary formula divides by zero and the answer needs an extra factor of $x$ (or $x^{2}$ ).

Cauchy-Euler (equidimensional) equations

In this equation each derivative $y^{(k)}$ is multiplied by a constant times $x^{k}$ . It goes by several names for the same form: Cauchy's (homogeneous linear) equation, Euler's equation, and the Euler-Cauchy or equidimensional equation. The substitution $x = e^{z}$ turns it into an equation with constant coefficients in $z$ , so all the methods of the previous two sections apply.

CAUCHY-EULER EQUATION AND SUBSTITUTION

x^{2} y^{''} + a x y^{'} + b y = r (x), x > 0 : x = e^{z}, z = ln x, θ \equiv \frac{d}{d z}

$a$ and $b$ are constants. For $x < 0$ use $x = - e^{z}$ . This amounts to replacing $ln x$ by $ln ∣ x ∣$ and $x^{m}$ by $∣ x ∣^{m}$ .

OPERATOR REPLACEMENTS

x D = θ, x^{2} D^{2} = θ (θ - 1), x^{3} D^{3} = θ (θ - 1) (θ - 2)

Here $D = \frac{d}{d x}$ and $θ = \frac{d}{d z}$ . Rewrite the right-hand side in $z$ as well: $x^{k} = e^{k z}$ and $ln x = z$ .

AUXILIARY EQUATION FROM Y = X^M

m (m - 1) + a m + b = 0

Obtained by trying $y = x^{m}$ directly (equivalently, $θ \to m$ ). The $x^{2} y^{''}$ term contributes $m (m - 1)$ , not $m^{2}$ .

Roots of $m (m - 1) + am + b = 0$	Complementary function ( $x > 0$ )
Real and distinct: $m_{1}, m_{2}$	$c_{1} x^{m_{1}} + c_{2} x^{m_{2}}$
Real, repeated: $m$	$(c_{1} + c_{2} ln x) x^{m}$
Complex: $α \pm j β$	$x^{α} [c_{1} cos (β ln x) + c_{2} sin (β ln x)]$

LEGENDRE'S LINEAR EQUATION

(p x + q)^{2} y^{''} + a (p x + q) y^{'} + b y = r (x) : p x + q = e^{z}, (p x + q) D = p θ, (p x + q)^{2} D^{2} = p^{2} θ (θ - 1)

The same idea, with $x$ replaced by the linear expression $p x + q$ (taken positive).

WORKED EXAMPLE

Find the general solution of $x^{2} y^{''} - 3 x y^{'} + 4 y = x^{2} ln x$ , $x > 0$ .

Solution

With $x = e^{z}$ : $θ (θ - 1) - 3 θ + 4 = θ^{2} - 4 θ + 4 = (θ - 2)^{2}$ . The right-hand side becomes $x^{2} ln x = z e^{2 z}$ .
The root $m = 2$ is double, so $y_{c} = (c_{1} + c_{2} z) e^{2 z} = (c_{1} + c_{2} ln x) x^{2}$ .
$y_{p} = \frac{1}{( θ - 2 ) ^{2}} [e^{2 z} z] = e^{2 z} \frac{1}{θ ^{2}} z = e^{2 z} \frac{z ^{3}}{6}$ , using the exponential shift.
Back in $x$ : $y_{p} = \frac{1}{6} x^{2} (ln x)^{3}$ .

Answer: $y = (c_{1} + c_{2} ln x) x^{2} + \frac{1}{6} x^{2} (ln x)^{3}$ .

EXAM TIP

Shortcut for a power on the right. Write $p (m) = m (m - 1) + am + b$ and let $r (x) = K x^{k}$ . If $p (k) \neq = 0$ , then $y_{p} = \frac{K x ^{k}}{p ( k )}$ , from trying $y = A x^{k}$ . If $k$ is a simple root, $y_{p} = \frac{K x ^{k} l n x}{p ^{'} ( k )}$ with $p^{'} (k) = 2 k - 1 + a$ . If $k$ is a double root, $y_{p} = \frac{K}{2} x^{k} (ln x)^{2}$ . These are the constant-coefficient rules for $e^{k z}$ with $z = ln x$ : the extra factor $x$ there becomes $ln x$ here.

Initial and boundary value problems

An initial value problem (IVP) gives all $n$ conditions at one point: $y (x_{0}), y^{'} (x_{0}), \dots, y^{(n - 1)} (x_{0})$ . Take a linear ODE whose coefficients and $r (x)$ are continuous, with a non-zero leading coefficient, on an interval containing $x_{0}$ . Its IVP has exactly one solution on that interval. A boundary value problem (BVP) gives conditions at two or more points. It can have one solution, none, or infinitely many.

First-order IVPs in general. For $y^{'} = f (x, y)$ with $y (x_{0}) = y_{0}$ : if $f$ and $\frac{\partial f}{\partial y}$ are continuous on a rectangle containing $(x_{0}, y_{0})$ , the IVP has exactly one solution on some interval around $x_{0}$ . Continuity of $f$ alone guarantees at least one solution, but not uniqueness. Example: $y^{'} = y^{1/3}$ , $y (0) = 0$ has $\frac{\partial f}{\partial y} = \frac{1}{3} y^{- 2/3}$ , which is not continuous at $y = 0$ , and for $x \geq 0$ it has two solutions, $y = 0$ and $y = (\frac{2 x}{3})^{3/2}$ .

Find the general solution $y = y_{c} + y_{p}$ first.
Apply the conditions to the full $y$ , not to $y_{c}$ alone, and solve the resulting linear equations for the constants.
For a linear BVP, there is a unique solution for every choice of boundary values exactly when the determinant of that linear system is non-zero. When it is zero, there is either no solution or infinitely many, depending on the boundary values.

WORKED EXAMPLE

Solve $y^{''} + 4 y^{'} + 13 y = 0$ with $y (0) = 1$ and $y^{'} (0) = 0$ , and find $y (π /3)$ .

Solution

$m^{2} + 4 m + 13 = 0 \Rightarrow m = - 2 \pm 3 j$ , so $y = e^{- 2 x} (c_{1} cos 3 x + c_{2} sin 3 x)$ .
$y (0) = c_{1} = 1$ .
$y^{'} = e^{- 2 x} [(3 c_{2} - 2 c_{1}) cos 3 x - (3 c_{1} + 2 c_{2}) sin 3 x]$ , so $y^{'} (0) = 3 c_{2} - 2 c_{1} = 0$ and $c_{2} = \frac{2}{3}$ .
$y = e^{- 2 x} (cos 3 x + \frac{2}{3} sin 3 x)$ . At $x = π /3$ , $cos π = - 1$ and $sin π = 0$ .

Answer: $y (π /3) = - e^{- 2 π /3} \approx - 0.123$ .

The three BVP outcomes for $y^{''} + y = 0$ , whose general solution is $y = c_{1} cos x + c_{2} sin x$ :
$y (0) = 0$ , $y (π /2) = 3$ : $c_{1} = 0$ and $c_{2} = 3$ . One solution, $y = 3 sin x$ .
$y (0) = 0$ , $y (π) = 0$ : $c_{1} = 0$ , and $y (π) = - c_{1} = 0$ holds for every $c_{2}$ . Infinitely many solutions, $y = c_{2} sin x$ .
$y (0) = 0$ , $y (π) = 2$ : $c_{1} = 0$ , so $y (π) = - c_{1} = 0$ , which cannot equal 2. No solution.

EIGENVALUE BVP (FIXED ENDS)

y^{''} + λ y = 0, y (0) = y (L) = 0 : λ_{n} = (\frac{nπ}{L})^{2}, y_{n} = sin \frac{nπ x}{L}, n = 1, 2, 3, \dots

For $λ \leq 0$ , only $y = 0$ satisfies both conditions. With $y^{'} (0) = y^{'} (L) = 0$ instead, $λ_{n} = (nπ / L)^{2}$ and $y_{n} = cos \frac{nπ x}{L}$ for $n = 0, 1, 2, \dots$ ( $n = 0$ , that is $λ = 0$ , gives a constant). These are the spatial modes in the PDE solutions below.

LAPLACE TRANSFORM METHOD FOR AN IVP

L {y^{'}} = s Y (s) - y (0), L {y^{''}} = s^{2} Y (s) - s y (0) - y^{'} (0)

One-sided (unilateral) transform, $Y (s) = \int_{0}^{\infty} y (t) e^{- s t} d t$ , with $t \geq 0$ the independent variable. In general, $L {y^{(n)}} = s^{n} Y (s) - s^{n - 1} y (0) - \dots - y^{(n - 1)} (0)$ . Transform the constant-coefficient ODE, solve for $Y (s)$ with the initial conditions already included, then invert by partial fractions. Example: $y^{'} + 3 y = 0$ , $y (0) = 2$ gives $s Y - 2 + 3 Y = 0$ , so $Y = \frac{2}{s + 3}$ and $y = 2 e^{- 3 t}$ .

EXAM TIP

Differentiate a product like $e^{α x} (\dots)$ carefully before using $y^{'} (0)$ . In the example, the $- 2 c_{1}$ that comes from differentiating $e^{- 2 x}$ is easy to drop. Also fix the constants only after adding $y_{p}$ . Using $y_{c}$ alone gives wrong constants whenever $y_{p} (0) \neq = 0$ or $y_{p}^{'} (0) \neq = 0$ .

Partial differential equations: classification and separation of variables

A second-order linear PDE in two variables is classified by its second-order terms only. Write it as $A u_{xx} + B u_{x y} + C u_{y y} + (lower-order terms) = 0$ , where $B$ is the whole coefficient of $u_{x y}$ . Some texts write $2 B$ for that coefficient, and the test then becomes $B^{2} - A C$ . If $A$ , $B$ and $C$ depend on $x$ and $y$ , the type is decided point by point and can change across the plane.

CLASSIFICATION OF SECOND-ORDER LINEAR PDES

B^{2} - 4 A C < 0 : elliptic, B^{2} - 4 A C = 0 : parabolic, B^{2} - 4 A C > 0 : hyperbolic

For $A u_{xx} + B u_{x y} + C u_{y y} + \dots = 0$ . Lower-order terms and the right-hand side do not affect the type.

Equation	Name and typical model	$A, B, C$	Type
$u_{t} = c^{2} u_{xx}$	Heat (diffusion): temperature in a rod; $c^{2}$ is the diffusivity	$A = c^{2}, B = 0, C = 0$ (in $x, t$ )	Parabolic ( $B^{2} - 4 A C = 0$ )
$u_{tt} = c^{2} u_{xx}$	Wave: vibrating string; $c$ is the wave speed	$A = c^{2}, B = 0, C = - 1$	Hyperbolic ( $4 c^{2} > 0$ )
$u_{xx} + u_{y y} = 0$	Laplace: steady-state temperature, electrostatic potential	$A = 1, B = 0, C = 1$	Elliptic ( $- 4 < 0$ )

Assume $u (x, t) = X (x) T (t)$ . Substitute and divide by $X T$ , so that a function of $x$ equals a function of $t$ . Both must equal the same constant.
Choose the constant as $- λ^{2}$ for the variable with the two homogeneous boundary conditions, so that $X = k_{1} cos λ x + k_{2} sin λ x$ . For insulated ends, also keep the case $λ = 0$ , which gives the constant mode.
Apply the boundary conditions to $X$ . This gives the allowed values $λ_{n}$ and the modes $X_{n}$ (see the eigenvalue BVP above).
Solve the $T$ (or $Y$ ) equation for each $λ_{n}$ and superpose: $u = \sum X_{n} T_{n}$ .
Fit the initial condition (or the remaining boundary condition) with a Fourier sine or cosine series.

HEAT EQUATION, ENDS HELD AT ZERO

u (x, t) = n = 1 \sum \infty b_{n} sin \frac{nπ x}{L} e^{- c^{2} n^{2} π^{2} t / L^{2}}, b_{n} = \frac{2}{L} \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x

For $u_{t} = c^{2} u_{xx}$ , $0 < x < L$ , with $u (0, t) = u (L, t) = 0$ and $u (x, 0) = f (x)$ . Every mode decays, and higher modes decay faster.

HEAT EQUATION, INSULATED ENDS

u (x, t) = a_{0} + n = 1 \sum \infty a_{n} cos \frac{nπ x}{L} e^{- c^{2} n^{2} π^{2} t / L^{2}}, a_{0} = \frac{1}{L} \int_{0}^{L} f (x) d x, a_{n} = \frac{2}{L} \int_{0}^{L} f (x) cos \frac{nπ x}{L} d x

For $u_{x} (0, t) = u_{x} (L, t) = 0$ . As $t \to \infty$ , $u \to a_{0}$ , the average initial temperature.

WAVE EQUATION, STRING FIXED AT BOTH ENDS

u (x, t) = n = 1 \sum \infty (B_{n} cos \frac{c nπ t}{L} + B_{n}^{*} sin \frac{c nπ t}{L}) sin \frac{nπ x}{L}

For $u_{tt} = c^{2} u_{xx}$ with $u (0, t) = u (L, t) = 0$ , $u (x, 0) = f (x)$ and $u_{t} (x, 0) = g (x)$ . Mode $n$ oscillates at $ω_{n} = \frac{c nπ}{L}$ rad/s, that is, $f_{n} = \frac{n c}{2 L}$ Hz.

WAVE EQUATION COEFFICIENTS

B_{n} = \frac{2}{L} \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x, B_{n}^{*} = \frac{2}{c nπ} \int_{0}^{L} g (x) sin \frac{nπ x}{L} d x

A string released from rest ( $g = 0$ ) has every $B_{n}^{*} = 0$ .

D'ALEMBERT'S SOLUTION

u (x, t) = \frac{1}{2} [f (x + c t) + f (x - c t)] + \frac{1}{2 c} \int_{x - c t}^{x + c t} g (s) d s

For the wave equation on an infinite string with $u (x, 0) = f (x)$ and $u_{t} (x, 0) = g (x)$ . It shows two waves travelling at speed $c$ in opposite directions.

LAPLACE EQUATION IN A RECTANGLE

u (x, y) = n = 1 \sum \infty A_{n} sin \frac{nπ x}{a} sinh \frac{nπ y}{a}, A_{n} = \frac{2}{a sinh ( nπ b / a )} \int_{0}^{a} f (x) sin \frac{nπ x}{a} d x

For $u_{xx} + u_{y y} = 0$ on $0 \leq x \leq a$ , $0 \leq y \leq b$ , with $u = 0$ on $x = 0$ , $x = a$ and $y = 0$ , and $u (x, b) = f (x)$ . The sine in $x$ comes from the two zero side conditions, and $sinh$ is the combination in $y$ that vanishes at $y = 0$ .

WORKED EXAMPLE

Solve $u_{t} = 2 u_{xx}$ , $0 < x < π$ , $t > 0$ , with $u (0, t) = u (π, t) = 0$ and $u (x, 0) = 3 sin x - 2 sin 3 x$ . Find $u (π /2, 0.5)$ .

Solution

Here $L = π$ and $c^{2} = 2$ , so the modes are $sin n x e^{- 2 n^{2} t}$ .
The initial condition is already a sine series: $b_{1} = 3$ , $b_{3} = - 2$ , and all other $b_{n} = 0$ . No integration is needed.
$u (x, t) = 3 e^{- 2 t} sin x - 2 e^{- 18 t} sin 3 x$ .
At $x = π /2$ , $t = 0.5$ : $sin (π /2) = 1$ and $sin (3 π /2) = - 1$ , so $u = 3 e^{- 1} + 2 e^{- 9} \approx 1.1036 + 0.0002$ .

Answer: $u (π /2, 0.5) \approx 1.104$ .

WORKED EXAMPLE

Separation of variables also solves first-order linear PDEs. Solve $u_{x} = 2 u_{t} + u$ with $u (x, 0) = 6 e^{- 3 x}$ .

Solution

Put $u = X (x) T (t)$ : $X^{'} T = 2 X T^{'} + X T$ . Dividing by $X T$ gives $\frac{X ^{'}}{X} = \frac{2 T ^{'}}{T} + 1 = k$ , a constant.
$X^{'} = k X$ gives $X = e^{k x}$ (up to a constant factor). $\frac{2 T ^{'}}{T} = k - 1$ gives $T = e^{(k - 1) t /2}$ . So $u = C e^{k x + (k - 1) t /2}$ .
$u (x, 0) = C e^{k x} = 6 e^{- 3 x}$ gives $C = 6$ and $k = - 3$ , so the exponent of $t$ is $\frac{- 3 - 1}{2} = - 2$ .
Check: $u = 6 e^{- 3 x - 2 t}$ has $u_{x} = - 3 u$ and $u_{t} = - 2 u$ , and $2 u_{t} + u = - 4 u + u = - 3 u = u_{x}$ .

Answer: $u (x, t) = 6 e^{- 3 x - 2 t}$ .

EXAM TIP

(1) If the initial condition is already a finite sum of modes, such as $u (x, 0) = 5 sin \frac{2 π x}{L}$ , the solution is those same terms, each multiplied by its own time factor. No Fourier integrals are needed. (2) The steady state of the heat equation solves $u_{xx} = 0$ . With end temperatures $T_{1}$ and $T_{2}$ , it is the straight line $u_{s} = T_{1} + (T_{2} - T_{1}) \frac{x}{L}$ . Solve for $u - u_{s}$ , which is zero at both ends and starts from $f (x) - u_{s} (x)$ . (3) Take the separation constant negative ( $- λ^{2}$ ) for the variable with two homogeneous boundary conditions. A positive constant gives $e^{\pm λ x}$ , which can vanish at both ends only if it is identically zero.

PREVIOUS-YEAR QUESTIONS

19 questions from the EC, EE and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2022 · GATE ECE 2020 · GATE ECE 2018 · GATE EE 2026 · GATE EE 2025 · GATE EE 2024 · GATE EE 2020 · GATE IN 2026 · GATE IN 2025 · GATE IN 2024 · GATE IN 2023 · GATE IN 2022 · GATE IN 2020

GATE 2026 EC Q.11

GATE 2026 · ECQ.11 · 1 mark · MCQ

Consider the differential equation $\dot{w} = A w$ , with $w (t = 0) = [11]$ .

If $w (t) = e^{t} u_{x} + e^{- 2 t} u_{y}$ be the solution to the equation where $u_{x}$ and $u_{y}$ are unit vectors along the positive $x$ and $y$ axes respectively, then which of the following options is the correct matrix representing $A$ ?

(A)
$[10 0 - 2]$
(B)
$[- 1 0 02]$
(C)
$[01 - 2 0]$
(D)
$[0 - 1 20]$

Answer and solution

GATE 2026 EC Q.11: answer and solution

See the question in the paper

Differential equations

Classifying an ODE: order, degree, linearity

First-order equations

Linear ODEs with constant coefficients: complementary function and Wronskian

Particular integral: operator method, undetermined coefficients, variation of parameters

Cauchy-Euler (equidimensional) equations

Initial and boundary value problems

Partial differential equations: classification and separation of variables

19 questions from the EC, EE and IN papers

6 questions to test yourself