3. Ordinary Differential Equations

Linear differential equations with constant coefficients

Linear ordinary differential equations with constant coefficients form the cornerstone of operational calculus, modeling harmonic oscillators and electrical circuits with exponential and sinusoidal responses.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Superposition and General Solution Structure for Constant Coefficient ODEs

Consider an $n$-th order linear homogeneous differential equation with constant coefficients: $$a_n \frac{d^n y}{dx^n} + a_{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \dots + a_1 \frac{dy}{dx} + a_0 y = 0$$ where $a_i \in \mathbb{R}$ for all $i = 0, 1, \dots, n$ and $a_n \neq 0$. If $y_1(x), y_2(x), \dots, y_n(x)$ form a fundamental set of solutions (i.e., linearly independent with a non-vanishing Wronskian $W(y_1, \dots, y_n) \neq 0$), then the general solution is given by $y(x) = \sum_{i=1}^n C_i y_i(x)$, where $C_i$ are arbitrary constants.

📐 Geometric Interpretation

The solutions define an $n$-dimensional vector space of smooth curves over $\mathbb{R}$. Geometrically, each independent solution represents a fundamental mode of system evolution, and the phase portrait is determined entirely by the eigenvalues of the associated companion matrix.

⚙️ Algebraic & Structural Meaning

The differential operator $L = \sum_{i=0}^n a_i D^i$ (where $D = \frac{d}{dx}$) factors algebraically according to the roots of the auxiliary polynomial $P(m) = \sum_{i=0}^n a_i m^i = 0$, transforming calculus problems into linear algebra eigenvalue problems.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Constant Coefficients ($a_i \in \mathbb{R}$)	`Entire real line $\mathbb{R}$`	Strictly Mandatory / Necessary	If coefficients are variable (e.g., Cauchy-Euler or general variable coefficients), the auxiliary polynomial method fails and power series or Frobenius methods must be invoked.
Linearity	`Domain of definition $I \subset \mathbb{R}$`	Strictly Mandatory	Nonlinear terms (like $y^2$ or $(\frac{dy}{dx})^2$) violate the superposition principle and prevent the formation of a characteristic auxiliary equation.
Non-zero Leading Coefficient ($a_n \neq 0$)	`Discrete condition`	Essential for order definition	If $a_n = 0$, the equation drops in order, altering the dimension of the solution space and reducing the number of independent integration constants.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Auxiliary Equation Roots to Complementary Function (CF)

$$P(m) = a_n m^n + \dots + a_1 m + a_0 = 0$$

Rule: For real distinct roots $m_1, m_2$, CF is $C_1 e^{m_1 x} + C_2 e^{m_2 x}$. For repeated root $m$, use $(C_1 + C_2 x)e^{mx}$. For complex conjugate roots $\alpha \pm i\beta$, use $e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))$.

Application: undefined

Exponential Shift Theorem (Inverse Operator Rule)

$$f(D)(e^{ax} V) = e^{ax} f(D+a)V$$

Rule: To evaluate $\frac{1}{f(D)}(e^{ax} V)$, replace $D$ with $D+a$ and bring $e^{ax}$ to the front. If $f(a) = 0$, factor out $(D-a)^k$ and integrate $k$ times.

Shortcut for Polynomial Right-Hand Side

$$\frac{1}{f(D)} x^k = [f(D)]^{-1} x^k$$

Rule: Expand $[f(D)]^{-1}$ in ascending powers of $D$ up to degree $k$ using binomial expansion and apply differential operators term by term.

Shortcut for Trigonometric RHS (Sine and Cosine)

$$\frac{1}{f(D^2)} \sin(ax) = \frac{1}{f(-a^2)} \sin(ax), \quad \text{provided } f(-a^2) \neq 0$$

Rule: Substitute $-a^2$ for $D^2$ directly. If $f(-a^2) = 0$, multiply by $x$ and differentiate the denominator with respect to $D$.

Application: Extensively tested in UP TGT/PGT for second-order differential equations.

Wronskian Evaluation for Constant Coefficients

$$W(y_1, y_2, \dots, y_n) = Ce^{-\int P_1(x) dx}$$

Rule: For constant coefficient equations, Abel's formula simplifies drastically. The Wronskian is a constant multiple of $e^{-\frac{a_{n-1}}{a_n} x}$.

Application: Used to test linear independence of solution sets in PGT screening tests.

Rapid Elimination via Initial Conditions

$$y(0) = y_0, \quad y'(0) = y_1$$

Rule: Substitute $x=0$ and compute $y'(0)$ directly into the option expressions to eliminate 2 or 3 incorrect options without full integration.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting Repeated Root Multipliers

Common Pitfall: Candidates write $y = (C_1 + C_2)e^{mx}$ or $C_1 e^{mx} + C_2 e^{mx}$ when the auxiliary equation has repeated real roots $m = m_1, m_1$.

Counterexample: $(D-2)^2 y = 0 \implies y \neq (C_1 + C_2)e^{2x}

Why it fails: The terms collapse into a single constant, reducing the order of the solution space from 2 to 1 and failing to form the general solution.

🛡️ Exam Defense: Always multiply the second and subsequent repeated roots by powers of $x$: $C_1 e^{mx} + C_2 x e^{mx} + C_3 x^2 e^{mx} + \dots$

⚠️ COMMON ERROR #02

Misapplying Trigonometric Shortcut when $f(-a^2) = 0$

Common Pitfall: Substituting $-a^2$ for $D^2$ when the denominator becomes zero, leading to division by zero error.

Counterexample: $(\frac{1}{D^2 + a^2}) \sin(ax)$

Why it fails: Division by zero invalidates the algebraic substitution. The failure indicates resonance.

🛡️ Exam Defense: When $f(-a^2) = 0$, multiply the operator expression by $x$ and differentiate the denominator: $\frac{x}{2D} \sin(ax) = -\frac{x}{2a} \cos(ax)$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Homogeneous Constant Coefficient ODE	Constant coefficients $a_i \in \mathbb{R}$, RHS = 0	$y = \sum C_i e^{m_i x}	Exponential growth/decay or oscillation	RHS is identically zero.
Non-Homogeneous (Undetermined Coefficients)	Constant coefficients, RHS is $e^{ax}, \sin(bx), \cos(bx)$ or polynomial	$y = CF + PI$	Forced oscillation / particular response	RHS has standard algebraic/trig/exponential form.
Cauchy-Euler Equidimensional Equation	Variable coefficients of form $x^n y^{(n)} + \dots + a_0 y = 0$	Substitute $x = e^t$ to get constant coefficients	Scaling invariant curves	Variable coefficients with matching derivative orders and powers of $x$.
Variation of Parameters	Linear ODE of any order, constant or variable coefficients	$PI = -y_1 \int \frac{y_2 R}{W} dx + y_2 \int \frac{y_1 R}{W} dx$	Integral formula using Wronskian	RHS is non-standard (e.g., $\tan x$, $\sec x$, $\ln x$).

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the general solution of the differential equation $\frac{d^2 y}{dx^2} - 4\frac{dy}{dx} + 4y = 0$.

(A) $y = (C_1 + C_2 x)e^{2x}$(B) $y = C_1 e^{2x} + C_2 e^{-2x}$(C) $y = (C_1 \cos 2x + C_2 \sin 2x)e^x$(D) $y = C_1 e^{2x} + C_2 x^2 e^{2x}$

📚 Method A: Formal Textbook Derivation

Step 1: Write auxiliary equation $m^2 - 4m + 4 = 0$. Step 2: Factor as $(m-2)^2 = 0$, giving repeated roots $m = 2, 2$. Step 3: Form general solution for repeated roots: $y = (C_1 + C_2 x)e^{2x}$.

⚡ Method B: 25-Second Exam Shortcut

Recognize repeated root $m=2$ immediately and check for the factor of $x$ in the linear combination term $(C_1 + C_2 x)$.

💡 Tip: Watch out for distractors omitting the factor of $x$ in repeated root cases.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Determine the particular integral (PI) for $(D^2 + 4)y = \sin(2x)$.

(A) $-\frac{x}{4} \cos(2x)$(B) $\frac{x}{2} \sin(2x)$(C) $\frac{1}{4} \sin(2x)$(D) $-\frac{x}{2} \cos(2x)$

📚 Method A: Formal Textbook Derivation

Step 1: Write $PI = \frac{1}{D^2 + 4} \sin(2x)$. Step 2: Substituting $D^2 = -(2)^2 = -4$ gives denominator $-4 + 4 = 0$ (failure). Step 3: Apply resonance rule: multiply by $x$ and differentiate denominator with respect to $D$: $PI = x \frac{1}{2D} \sin(2x) = \frac{x}{2} \left(-\frac{\cos(2x)}{2}\right) = -\frac{x}{4} \cos(2x)$.

⚡ Method B: 25-Second Exam Shortcut

Use the derivative shortcut: $\frac{1}{f(D^2)}\sin(ax) \implies x \frac{1}{f'(D^2)}\sin(ax)$ when $f(-a^2)=0$. Here $f'(D^2) = 2D$, so $\frac{x}{2D}\sin(2x) = -\frac{x}{4}\cos(2x).$

💡 Tip: This is a recurring UP PGT problem testing failure of the standard trigonometric operator rule.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The auxiliary equation of $(D^3 - 3D^2 + 4D - 2)y = 0$ has roots:

A $1, 1 \pm i$

B $-1, 1 \pm i$

C $1, 1, 2$

D $1, -1, -2$

Derivation & Analysis:

Testing $m=1$ in $m^3 - 3m^2 + 4m - 2 = 0$ yields $1 - 3 + 4 - 2 = 0$. Synthetic division by $(m-1)$ leaves $m^2 - 2m + 2 = 0$, whose roots are $1 \pm i$. Thus roots are $1, 1 \pm i$.

MCQ 2: The complementary function of $\frac{d^3 y}{dx^3} - y = 0$ is:

A $C_1 e^x + e^{-x/2}(C_2 \cos(\frac{\sqrt{3}}{2}x) + C_3 \sin(\frac{\sqrt{3}}{2}x))$

B $C_1 e^x + C_2 e^{-x} + C_3 e^{2x}$

C $C_1 e^x + (C_2 + C_3 x)e^{-x}$

D $C_1 \cos x + C_2 \sin x + C_3 e^x$

Derivation & Analysis:

Auxiliary equation is $m^3 - 1 = 0 \implies (m-1)(m^2+m+1) = 0$. Roots are $m = 1$ and $m = -1/2 \pm i\sqrt{3}/2$. This gives CF: $C_1 e^x + e^{-x/2}(C_2 \cos(\frac{\sqrt{3}}{2}x) + C_3 \sin(\frac{\sqrt{3}}{2}x))$. Accuracy in complex root formulation is vital.

MCQ 3: The particular integral (PI) of $(D^2 - 4)y = e^{2x}$ is:

A $\frac{x}{4} e^{2x}$

B $\frac{1}{4} e^{2x}$

C $x e^{2x}$

D $\frac{x^2}{2} e^{2x}$

Derivation & Analysis:

Substitute $D = 2$ into $\frac{1}{D^2-4} e^{2x}$. Since $2^2 - 4 = 0$, we apply the exponential failure rule: $PI = x \frac{1}{2D} e^{2x} = \frac{x}{2(2)} e^{2x} = \frac{x}{4} e^{2x}$.

MCQ 4: If the auxiliary equation of a second-order linear homogeneous ODE has equal roots $m_1 = m_2 = 3$, the general solution is:

A $C_1 e^{3x} + C_2 x e^{3x}$

B $(C_1 + C_2 x^2) e^{3x}$

C $C_1 e^{3x} + C_2 e^{-3x}$

D $(C_1 \cos 3x + C_2 \sin 3x)e^x$

Derivation & Analysis:

For repeated real roots $m$, the independent solutions are $e^{mx}$ and $x e^{mx}$. Thus $y = (C_1 + C_2 x)e^{3x}$.

MCQ 5: The particular integral of $(D^2 + 1)y = x^2$ is:

A $x^2 - 2$

B $x^2 + 2$

C $x^2 - 1$

D $x^2$

Derivation & Analysis:

$PI = (1 + D^2)^{-1} x^2 = (1 - D^2 + D^4 - \dots)x^2 = x^2 - \frac{d^2}{dx^2}(x^2) = x^2 - 2$.

MCQ 6: The Wronskian of the solutions $e^{2x}$ and $e^{-2x}$ of $\frac{d^2 y}{dx^2} - 4y = 0$ is:

A $-4$

B $4$

C $0$

D $2e^{2x}$

Derivation & Analysis:

$W = \begin{vmatrix} e^{2x} & e^{-2x} \\ 2e^{2x} & -2e^{-2x} \end{vmatrix} = e^{2x}(-2e^{-2x}) - e^{-2x}(2e^{2x}) = -2 - 2 = -4$.

MCQ 7: Which of the following differential equations has $y = C_1 \cos(3x) + C_2 \sin(3x)$ as its general solution?

A $\frac{d^2 y}{dx^2} + 9y = 0$

B $\frac{d^2 y}{dx^2} - 9y = 0$

C $\frac{d^2 y}{dx^2} + 3y = 0$

D $\frac{d^2 y}{dx^2} - 3y = 0$

Derivation & Analysis:

Complex roots $\alpha \pm i\beta$ with $\alpha = 0, \beta = 3$ correspond to auxiliary roots $m = \pm 3i$, yielding $m^2 + 9 = 0$, which gives the differential equation $\frac{d^2 y}{dx^2} + 9y = 0$.

MCQ 8: Assertion (A): The method of undetermined coefficients works when the RHS is a sum of products of polynomials, exponentials, and sines/cosines. Reason (R): Such functions have a finite number of linearly independent derivatives under differentiation.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

The finite-dimensional invariant subspace generated by repeated differentiation of exponential, trigonometric, and polynomial terms justifies the method of undetermined coefficients.

MCQ 9: The particular integral of $(D - 1)e^x = $

A $x e^x$

B $\frac{1}{2} x^2 e^x$

C $e^x$

D $x^2 e^x$

Derivation & Analysis:

Using the exponential shift rule or formula for $1/(D-a) e^{ax} = x e^{ax}$, we get $x e^x$.

Q10

MCQ 10: For the Cauchy-Euler equation $x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} - y = 0$, substituting $x = e^t$ transforms it into:

A $\frac{d^2 y}{dt^2} - y = 0$

B $\frac{d^2 y}{dt^2} + \frac{dy}{dt} - y = 0$

C $\frac{d^2 y}{dt^2} + y = 0$

D $\frac{d^2 y}{dt^2} - \frac{dy}{dt} - y = 0$

Derivation & Analysis:

With $x = e^t$, $x \frac{dy}{dx} = D_t y$ and $x^2 \frac{d^2 y}{dx^2} = D_t(D_t - 1)y$. Substituting into the equation gives $D_t(D_t - 1)y + D_t y - y = 0 \implies (D_t^2 - 1)y = 0$, which is $\frac{d^2 y}{dt^2} - y = 0$.

Common questions

Use Undetermined Coefficients when the right-hand side (RHS) is a standard combination of polynomials, exponentials, sines, or cosines. Use Variation of Parameters as a universal fallback when the RHS involves non-standard functions like $\tan x$, $\sec x$, $\ln x$, or reciprocal algebraic terms.

Failing to check whether $f(-a^2) = 0$ before substituting $D^2 = -a^2$. If the denominator vanishes, direct substitution results in division by zero; you must multiply by $x$ and differentiate the denominator with respect to $D$.

Master the auxiliary root patterns and use initial/boundary conditions or derivative checks at $x=0$ to eliminate 2 or 3 incorrect options instantly without calculating the full particular integral.

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