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These notes provide a comprehensive and structured coverage of Differential Equations, designed for undergraduate and competitive-exam preparation. The content is written in clear, exam-oriented language with step-by-step explanations and solved examples. Contents include: Introduction to Ordinary Differential Equations (ODEs) Order and degree of differential equations Formation of differential equations First-order differential equations: Variables separable Homogeneous and non-homogeneous equations Exact and non-exact equations Linear differential equations Higher-order linear differential equations with constant coefficients Complementary function (CF) and particular integral (PI) Initial Value Problems (IVP) Boundary Value Problems (BVP) Applications in physics, engineering, and applied mathematics Important formulas, shortcuts, and exam tips Features: Neatly typed PDF notes Well-explained theory + solved problems Suitable for Class 12, undergraduate (BSc / BTech
Typology: Lecture notes
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Definition: A differential equation is an equation involving a function y(x) and its derivatives: F (x, y, y′, y′′,... , y(n)) = 0 Properties:
xy = 0 Linear? Answer: Linear
Definition: Process of eliminating arbitrary constants from a family of curves to obtain an ODE. Key Steps:
x^2 + C ⇒ yy′^ − x = 0
Definition: - Linearly independent: no non-trivial combination gives zero - Linearly dependent: non-trivial combination gives zero Test: Wronskian W (f 1 ,... , fn) ̸= 0 ⇒ independent Hidden Concept: Zero Wronskian may not always imply dependence (check con- text) MSQs:
Definition: A set of linearly independent solutions {y 1 ,... , yn} such that
y = C 1 y 1 + · · · + Cnyn
is the general solution. Hidden Concept: Order = number of fundamental solutions, Wronskian ̸= 0 con- firms independence MSQs:
Definition: If y 1 ,... , yn are solutions of a linear homogeneous ODE, any linear combi- nation is also a solution: y = C 1 y 1 + · · · + Cnyn Hidden Concept: Applies only to homogeneous linear ODEs. For nonhomogeneous: y = yh + yp MSQs:
Statement: For n-th order linear homogeneous ODE
y(n)^ + an− 1 (x)y(n−1)^ + · · · + a 0 (x)y = 0
with continuous coefficients ai(x) on interval I, given initial conditions
y(x 0 ) = y 0 ,... , y(n−1)(x 0 ) = yn− 1
there exists a unique solution on I. Hidden Concept: - Linearity + continuity ensures existence and uniqueness - Wron- skian ̸= 0 confirms independent solutions MSQs:
Let y 1 (x), y 2 (x),... , yn(x) be n functions that are n − 1 times differentiable on an interval I. Then the Wronskian W (y 1 ,... , yn)(x) is defined as:
W (y 1 , y 2 ,... , yn)(x) =
y 1 y 2 · · · yn y′ 1 y′ 2 · · · y n′ .. .
y 1 (n −1) y 2 (n −1) · · · y n(n−1)
It is a determinant involving the functions and their derivatives up to order n − 1. —
Fundamental solutions: { 1 , sin x, cos x}. At x 0 = 0:
So W (x) = −1, constant and nonzero. — Example 3. Verify independence for y 1 = ex, y 2 = e−x.
a 1 (x) = 0 ⇒ W (x) = W (0)
Hence W (x) = −2, constant and nonzero, so the functions are linearly independent. —
— Summary:
R (^) a n− 1 (x)dx.
Question: Let P (x), Q(x) be continuous real–valued functions defined on [− 1 , 1], and let u 1 , u 2 : [− 1 , 1] → R be solutions of the differential equation
d^2 u dx^2
du dx
satisfying the conditions
u 1 (x) ≥ 0 , u 2 (x) ≤ 0 , and u 1 (0) = u 2 (0) = 0.
Let W (x) denote the Wronskian of u 1 and u 2. Then which of the following statements are true?
(A) u 1 and u 2 are linearly independent.
(B) u 1 and u 2 are linearly dependent.
(C) W (x) = 0 for all x ∈ [− 1 , 1].
(D) W (x) ̸= 0 for some x ∈ [− 1 , 1].
Detailed Solution: The given ODE is u′′^ + P (x)u′^ + Q(x)u = 0,
where P and Q are continuous functions on [− 1 , 1].
The Wronskian of two solutions u 1 , u 2 is defined as:
W (x) =
u 1 (x) u 2 (x) u′ 1 (x) u′ 2 (x)
= u 1 (x)u′ 2 (x) − u′ 1 (x)u 2 (x).
At x = 0, u 1 (0) = u 2 (0) = 0 =⇒ W (0) = u 1 (0)u′ 2 (0) − u′ 1 (0)u 2 (0) = 0.
By the Abel’s formula, the Wronskian of any two solutions of a second–order linear homogeneous ODE satisfies:
W (x) = W (x 0 ) e−^
R (^) x x 0 P^ (t)^ dt.
Choosing x 0 = 0, we have: W (x) = W (0) e−^
R (^) x 0 P^ (t)^ dt.
Since W (0) = 0, it follows that
W (x) ≡ 0 for all x ∈ [− 1 , 1].
The linear dependence criterion states that if the Wronskian of two solutions is iden- tically zero on an interval where the coefficients are continuous, the solutions are linearly dependent on that interval.
⇒ u 1 and u 2 are linearly dependent.
Final Answer:
Correct options: (B) and (C)
u 1 , u 2 are linearly dependent and W (x) = 0 for all x ∈ [− 1 , 1].
Problem. Let P be a continuous function on R. Consider the second order linear ODE
(1 + x^2 ) y′′^ + P (x) y′^ + x y = 0, x ∈ R,
and let y 1 , y 2 be two linearly independent solutions. Let W (x) denote the Wronskian of y 1 , y 2. Suppose W (1) = a, W (2) = b, W (3) = c.
Which of the following statements must hold?
y = yh + yp,
where yh is the complementary (homogeneous) solution and yp is a particular solu- tion.
Solve: dy dx
Solution: This is linear in y. Integrating factor (I.F.) = e
R (^) P (x) dx . Multiplying through by I.F., we get:
d dx
y e
R P (x) dx
= Q(x)e
R P (x) dx.
Integrating both sides,
ye
R P (x) dx (^) =
Q(x)e
R P (x) dx (^) dx + C.
y = e−^
R P (x) dx
h Z Q(x)e
R P (x) dx (^) dx + C
i .
Solve: x^2 y′′^ − 3 xy′^ + 4y = 0, (x > 0). Step 1: Divide by x^2 : y′′^ −
x
y′^ +
x^2
y = 0.
Step 2: Let y = xm. Then y′^ = mxm−^1 , y′′^ = m(m − 1)xm−^2. Substitute: m(m − 1)xm−^2 −
3 m x
xm−^1 +
x^2
xm^ = 0.
Simplify: [m(m − 1) − 3 m + 4]xm−^2 = 0.
m^2 − 4 m + 4 = 0 ⇒ (m − 2)^2 = 0 ⇒ m = 2. Hence, the two linearly independent solutions are:
y 1 = x^2 , y 2 = x^2 ln x.
y = C 1 x^2 + C 2 x^2 ln x.
Given one solution y 1 , to find the second solution y 2 of
y′′^ + P (x)y′^ + Q(x)y = 0,
use the formula:
y 2 = y 1
e−^
R (^) P (x) dx
(y 1 )^2
dx.
Type Equation Method 1st order linear y′^ + P (x)y = Q(x) Integrating factor 2nd order homogeneous y′′^ + P (x)y′^ + Q(x)y = 0 Reduction of order 2nd order non-homogeneous y′′^ + P (x)y′^ + Q(x)y = R(x) Variation of parameters Cauchy–Euler type x^2 y′′^ + axy′^ + by = 0 Substitution y = xm
If one solution y 1 (x) of a second-order linear homogeneous differential equation is known,
y′′^ + P (x)y′^ + Q(x)y = 0,
then the reduction of order method allows us to find a second linearly independent solution y 2 (x), without solving the entire differential equation from scratch. This method “reduces” the second-order ODE to a first-order equation using the substitution y = v(x)y 1 (x). —
Solve using reduction of order:
y′′^ − y = 0, given one solution y 1 = ex.
Step 1: Compare with standard form:
y′′^ + P (x)y′^ + Q(x)y = 0 ⇒ P (x) = 0, Q(x) = − 1.
Step 2: Use the formula
y 2 = y 1
e−^
R P (x) dx (y 1 )^2
dx.
Here, P (x) = 0 ⇒ e−^
R P (x)dx (^) = 1.
y 2 = ex
(ex)^2
dx = ex
e−^2 x^ dx.
y 2 = ex
e−^2 x
e−x.
Thus the two independent solutions are y 1 = ex^ and y 2 = e−x.
y = C 1 ex^ + C 2 e−x. —
For the equation: y′′^ + P (x)y′^ + Q(x)y = 0,
if y 1 (x) is a known nonzero solution, then another linearly independent solution is:
y 2 = y 1 (x)
e−^
R (^) P (x) dx
(y 1 (x))^2
dx.
Consider a second-order linear differential equation of the form:
y′′^ + P (x)y′^ + Q(x)y = R(x),
where P (x), Q(x), R(x) are continuous functions on an interval I. Suppose one part of the complementary function (C.F.), say y 1 (x), is already known. Then the following rules help to determine the second solution or the complete solution.
First, consider the associated homogeneous equation:
y′′^ + P (x)y′^ + Q(x)y = 0.
If y 1 (x) is a known solution, then the second solution y 2 (x) is found by the Reduction of Order method. —
Let y 2 (x) = v(x) y 1 (x),
where v(x) is an unknown function to be determined. Compute derivatives:
y′ 2 = v′y 1 + vy′ 1 , y 2 ′′ = v′′y 1 + 2v′y 1 ′ + vy 1 ′′.
Substitute into the homogeneous equation and simplify using that y 1 satisfies it. You obtain: v′′y 1 + v′(2y 1 ′ + P (x)y 1 ) = 0. —
Let w = v′, then: w′y 1 + w(2y 1 ′ + P (x)y 1 ) = 0.
This is a first-order linear ODE in w:
dw dx
(2y 1 ′ + P (x)y 1 ) y 1
w.
Integrate to find:
w = C exp
(2y′ 1 /y 1 + P (x)) dx
Consider a second-order linear homogeneous ODE:
y′′^ + P (x)y′^ + Q(x)y = 0.
If one solution y is already known, the following 7 rules give a direct condition for P (x) and Q(x) or for the exponent m.
Known Solution y Derivatives Condition / Formula
y = x y′^ = 1, y′′^ = 0 P + Qx = 0
y = x^2 y′^ = 2x, y′′^ = 2 2 + 2P x + Qx^2 = 0
y = xm^ y′^ = mxm−^1 , y′′^ = m(m − 1)xm−^2 m(m − 1) + mxP (x) + x^2 Q(x) = 0
y = ex^ y′^ = ex, y′′^ = ex^ 1 + P + Q = 0
y = ekx^ y′^ = kekx, y′′^ = k^2 ekx^ k^2 + kP + Q = 0
y = sin x y′^ = cos x, y′′^ = − sin x P = 0, Q = 1
y = cos x y′^ = − sin x, y′′^ = − cos x P = 0, Q = 1
Notes:
Problem: Find the relation between P (x) and Q(x) if y = x is a solution of
y′′^ + P (x)y′^ + Q(x)y = 0.
Solution: For y = x, we have y′^ = 1, y′′^ = 0. Substitute into the ODE:
0 + P (x)(1) + Q(x)(x) = 0 ⇒ P (x) + Q(x)x = 0.
Answer: P (x) + xQ(x) = 0. —
Problem: Determine the relation between P and Q if y = x^2 is a solution of
y′′^ + P (x)y′^ + Q(x)y = 0.
Solution: For y = x^2 , y′^ = 2x, y′′^ = 2. Substitute:
2 + P (x) · 2 x + Q(x) · x^2 = 0 ⇒ 2 + 2xP (x) + x^2 Q(x) = 0.
Answer: 2 + 2xP (x) + x^2 Q(x) = 0. —
Problem: Solve the ODE x^2 y′′^ − 3 xy′^ + 4y = 0
using the shortcut for y = xm. Solution: Write in standard form: y′′^ −
x
y′^ +
x^2
y = 0.
Euler/Cauchy form: x^2 y′′^ + axy′^ + by = 0, with a = − 3 , b = 4. Characteristic (indicial) equation:
m(m − 1) + am + b = 0 ⇒ m(m − 1) − 3 m + 4 = 0
m^2 − 4 m + 4 = 0 ⇒ (m − 2)^2 = 0 ⇒ m = 2 (repeated root). General solution: y = C 1 x^2 + C 2 x^2 ln x. Answer: y = C 1 x^2 + C 2 x^2 ln x.
Consider the general second-order linear homogeneous ODE:
y′′^ + P (x)y′^ + Q(x)y = 0.
The normal form of a second-order linear ODE is:
y′′^ + r(x)y = 0,
where the first-derivative term is eliminated. —
Consider a second-order linear nonhomogeneous ODE:
y′′^ + P (x)y′^ + Q(x)y = f (x),
with a known complementary function (C.F.)
yc = C 1 y 1 (x) + C 2 y 2 (x),
where y 1 , y 2 are linearly independent solutions of the homogeneous equation.
The particular solution yp is given by
yp = −y 1
y 2 f (x) W (y 1 , y 2 )
dx + y 2
y 1 f (x) W (y 1 , y 2 )
dx ,
where W (y 1 , y 2 ) = y 1 y′ 2 − y′ 1 y 2 is the Wronskian.
y = yc + yp = C 1 y 1 + C 2 y 2 + yp. —
Step 1: Solve homogeneous ODE:
y′′^ − y = 0 ⇒ r^2 − 1 = 0 ⇒ r = ± 1
yc = C 1 ex^ + C 2 e−x, y 1 = ex, y 2 = e−x. Step 2: Wronskian:
W = ex(−e−x) − (ex)(e−x) = − 2.
Step 3: Particular solution:
yp = −y 1
y 2 f W
dx + y 2
y 1 f W
dx = −ex
e−xex − 2
dx + e−x
exex − 2
dx
yp = ex
dx − e−x
e^2 x 2
dx =
xex^ −
ex^ =
xex^ −
ex.
Step 4: General solution:
y = C 1 ex^ + C 2 e−x^ +
xex^ −
ex^.
—
Step 1: Solve homogeneous ODE:
y′′^ + y = 0 ⇒ yc = C 1 cos x + C 2 sin x, y 1 = cos x, y 2 = sin x.
Step 2: Wronskian:
W = y 1 y′ 2 − y′ 1 y 2 = cos x · cos x − (− sin x) · sin x = cos^2 x + sin^2 x = 1.
Step 3: Particular solution:
yp = − cos x
sin x · tan x 1
dx + sin x
cos x · tan x 1
dx
yp = − cos x
sin^2 x/ cos xdx+sin x
sin xdx = − cos x
(sin^2 x/ cos x)dx+sin x(− cos x)?
Compute carefully: sin^2 x/ cos x = (1 − cos^2 x)/ cos x =
cos x
− cos x = sec x − cos x Z (sec x − cos x)dx = ln | sec x + tan x| − sin x
yp = − cos x[ln | sec x + tan x| − sin x] + sin x
cos x tan xdx Z cos x tan xdx =
cos x ·
sin x cos x
dx =
sin xdx = − cos x
yp = − cos x ln | sec x + tan x| + cos x sin x − sin x cos x = − cos x ln | sec x + tan x| Step 4: General solution:
y = C 1 cos x + C 2 sin x − cos x ln | sec x + tan x|.
—
Step 1: Homogeneous solution:
y′′^ − 2 y′^ + y = 0 ⇒ r^2 − 2 r + 1 = 0 ⇒ r = 1 (double root)
yc = (C 1 + C 2 x)ex, y 1 = ex, y 2 = xex Step 2: Wronskian:
y′ 1 = ex, y′ 2 = ex^ + xex^ = (1 + x)ex
W = y 1 y 2 ′ − y 1 ′y 2 = ex(1 + x)ex^ − ex^ · xex^ = e^2 x Step 3: Particular solution:
yp = −y 1
y 2 f W
dx + y 2
y 1 f W
dx = −ex
xex^ · ex/x e^2 x^
dx + xex
ex^ · ex/x e^2 x^
dx
