Cauchy–Euler Equations and Variation of Parameters
थोडक्यात सारांश
Some linear differential equations have variable coefficients. The Cauchy–Euler (homogeneous linear) equation, with terms like x²y″ and xy′, becomes an equation with constant coefficients when we put x = eᶻ. The method of variation of parameters gives a particular integral for any right-hand side once the complementary function is known.
महत्त्वाचे मुद्दे
Cauchy–Euler equation: x²y″ + a₁xy′ + a₂y = X(x), where a₁, a₂ are constants.
Put x = eᶻ (z = log x) and D = d/dz. Then xy′ = Dy, x²y″ = D(D − 1)y, x³y‴ = D(D − 1)(D − 2)y.
The equation becomes linear with constant coefficients in z; solve as usual and then put z = log x, eᶻ = x.
Legendre's linear equation: (a + bx)²y″ + a₁(a + bx)y′ + a₂y = X. Put a + bx = eᶻ; then (a + bx)y′ = bDy and (a + bx)²y″ = b²D(D − 1)y.
Variation of parameters for y″ + P(x)y′ + Q(x)y = R(x): if y₁, y₂ are independent solutions of the homogeneous equation, take y = u y₁ + v y₂.
Wronskian W = y₁y₂′ − y₂y₁′ (non-zero for independent solutions).
u = −∫ y₂R/W dx and v = ∫ y₁R/W dx. The PI is u y₁ + v y₂.
Variation of parameters works even for R(x) = sec x, tan x, cosec x, where the short operator methods fail.
महत्त्वाच्या संज्ञा
Complementary function (CF) – general solution of the equation with right side 0.
Particular integral (PI) – any one solution of the full equation.
Wronskian (रॉन्स्कियन) – the determinant used to test independence of solutions.
सूत्रे
x = eᶻ: x dy/dx = Dy, x² d²y/dx² = D(D − 1)y
W = y₁y₂′ − y₂y₁′
PI = −y₁∫(y₂R/W) dx + y₂∫(y₁R/W) dx
५ महत्त्वाचे प्रश्न
Q1. What substitution reduces a Cauchy–Euler equation to constant coefficients?
Put x = eᶻ, i.e. z = log x, and write D = d/dz. Then x dy/dx = Dy and x² d²y/dx² = D(D − 1)y.
Q2. Define the Wronskian of y₁ and y₂.
W(y₁, y₂) = y₁y₂′ − y₂y₁′. If W ≠ 0, the two solutions are linearly independent. For y₁ = cos x, y₂ = sin x, W = cos²x + sin²x = 1.
Q3. Solve x²y″ − 3xy′ + 4y = 0.
Put x = eᶻ: D(D − 1)y − 3Dy + 4y = 0, i.e. (D² − 4D + 4)y = 0.
Auxiliary equation m² − 4m + 4 = 0 gives m = 2, 2 (repeated).
y = (c₁ + c₂z)e²ᶻ.
Back to x: y = (c₁ + c₂ log x)x².
Check y = x²: x²(2) − 3x(2x) + 4x² = 0 ✓.
Q4. Solve x²y″ + xy′ − y = x².
Put x = eᶻ: D(D − 1)y + Dy − y = e²ᶻ, i.e. (D² − 1)y = e²ᶻ.
CF: m = ±1, so c₁eᶻ + c₂e⁻ᶻ = c₁x + c₂/x.
PI = e²ᶻ/(2² − 1) = e²ᶻ/3 = x²/3.
y = c₁x + c₂/x + x²/3.
Check the PI: x²(2/3) + x(2x/3) − x²/3 = x² ✓.
Q5. Solve y″ + y = sec x by variation of parameters.
CF: y₁ = cos x, y₂ = sin x; W = 1. R = sec x.
u = −∫ sin x sec x dx = −∫ tan x dx = log(cos x).
v = ∫ cos x sec x dx = ∫ 1 dx = x.
PI = cos x log(cos x) + x sin x.
y = c₁cos x + c₂sin x + cos x log(cos x) + x sin x.
परीक्षेसाठी टीप
After solving in z, never forget to change back to x (eᶻ = x, z = log x) — leaving the answer in z loses marks. In variation of parameters, compute W first and write u and v formulas before integrating.
Some problems involve more than one unknown function or three variables at once. Simultaneous equations of the form dx/P = dy/Q = dz/R are solved by grouping terms or by using multipliers to find two independent integrals. A total differential equation P dx + Q dy + R dz = 0 can be solved when it satisfies an integrability condition. Systems of linear equations in t are solved by eliminating one variable.
महत्त्वाचे मुद्दे
Form: dx/P = dy/Q = dz/R, where P, Q, R are functions of x, y, z. The solution is two independent relations u = c₁, v = c₂.
Method of grouping: if two of the fractions involve only two variables, solve them directly to get one integral.
Method of multipliers: if l, m, n are chosen so that lP + mQ + nR = 0, then l dx + m dy + n dz = 0; if this is exact, integrate it. Each fraction also equals (l dx + m dy + n dz)/(lP + mQ + nR).
Common multipliers: (1, 1, 1) and (x, y, z).
Total DE: P dx + Q dy + R dz = 0 is integrable iff P(∂Q/∂z − ∂R/∂y) + Q(∂R/∂x − ∂P/∂z) + R(∂P/∂y − ∂Q/∂x) = 0.
If the left side is an exact differential, the solution is found directly: e.g. y dx + x dy = d(xy).
Linear systems such as dx/dt = ax + by, dy/dt = cx + dy are solved by eliminating y (or x) to get one second-order equation with constant coefficients.
महत्त्वाच्या संज्ञा
Integral of a system – a relation u(x, y, z) = c satisfied along every solution.
Condition of integrability (समाकलनक्षमता अट) – the condition that P dx + Q dy + R dz = 0 has a solution f(x, y, z) = c.
Exact differential – an expression that equals du for some function u.
सूत्रे
dx/P = dy/Q = dz/R = (l dx + m dy + n dz)/(lP + mQ + nR)
Many important differential equations with variable coefficients (Legendre, Bessel, Hermite) cannot be solved in closed form. We assume the solution is a power series y = Σ aₙxⁿ, substitute, and find a recurrence relation for the coefficients. Near a regular singular point we use the Frobenius method, y = Σ aₙx^(n+r), where r comes from the indicial equation.
महत्त्वाचे मुद्दे
Write the equation as y″ + P(x)y′ + Q(x)y = 0.
x = x₀ is an ordinary point if P and Q are analytic there (have Taylor series). Otherwise it is a singular point.
A singular point x₀ is regular if (x − x₀)P and (x − x₀)²Q are analytic there; otherwise irregular.
Near an ordinary point: put y = Σ aₙxⁿ, y′ = Σ naₙxⁿ⁻¹, y″ = Σ n(n − 1)aₙxⁿ⁻². Shift indices so every sum has xⁿ, and equate coefficients to zero.
This gives a recurrence relation for aₙ; a₀ and a₁ stay arbitrary, giving two independent solutions.
Frobenius method (regular singular point at 0): put y = Σ aₙx^(n+r), a₀ ≠ 0. The lowest power gives the indicial equation for r.
If the indicial roots differ by a non-integer, both give independent series solutions. Equal roots or roots differing by an integer need extra work (a log term may appear).
महत्त्वाच्या संज्ञा
Ordinary point (सामान्य बिंदू) – a point where the coefficients are analytic.
Recurrence relation – a formula giving aₙ₊₂ (or aₙ₊₁) in terms of earlier coefficients.
Indicial equation – the quadratic in r from the lowest power of x.
Q1. Define an ordinary point and a regular singular point.
For y″ + Py′ + Qy = 0, x₀ is ordinary if P and Q are analytic at x₀. It is a regular singular point if it is not ordinary but (x − x₀)P and (x − x₀)²Q are analytic there.
Q2. Solve y′ = y by the power series method.
Put y = Σ aₙxⁿ: Σ (n + 1)aₙ₊₁xⁿ = Σ aₙxⁿ, so aₙ₊₁ = aₙ/(n + 1). Then aₙ = a₀/n!, and y = a₀(1 + x + x²/2! + …) = a₀eˣ.
The first series is e^(x²/2), since x^(2k)/(2ᵏk!) gives 1, ½, ⅛, 1/48 ✓.
Q5. Find the indicial equation and its roots for 2xy″ + y′ + y = 0.
x = 0 is a regular singular point. Put y = Σ aₙx^(n+r).
2xy″ gives Σ 2(n + r)(n + r − 1)aₙx^(n+r−1); y′ gives Σ (n + r)aₙx^(n+r−1); y gives higher powers.
The lowest power x^(r−1): a₀[2r(r − 1) + r] = a₀ r(2r − 1) = 0.
Indicial equation r(2r − 1) = 0, roots r = 0 and r = ½. They differ by a non-integer, so there are two independent Frobenius series.
परीक्षेसाठी टीप
Before equating coefficients, shift every sum so it has the same power xⁿ and the same starting index. Write a few coefficients and then spot the pattern.
Legendre and Bessel functions are special functions that come from solving Legendre's and Bessel's differential equations by series. They appear in physics problems with spherical and cylindrical symmetry. We study their definitions, Rodrigues' formula, generating functions, recurrence relations and orthogonality.
महत्त्वाचे मुद्दे
Legendre's equation: (1 − x²)y″ − 2xy′ + n(n + 1)y = 0. For integer n, one solution is the polynomial Pₙ(x).
Q5. Given J_(1/2) = √(2/(πx)) sin x and J_(−1/2) = √(2/(πx)) cos x, find J_(3/2).
Use Jₙ₋₁ + Jₙ₊₁ = (2n/x)Jₙ with n = ½: J_(−1/2) + J_(3/2) = (1/x)J_(1/2).
J_(3/2) = (1/x)J_(1/2) − J_(−1/2).
= √(2/(πx)) (sin x/x − cos x).
परीक्षेसाठी टीप
Learn the first four Legendre polynomials and the two main Bessel recurrences by heart — most short questions use only these. In "express in Legendre polynomials" problems, start from the highest power.
The Laplace transform changes a function of time f(t) into a function of s, F(s). It turns differential equations into algebraic equations, which are easier to solve. This topic covers the definition, standard transforms, properties (linearity, shifting, derivatives), inverse transforms and solving initial value problems.
महत्त्वाचे मुद्दे
Definition: L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt, for those s where the integral converges.
Existence: f should be piecewise continuous and of exponential order (|f(t)| ≤ Me^(at)).
Linearity: L{af + bg} = aL{f} + bL{g}.
First shifting theorem: L{e^(at) f(t)} = F(s − a).
Second shifting theorem: L{f(t − a) u(t − a)} = e^(−as)F(s), where u is the unit step function.
Learn the table of seven standard transforms and the derivative formula with initial values. In inverse problems, partial fractions plus the first shifting theorem solve almost every question.
The inverse Laplace transform takes a function F(s) back to the function f(t) whose transform it is. We use standard tables, partial fractions, the shifting theorem and the convolution theorem. The main application is solving linear differential equations with initial conditions: the equation becomes algebraic in s, we solve for Y(s), and invert.
महत्त्वाचे मुद्दे
If L{f(t)} = F(s), then f(t) = L⁻¹{F(s)}. The inverse is linear: L⁻¹{aF + bG} = aL⁻¹{F} + bL⁻¹{G}.
If L⁻¹{F(s)} = f(t) and L⁻¹{G(s)} = g(t), then L⁻¹{F(s)G(s)} = ∫₀ᵗ f(u)g(t − u) du, the convolution f ∗ g. Also f ∗ g = g ∗ f.
परीक्षेसाठी टीप
Always check your partial-fraction constants by the "cover-up" method and verify the final y(t) with the initial conditions — it takes 30 seconds and catches most mistakes.
A Fourier series represents a periodic function as a sum of a constant and sine and cosine terms. This topic covers Euler's formulas for the coefficients, Dirichlet's conditions, series over (−π, π) and (0, 2π), even and odd functions, half-range sine and cosine series, and change of interval to (−l, l).
महत्त्वाचे मुद्दे
f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx), n = 1 to ∞, for a function of period 2π.
The coefficients come from Euler's formulas, using orthogonality of sin nx and cos nx over a period.
Dirichlet's conditions: in one period f is bounded, single-valued, has finitely many discontinuities and finitely many maxima and minima. Then the series converges to f(x) at points of continuity and to ½[f(x⁺) + f(x⁻)] at a jump.
Even function on (−π, π): bₙ = 0 (cosine series). Odd function: a₀ = aₙ = 0 (sine series).
Half-range series: a function given only on (0, π) can be expanded as a pure sine series (odd extension) or a pure cosine series (even extension).
For period 2l (interval (−l, l)), replace nx by nπx/l.
Putting a special value of x in a series gives sums of numerical series, e.g. Σ1/n² = π²/6.
Useful integrals: sin nπ = 0, cos nπ = (−1)ⁿ.
महत्त्वाच्या संज्ञा
Periodic function (आवर्ती फलन) – f(x + T) = f(x).
Fourier coefficients – a₀, aₙ, bₙ.
Half-range expansion – series using only sines or only cosines on half the period.
Orthogonality – integral of the product of two different basis functions over a period is zero.
Q1. Write Euler's formulas for the Fourier coefficients in (−π, π).
a₀ = (1/π)∫₋π^π f(x) dx, aₙ = (1/π)∫₋π^π f(x) cos nx dx, bₙ = (1/π)∫₋π^π f(x) sin nx dx, and f(x) = a₀/2 + Σ(aₙ cos nx + bₙ sin nx).
Q2. What happens to the Fourier coefficients of an even function in (−π, π)?
For even f, f(x) sin nx is odd, so bₙ = 0. Also a₀ = (2/π)∫₀^π f dx and aₙ = (2/π)∫₀^π f cos nx dx. The series contains only the constant and cosine terms.
Q3. Find the Fourier series of f(x) = x² in (−π, π) and deduce that Σ1/n² = π²/6.
f is even, so bₙ = 0.
a₀ = (2/π)∫₀^π x² dx = (2/π)(π³/3) = 2π²/3.
∫₀^π x² cos nx dx = [x² sin nx/n]₀^π − (2/n)∫₀^π x sin nx dx = 0 − (2/n)(π(−1)ⁿ⁺¹/n) = 2π(−1)ⁿ/n².
aₙ = (2/π) × 2π(−1)ⁿ/n² = 4(−1)ⁿ/n².
x² = π²/3 + 4Σ(−1)ⁿ cos nx/n² = π²/3 − 4[cos x − cos 2x/4 + cos 3x/9 − …].
Put x = π (cos nπ = (−1)ⁿ, so (−1)ⁿ(−1)ⁿ = 1): π² = π²/3 + 4Σ1/n² → Σ1/n² = π²/6.
Q4. Find the half-range cosine series of f(x) = x in (0, π).
a₀ = (2/π)∫₀^π x dx = (2/π)(π²/2) = π.
∫₀^π x cos nx dx = [x sin nx/n + cos nx/n²]₀^π = ((−1)ⁿ − 1)/n².
aₙ = (2/π)((−1)ⁿ − 1)/n²: zero for even n, −4/(πn²) for odd n.
x = π/2 − (4/π)[cos x + cos 3x/3² + cos 5x/5² + …].
(Putting x = 0 gives 1 + 1/9 + 1/25 + … = π²/8.)
Q5. Find the half-range sine series of f(x) = 1 in (0, π).
Check even/odd first; it saves half the work. In integration by parts, use the shortcut ∫uv = uv₁ − u'v₂ + u''v₃ − … and write cos nπ = (−1)ⁿ, sin nπ = 0 straight away.
A partial differential equation (PDE) involves partial derivatives of a function of two or more variables. This topic covers forming PDEs by eliminating arbitrary constants or functions, Lagrange's linear equation Pp + Qq = R, standard non-linear forms, classification of second-order equations, and the method of separation of variables.
महत्त्वाचे मुद्दे
Notation: z = f(x, y); p = ∂z/∂x, q = ∂z/∂y, r = ∂²z/∂x², s = ∂²z/∂x∂y, t = ∂²z/∂y².
Order = order of the highest derivative; degree = power of that derivative.
Formation by eliminating arbitrary constants: differentiate partially w.r.t. x and y and eliminate the constants.
Formation by eliminating an arbitrary function: differentiate and eliminate the function's derivative.
Lagrange's linear equation: Pp + Qq = R (P, Q, R functions of x, y, z). Auxiliary (subsidiary) equations: dx/P = dy/Q = dz/R. If u = c₁ and v = c₂ are two independent solutions, the general solution is φ(u, v) = 0.
Methods for the auxiliary equations: grouping, and using multipliers (l, m, n) with lP + mQ + nR = 0.
Non-linear first order standard forms: f(p, q) = 0; z = px + qy + f(p, q) (Clairaut's form, solution z = ax + by + f(a, b)); f(z, p, q) = 0; f₁(x, p) = f₂(y, q). Charpit's method is the general method.
Classification of A uₓₓ + B uₓᵧ + C uᵧᵧ + … = 0: B² − 4AC < 0 elliptic; = 0 parabolic; > 0 hyperbolic.
Laplace's equation uₓₓ + uᵧᵧ = 0 is elliptic; heat equation uₜ = c²uₓₓ is parabolic; wave equation uₜₜ = c²uₓₓ is hyperbolic.
Separation of variables: assume u = X(x)T(t), separate into ordinary differential equations equal to a constant, solve, then apply boundary conditions.
महत्त्वाच्या संज्ञा
Complete integral – solution containing as many arbitrary constants as independent variables.
General solution – solution containing an arbitrary function.
Boundary conditions (सीमा अटी) – values of the solution prescribed at the edges.
सूत्रे
Lagrange: dx/P = dy/Q = dz/R → φ(u, v) = 0
Clairaut: z = px + qy + f(p, q) → z = ax + by + f(a, b)
For Lagrange's equation, always write the auxiliary equations first and find two independent integrals; the answer is φ(u, v) = 0. In separation of variables, show clearly where the separation constant is used to match the given condition.
Calculus of variations finds the function y(x) that makes an integral (a functional) as small or large as possible, such as the shortest curve between two points or the curve of quickest descent. The necessary condition is Euler's equation (the Euler–Lagrange equation). Its solutions are called extremals.
महत्त्वाचे मुद्दे
A functional I[y] = ∫ₐᵇ F(x, y, y′) dx assigns a number to each function y(x) with fixed end values y(a) = A, y(b) = B.
Euler's equation: ∂F/∂y − d/dx (∂F/∂y′) = 0. Any y that makes I stationary must satisfy it.
Solutions of Euler's equation satisfying the end conditions are extremals.
Special case 1: F does not contain y. Then ∂F/∂y′ = constant.
Special case 2: F does not contain x. Then F − y′ ∂F/∂y′ = constant (Beltrami identity).
Special case 3: F does not contain y′. Then ∂F/∂y = 0, an ordinary equation.
Shortest distance between two points in a plane: F = √(1 + y′²) gives a straight line.
Brachistochrone (quickest descent under gravity): the extremal is a cycloid.
Minimum surface of revolution: the extremal is a catenary y = c cosh((x − d)/c).
For isoperimetric problems (an extra integral condition), use F + λG in Euler's equation.
महत्त्वाच्या संज्ञा
Functional (फलनक) – a rule assigning a number to a function.
Extremal – a curve satisfying Euler's equation.
Variation δy – a small change in the function y.
सूत्रे
∂F/∂y − d/dx (∂F/∂y′) = 0
F free of y: ∂F/∂y′ = c
F free of x: F − y′(∂F/∂y′) = c
५ महत्त्वाचे प्रश्न
Q1. Write Euler's equation for the functional ∫ F(x, y, y′) dx.
∂F/∂y − (d/dx)(∂F/∂y′) = 0.
Q2. Show that the shortest curve between two points in a plane is a straight line.
Length = ∫ √(1 + y′²) dx. F has no y, so ∂F/∂y′ = y′/√(1 + y′²) = c. This gives y′ = constant = m, so y = mx + k, a straight line.
Q3. Find the extremal of ∫₀¹ (y′² + 12xy) dx with y(0) = 0, y(1) = 1.
By Euler's equation the bracket is 0, so d/dx (F − y′F_y′) = 0.
Hence F − y′ ∂F/∂y′ = constant.
परीक्षेसाठी टीप
First check which variable is missing from F — it decides which short form to use and saves time. Always apply the end conditions at the last step to get the particular extremal.
Numerical methods give approximate answers when exact methods are hard or impossible. This topic covers finding roots of equations (bisection, regula falsi, Newton–Raphson), finite differences and interpolation (Newton's and Lagrange's formulas), numerical integration (trapezoidal and Simpson's rules), and solving ODEs (Euler and Runge–Kutta methods).
महत्त्वाचे मुद्दे
Bisection method: if f(a) and f(b) have opposite signs, a root lies in (a, b). Take the midpoint and keep the half where the sign changes. Always converges, but slowly.
Regula falsi (false position): uses the chord joining (a, f(a)) and (b, f(b)) instead of the midpoint.
Newton–Raphson method: fast (quadratic convergence) when the starting value is close to the root; fails if f'(x) is near 0.
Finite differences: forward Δyᵢ = yᵢ₊₁ − yᵢ; backward ∇yᵢ = yᵢ − yᵢ₋₁. For a polynomial of degree n, the n-th differences are constant.
Newton's forward interpolation is used near the beginning of a table; backward near the end. Both need equal intervals.
Lagrange's interpolation works for unequal intervals.
Trapezoidal rule: joins points by straight lines. Simpson's 1/3 rule: fits parabolas; needs an even number of intervals; more accurate. Simpson's 3/8 rule needs intervals in multiples of 3.
The interval must be divided into an even number of equal sub-intervals (an odd number of ordinates), because the rule fits a parabola through each set of three points.
Q2. Write the Newton–Raphson formula and one limitation.
xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ). It fails or converges slowly if f'(x) is zero or very small near the root, or if the starting value is far from the root.
Q3. Find √2 by the Newton–Raphson method, starting from x₀ = 1.5.
Q5. Use Euler's method with h = 0.1 to find y(0.2) for dy/dx = x + y, y(0) = 1.
f(x, y) = x + y; x₀ = 0, y₀ = 1.
y₁ = y₀ + h f(x₀, y₀) = 1 + 0.1(0 + 1) = 1.1 at x = 0.1.
y₂ = y₁ + h f(x₁, y₁) = 1.1 + 0.1(0.1 + 1.1) = 1.1 + 0.12 = 1.22 at x = 0.2.
Exact solution y = 2eˣ − x − 1 gives y(0.2) ≈ 1.2428; Euler's error shows why RK4 is preferred.
परीक्षेसाठी टीप
Make neat tables of x and y values in integration and interpolation questions — it reduces errors and earns method marks. Keep at least four or five decimal places in working and round only the final answer.