2. Further Pure Mathematics 2

sinh x=(e^x−e^{−x})/2 and cosh x=(e^x+e^{−x})/2. Their quotient is tanh x=sinh x/cosh x. These definitions make exponential methods available for hyperbolic equations.

The key identity is cosh²x−sinh²x=1, analogous to a trigonometric identity with a sign change. sinh is odd, cosh is even, and tanh is odd.

At x=0, sinh0=0, cosh0=1 and tanh0=0. For large positive x, tanh x approaches 1 because e^{−x} becomes negligible relative to e^x.

Hyperbolic functions are not ordinary sine and cosine with a different name; their signs, domains and identities differ.

sinh x is an odd increasing curve through the origin; cosh x is an even U-shaped curve with minimum 1; tanh x is odd, increasing and bounded between −1 and 1.

The limits as x→±∞ explain the horizontal asymptotes of tanh. Evenness gives symmetry about the y-axis, while oddness gives rotational symmetry about the origin.

cosh(−x)=cosh x, so its graph mirrors across the y-axis. tanh x approaches 1 from below as x grows and −1 from above as x decreases.

Do not transfer the sine/cosine range or periodicity to hyperbolic functions; none of these basic hyperbolic graphs is periodic.

The inverse functions are defined on restricted domains: asinh x=ln(x+√(x²+1)), acosh x=ln(x+√(x²−1)) for x≥1, and atanh x=½ln((1+x)/(1−x)) for |x|<1.

The domain restrictions make the square roots real and select a one-to-one branch. Differentiate or substitute back to verify a result, and preserve absolute-value conditions when integrating logarithms.

asinh 0=0. For x=1, acosh1=0 because ln(1+0)=0; atanh x cannot accept x=1 because its denominator in the logarithmic form vanishes.

Inverse hyperbolic notation means inverse function, not reciprocal, and the domains are not optional.

Use cosh²x−sinh²x=1, tanh x=sinh x/cosh x and the exponential definitions to rewrite an equation in a form that can be factored or solved with logarithms.

Check the domain after squaring or taking square roots, and test candidates in the original equation. For expressions such as a cosh x+b sinh x, exponential substitution turns the problem into a quadratic in e^x.

Setting t=e^x>0 converts cosh x= (t+t^{-1})/2 and sinh x=(t−t^{-1})/2; solve the resulting quadratic and retain only positive t before taking ln t.

Algebraic roots of the transformed equation are not all valid x-values; positivity of e^x and the original domain must be checked.

A matrix equation such as AX=B is solved by multiplying by A⁻¹ on the correct side when A is square and invertible: X=A⁻¹B. For systems, row reduction can avoid forming an inverse.

Check dimensions before multiplying, and interpret a singular matrix as a possible no-solution or many-solution case rather than forcing an inverse.

If A is 2×2 and AX=B, compute X=A⁻¹B and verify by multiplying AX back to B. If det A=0, use row reduction to test consistency instead.

Matrix division is not scalar division, and multiplying by A⁻¹ on the right generally solves a different equation.

For an invertible square matrix A, A⁻¹=adj(A)/det(A). The determinant must be non-zero; for a 2×2 matrix [[a,b],[c,d]], the inverse is (1/(ad−bc))[[d,−b],[−c,a]].

The adjugate is the transpose of the cofactor matrix. Use a determinant test first, then verify the inverse by multiplication or by checking the formula’s dimensions.

For [[2,1],[1,1]], det=1 and the inverse is [[1,−1],[−1,2]]. Multiplying the two matrices gives the identity.

Changing one sign in the 2×2 formula can invalidate every entry, and det=0 means the formula is undefined rather than an inverse with infinite values.

An eigenvector v of A is a non-zero vector satisfying Av=λv; λ is its eigenvalue. The vector’s direction is preserved, though it may be stretched, reversed or collapsed.

Eigenvectors are directions, not arbitrary points. The zero vector is excluded because it would satisfy every λ and carry no directional information.

A diagonal matrix [[3,0],[0,−1]] has eigenvectors along the coordinate axes: the x-direction is scaled by 3 and the y-direction is reversed by factor −1.

An eigenvector is not unique in length, and a repeated eigenvalue can have one or many independent eigendirections.

Eigenvalues satisfy det(A−λI)=0. For each root λ, solve (A−λI)v=0 to find the non-zero eigenvectors in its eigenspace.

The characteristic polynomial may have repeated roots. A repeated algebraic root does not automatically provide the same number of independent eigenvectors; solve the nullspace and count its dimension.

For a diagonal matrix with diagonal entries 3 and −1, det(A−λI)=(3−λ)(−1−λ), giving λ=3 and −1; the corresponding eigenspaces are the coordinate axes.

Do not solve det(A−λ)=0 without the identity matrix, and do not treat λ itself as an eigenvector.

If A has a basis of independent eigenvectors, place them as columns of P and their eigenvalues in matching diagonal positions of D. Then A=PDP⁻¹ and A^n=PD^nP⁻¹.

The order of columns in P must match the order of eigenvalues in D. Diagonal powers are easy, which is why diagonalisation helps with recurrences and repeated transformations.

If P=[v₁ v₂] and Av₁=3v₁, Av₂=−v₂, then D=diag(3,−1); A^n acts by multiplying the two eigen-components by 3^n and (−1)^n.

A matrix with eigenvalues is not automatically diagonalizable; it needs enough linearly independent eigenvectors.

For a square matrix, the sum of eigenvalues equals the trace and the product of eigenvalues equals the determinant, counting algebraic multiplicity.

These invariants quickly check a characteristic polynomial or numerical answer, but they do not determine eigenvectors or guarantee diagonalisation.

For a 2×2 matrix with trace 5 and determinant 6, possible eigenvalues 2 and 3 pass both checks; an alleged pair 1 and 5 fails the determinant check.

Matching trace and determinant is necessary, not sufficient, for a complete eigen-analysis; the eigenspaces still need to be solved.

The core derivatives are d(sinh x)/dx=cosh x, d(cosh x)/dx=sinh x and d(tanh x)/dx=sech²x. Their integrals follow by reversing these relationships, with a constant of integration.

Use the chain rule for sinh(ax+b), and remember that ∫sech²x dx=tanh x. For logarithmic forms, check the domain and absolute-value convention where relevant.

∫3sinh(3x+1)dx=cosh(3x+1)+C, because the inner derivative 3 cancels the coefficient.

The derivative of cosh x is +sinh x, not −sinh x, and tanh is not differentiated like ordinary tan x.

When an integral contains a combination such as sinh x and cosh x, set u to a useful hyperbolic expression or use t=e^x. The identities cosh²x−sinh²x=1 and dx=dt/t can reduce the problem to algebraic or logarithmic terms.

Choose a substitution that matches the differential factor; do not expand blindly. Back-substitute and differentiate the result to verify the antiderivative.

With t=e^x, sinh x=(t−t⁻¹)/2 and cosh x=(t+t⁻¹)/2, so an integral in sinh and cosh becomes a rational expression in t.

A substitution is not complete until dx and the limits, if definite, are transformed consistently.

The Maclaurin series of f is f(0)+f′(0)x+f″(0)x²/2!+… . It is the Taylor expansion about x=0 and is useful when the series converges and a finite truncation gives the required accuracy.

Find a pattern in derivatives or use known expansions, state the interval or radius of convergence when relevant, and estimate the remainder if an error bound is required.

e^x=1+x+x²/2!+x³/3!+…; for small x, retaining terms through x² gives a controlled approximation whose error is of the order of the next term.

A formal power series is not automatically valid for every x, and truncating after a convenient term does not prove a numerical accuracy.

Use cosh²x−sinh²x=1, 1−tanh²x=sech²x and coth²x−1=csch²x to replace a difficult power or quotient by a form with a known derivative.

Choose the identity that leaves a differential factor. For even powers, separate one factor when integrating; for equations, isolate a single hyperbolic function before applying an inverse.

∫sinh²x dx can use sinh²x=(cosh2x−1)/2, giving a sum of an elementary term and a linear term in x.

The sign in cosh²−sinh² is opposite to the circular identity, and applying an identity without preserving the differential can make the integral harder.

A reduction formula expresses an integral I_n in terms of I_{n−1} or I_{n−2}, usually by integration by parts. Repeating the relation eventually reaches a base integral that can be evaluated directly.

Define the indexed integral and its limits clearly, derive the recurrence once, then apply it with the correct starting value. Track boundary terms and signs at each step.

For I_n=∫sin^n x dx on a fixed interval, integration by parts can relate I_n to I_{n−2}; repeated use reduces an even or odd power to I_0 or I_1.

A recurrence is not the final numerical answer: the base case and the index range are part of the proof.

The signed area between y=f(x) and the x-axis from a to b is ∫_a^b f(x)dx. Geometric area requires splitting where f changes sign or integrating top minus bottom between curves.

Find intersections first and use the interval that traces the intended region once. In parametric or polar forms, change the area formula rather than forcing Cartesian dx.

If f is below the x-axis on [a,b], the geometric area is −∫_a^b f(x)dx. For two curves, integrate (upper−lower) only after checking which is larger throughout.

A definite integral can be zero while geometric area is positive because positive and negative signed regions cancel.

For y=f(x), arc length from a to b is ∫√(1+(dy/dx)²)dx. Rotating the curve about an axis gives surface area 2π∫(radius)×(arc-length element), with the radius and variable chosen consistently.

Parametric and polar curves require their own ds formula. Find endpoints and avoid counting a self-intersecting curve or using signed radius without considering the geometry.

For y=x² from 0 to 1, ds=√(1+4x²)dx; surface area about the x-axis uses radius y=x², so the integrand is 2πx²√(1+4x²).

Arc length is not ∫dy or ∫|f|dx, and surface area is not volume; use the local slant factor.

If z=r(cosθ+i sinθ), then z^n=r^n(cos nθ+i sin nθ). The theorem also gives a systematic route to trigonometric identities and roots of complex numbers.

Convert to modulus–argument form, multiply the argument by n and convert back only at the end. Arguments are defined modulo 2π, so equivalent angles represent the same complex number.

(cosθ+i sinθ)^3=cos3θ+i sin3θ. Equating real parts gives cos3θ=4cos³θ−3cosθ.

The modulus is raised to n as well as the angle; multiplying only the angle gives the wrong magnitude.

To solve w^n=z=r(cosθ+i sinθ), the roots have modulus r^(1/n) and arguments (θ+2πk)/n for k=0,1,…,n−1.

The n values are equally spaced on a circle and are distinct modulo 2π. Give them in polar form or convert each to Cartesian form as requested.

The cube roots of 8 are modulus 2 with arguments 0, 2π/3 and 4π/3, giving 2, −1+i√3 and −1−i√3.

Using only θ/n misses the other roots; the 2πk term is what enumerates the full set.

Expand (cosθ+i sinθ)^n using the binomial theorem, then equate its real and imaginary parts with cos nθ+i sin nθ to obtain trigonometric identities.

Separate even powers of i for the real part and odd powers for the imaginary part. Keep the combinatorial coefficients and signs organised before simplifying.

For n=2, the real part of (cosθ+i sinθ)^2 gives cos2θ=cos²θ−sin²θ, while the imaginary part gives sin2θ=2sinθcosθ.

The identity follows from equality of complex numbers, not from treating i as an ordinary positive number.

Write the equation as dy/dx+P(x)y=Q(x). The integrating factor is μ=e^{∫P(x)dx}; multiplying through makes the left side d(μy)/dx, so μy=∫μQ dx+C.

Put every y term on the left and divide by the coefficient of dy/dx before choosing μ. Apply an initial condition only after the general solution is obtained.

For dy/dx+2y=e^x, μ=e^{2x}; then d(e^{2x}y)/dx=e^{3x}, so y=(1/3)e^x+Ce^{−2x}.

The integrating factor multiplies the whole equation, not just y, and P must be read after standardising the derivative coefficient.

The general solution of a linear ODE is complementary function (the homogeneous equation) plus a particular integral (one solution for the forcing term). Initial conditions determine the constants.

For constant coefficients, solve the auxiliary equation for the complementary function. Choose a particular form matching the forcing, multiplying by x when it duplicates a complementary term.

For y″−3y′+2y=e^x, the auxiliary roots are 1 and 2; because e^x duplicates a complementary term, try Axe^x for the particular solution.

The complementary function alone cannot satisfy a non-zero forcing term, and a guessed particular form must be adjusted when resonance occurs.

For ay″+by′+cy=0, try y=e^{mx}; the auxiliary equation am²+bm+c=0 determines the complementary function. Distinct real roots give two exponentials, a repeated root gives (C₁+C₂x)e^{mx}, and complex roots give e^{αx}(C₁cosβx+C₂sinβx).

The form follows from the root type, not from memorised labels. Substitute the proposed solution back into the differential equation to check the signs and coefficients.

For y″−4y′+13y=0, m²−4m+13=0 gives m=2±3i, so y=e^{2x}(C₁cos3x+C₂sin3x).

Complex roots do not make y complex when real initial data are used; the conjugate pair combines into real sine and cosine terms.

For ay″+by′+cy=F(x), write y=yc+yp. The complementary function yc solves the homogeneous equation; a particular integral yp accounts for the forcing term.

Choose a trial form matching F(x): exponentials, polynomials, sines/cosines or their combinations. If the trial duplicates part of yc, multiply by enough powers of x to make it independent.

For y″+y=cos x, cos x and sin x already belong to yc, so a trial such as Ax sin x is required rather than A cos x+B sin x.

The complementary function alone solves only the zero-forcing equation, and a trial form that duplicates yc cannot determine a particular solution.

Choose a new variable u=g(y) or u=g(x,y) so that the derivatives in the equation combine into du/dx or a simpler separable form. The substitution is useful only when it reduces the order or nonlinearity.

Differentiate the substitution explicitly, replace every occurrence consistently and recover y only after solving for u. Check any lost solutions introduced by division or square roots.

For an equation containing y′/y, set u=ln|y| so u′=y′/y; the transformed equation may become linear or directly integrable.

A substitution is not just a change of symbol: forgetting the chain-rule factor produces a different differential equation.

Solving an ODE gives a family containing arbitrary constants. Initial conditions such as y(0)=a and y′(0)=b provide simultaneous equations that determine those constants.

Keep the general solution and its derivative visible, substitute the conditions at the same point, and solve the resulting linear system. For a second-order equation, two independent conditions are normally needed.

If y=C₁e^x+C₂e^{2x}, y(0)=3 gives C₁+C₂=3; y′(0)=4 gives C₁+2C₂=4, so C₁=2 and C₂=1.

One condition cannot determine two arbitrary constants, and applying a condition before differentiating can lose necessary information.