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In the homogeneous equation, substitute the exponential function $e^{kx}$. The characteristic equation isIts distinct roots are $0,1,-1,i,-i$. Combining the complex-conjugate pair into real sine and cosine gives the complementary functionFive linearly independent terms supply the required five homogeneous degrees of freedom.

Impose the five initial conditions at zero on the preceding general solution. They giveThe second and fourth equations give $C_1=C_2$ and $C_4=0$. The third and fifth then yield $C_1=C_2=1/4$ and $C_3=-1/2$, while the first gives $C_0=1$. ThusFor a direct check, the difference of its fifth and first derivatives is $x$. At zero its value is one; its first, second, third and fourth derivatives vanish. In particular the quadratic term cancels the otherwise nonzero second derivative of the hyperbolic-cosine and cosine combination.

The required property is odd symmetry, $\boxed{f(-x)=-f(x)}$, up to irrelevant changes on a set of measure zero when discussing Fourier coefficients. An odd function has zero mean, and its products with cosine are odd, so every cosine coefficient vanishes. Its sine coefficients may remain nonzero.

For pointwise representation, the Fourier series takes the average of the one-sided limits at a jump. In particular a pure Fourier sine series has value zero at the origin and at the endpoints; arbitrary assigned values there need not equal the series sum.

Write the translated Fourier series in the convention $a_0^{(g)}/2+{\sum }_{n\ge 1}(a_n^{(g)}\cos k_{n}x+b_n^{(g)}\sin k_{n}x)$. Comparing with (i) givesThe minus sign in the cosine coefficient records translation in the positive $x$ direction.

If $\ell =mL$ with integer $m>0$, then $k_n\ell =2\pi nm$. All sine phase factors vanish and all cosine phase factors equal one. Thereforefor the periodic extension, and every Fourier coefficient is unchanged.

Use the periodic extension of $f$ when the translated argument falls outside the original interval. With $k_n=2\pi n/L$, the sine subtraction identity givesThis translation of a Fourier series generally introduces cosine terms: a translated odd function need not be odd about the original origin. Equality is understood wherever the original Fourier series represents its periodic extension, with midpoint values at jumps.

Apart from its endpoint convention, the ramp is odd, so its constant and cosine Fourier coefficients vanish. For the sine coefficient, evenness of the integrand and integration by parts giveHenceThis is the Fourier series of a periodic sawtooth. At the periodic jump, the series sums to zero rather than the assigned endpoint value $K/2$; changing a value at one endpoint does not change the coefficients.

A trigonometric simplification avoids repeated differentiation:for sufficiently small nonzero $x$. Continuity of cosine then gives

At $x=0$ and $x=L$, the fixed-end boundary conditions require $X(0)=X(L)=0$. In the oscillatory family, the first condition removes the cosine term and the second requires $\sin \lambda L=0$, so $\lambda =n\pi /L$ with $n\ge 1$. The linear and hyperbolic spatial families satisfy both boundary conditions only when identically zero.

Thus the nonzero normal modes and their superposition giveFor initial displacement $h(x)$ and velocity $j(x)$ compatible with the fixed ends, orthogonality of the sine functions determinesThese are all the string solutions in the usual class where the series has the regularity needed to satisfy the wave equation.

Use the double-angle identity $2\sin x\cos x=\sin 2x$. The Fourier series is already a finite trigonometric sum:In the usual Fourier sine series convention, $b_1=b_2=b_3=1$, all other $b_n$ vanish, and all cosine coefficients and the constant coefficient vanish.

Use $u=\cos x$, so $du=-\sin xdx$. Up to an overall minus sign, the differential form becomesSeeking a monomial integrating factor $u^{\alpha }{y}^{\beta }$, equality of mixed derivatives givesThese give $\alpha =1,\beta =0$, so the integrating factor in the original variables isOne may verify exactness directly without relying on invertibility of the cosine substitution:Thus the implicit first integral isThe equivalence with the differential equation holds directly where $\cos x\ne 0$. The zero solution $y=0$ is included at $C=0$. At a zero of $\cos x$, the original graph equation requires $y=0$ for any finite differentiable continuation, so no extra finite graph with nonzero $y$ is created by multiplying by the vanishing factor. If one allows vertical parametrized integral curves of the differential form, $x=\pi /2+n\pi$ is also such a curve and is already contained in the level set $C=0$.

The function is odd, so its constant and cosine coefficients are zero. Its sine coefficient isThe polynomial vanishes at both integration endpoints. Two integrations by parts therefore giveOne more integration yieldsConsequently $I_n=6\pi (-1{)}^n/n^3$ and $b_n=12(-1{)}^n/n^3$. HenceThe periodic continuation is continuous at $\pm \pi$, where its values are zero, so the series represents it there as well as in the open interval.

Apply the cosine double-angle identity twice:This finite trigonometric sum is already the Fourier series. In increasing harmonic order, including the constant term, the first two nonzero terms are

The new angular interval has width $\pi /2$, so its probability density function is $2/\pi$ there. Its endpoints stay away from the poles of the tangent function, making $Y$ bounded and its expected value finite. Integrate over the continuous uniform distribution of angles:The two cosines are $\sqrt{3}/2$ and $1/2$. HenceIt has units of length and is positive because the permitted positive angles extend farther than the negative ones.

Matching the two parametric equations at an intersection gives $\cos \varphi =-t$ and $\sin \varphi =1+t$. The Pythagorean trigonometric identity therefore requiresFor $t=0$, the sine is one and the cosine is zero; for $t=-1$, the cosine is one and the sine is zero. Keeping the stated half-open angular interval givesThe corresponding points are $(0,1)$ and $(2,0)$; both satisfy the original parametric equations.

The complex conjugate is $z^{\ast }=x-iy$. From the complex sine and cosine identities,Divide these expressions and multiply by the complex conjugate of the denominator. Its squared modulus is ${\cos }^{2}x{\cosh }^{2}y+{\sin }^{2}x{\sinh }^{2}y=(\cos 2x+\cosh 2y)/2$. The numerator simplifies using ${\cosh }^{2}y-{\sinh }^{2}y=1$ and the double-angle formulas. ThusThese expressions apply except at the poles $y=0$, $x=\pi /2+k\pi$ with integer $k$, where the original tangent function is undefined.

In spherical coordinates, the cone condition forces $z\ge 0$ and becomes $0\le \vartheta \le \alpha$. The unit ball gives $0\le \rho \le 1$. HenceFor an independent calculation in cylindrical coordinates, the lower boundary is $z=r\cot \alpha$ and the upper boundary is $z=\sqrt{1-r^2}$. They meet at $r=\sin \alpha$, since $r^2{\cot }^2\alpha =1-r^2$. ThusThe equality uses ${\sin }^3\alpha \cot \alpha ={\sin }^2\alpha \cos \alpha$. The positive cosine on the stated range removes any square-root sign ambiguity.

The superposition principle gives the general solution as $y_c+2y_{p,2}+y_{p,1}$:The first boundary condition gives $A=2/(3\sqrt{3})-1/3$. At $x=\pi$, the cosine vanishes and the sine equals $-1$. Comparing with the second boundary condition then givesThe unique solution satisfying both boundary conditions is thereforeUniqueness follows because a homogeneous solution zero at both endpoints would have $A=0$ from the first and $B=0$ from the second.

Choose the trigonometric substitution $x=\sin u$ with $-\pi /2\le u\le \pi /2$. The cosine is nonnegative there, so $\sqrt{1-x^2}=\cos u$ and $dx=\cos udu$. The double-angle formula then givesSince $u=\arcsin x$ and $\sin 2u=2x\sqrt{1-x^2}$,The formula is an antiderivative on $(-1,1)$ and has continuous endpoint values at $\pm 1$.

Use the method of undetermined coefficients, choosing a particular solution $y_p=A+B\cos 2x$. Substitution givesMatching the constant and cosine coefficients to $\frac{1}{2}-\frac{1}{2}\cos 2x$ gives $A=1/2$, $B=1/6$. ThusNeither forcing term is resonant with the homogeneous angular frequency one.