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.

For a real $2\pi$-periodic square-integrable function, use the Fourier series conventionwith Fourier coefficientsThe trigonometric orthogonality relations are ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$ for positive $n,m$, with the same formula for sines; sine-cosine products integrate to zero. The constant has squared integral $2\pi$ and is orthogonal to every positive harmonic. Thus the squared norm of the $N$th partial sum isTaking the limit in the L2 space gives the Fourier-series Parseval identity:Square integrability is the natural hypothesis for this finite norm identity; additional smoothness can be used when pointwise convergence is required.

Take the switch-off time to be $t=0$. The state remains continuous through the source step, so $\mathbf{y}(0)=(1,1,1/4{)}^T$. Its eigendirection decomposition isThe antisymmetric mode has zero amplitude because the first two initial components are equal. Applying the decay factors from part (b),The first two responses decay at rate one and the third at rate four, approaching the new zero-source steady state.

At a steady state, $K{\mathbf{y}}_0={\mathbf{f}}_0$. All eigenvalues of $K$ are nonzero, so it is invertible andFor the specified constant source this givesIndeed the first two rows of $K{\mathbf{y}}_0$ each give one, and the third gives $4(1/4)=1$.

Substitute $\mathbf{y}=\mathbf{v}{e}^{-\gamma t}$ into the unforced system. It gives $(K-\gamma I)\mathbf{v}=0$, so a nonzero single-exponential solution must use one of the eigendirections in part (a):These are all nonzero solutions having one common decay constant. Their arbitrary linear combination is the general unforced solution; all decay because the three eigenvalues are positive.

First divide each of the first two equations by two. ThenThe equal-component and opposite-component directions diagonalize the upper block, while the third coordinate is independent. Direct multiplication gives the eigenvalues and unit eigenvectorsThey form an orthonormal eigenbasis because $K$ is a real symmetric matrix.

Put $p=y^{\prime }$ and differentiate the general form of Clairaut's equation:Cancellation givesFor $f(p)=1/(p-1)$ the equation is defined only when $p\ne 1$. There are two types of smooth branch.

First, constant $p=C\ne 1$ gives the straight-line familySecond, $x+f^{\prime }(p)=0$ gives $x=1/(p-1{)}^2$, so $x>0$ and $p=1\pm x^{-1/2}$. Substituting in the original equation gives its two singular envelope branches,Their derivatives are indeed $1\pm 1/\sqrt{x}$, verifying the required value of $p$. There is no finite-slope continuation of either envelope through $x=0$, and no real envelope branch for $x<0$.

For completeness, if “solution” requires only a continuous first derivative, one can join a line from the family to the corresponding envelope at their tangency point $x=(C-1{)}^{-2}$. The function and its first derivative match there, so such piecewise branches also solve the original first-order equation. They are not twice continuously differentiable at a nontrivial join. The lines and envelope branches above exhaust the smooth branches; allowing these tangent joins supplies the additional $C^1$ solutions.