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For the first-order ordinary differential equation $y^{\prime }=f(x,y)$ defined on an open subset $D$ of the plane, an isocline of slope $k$ is the level setHere $k$ is a constant. At every point of this isocline, the direction field prescribes a tangent vector proportional to $(1,k)$ for a solution of a differential equation. It does not generally prescribe the tangent line of the isocline itself. The level set can have several components or fail to be a smooth curve, and points outside $D$ are excluded.

For example, take $y^{\prime }=x-y$ on the whole plane. Its isoclines are the parallel straight linesEach line has slope $1$, while the direction field along it has slope $k$. Thus the isocline $y=x$ carries horizontal segments even though the line itself rises. A function graph of a solution of a differential equation follows the segments, rather than following this line. Only the isocline with $k=1$, namely $y=x-1$, is also a solution of a differential equation.^[ref]

The first-order linear differential equation has the exact familyThe initial condition $y(x_0)=y_0$ selects $A=(y_0-x_0+1)e^{x_0}$ and hence a unique solution of a differential equation for every real $x$. The picture shows $y(0)=-1,0,1,2$. For $A=0$ the solution of a differential equation stays on the $k=1$ isocline; for other $A$ it moves through different isoclines. These formulas check the pictured trajectories by direct differentiation, independently of how the direction field was sampled.

The zero-slope isocline $f(x,y)=0$ is called a nullcline. Below the nullcline $y=x$ in this example, $y^{\prime }>0$; above it, $y^{\prime }<0$. For $A>0$, the solution of a differential equation meets it at $x=\log A$, where $y^{\prime }=0$ and $y^{\prime \prime }=1$, a strict local minimum. The nullcline is therefore a useful guide to turning points, although it is not itself a solution of a differential equation. In the equivalent planar autonomous differential equation $\dot{x}=1$, $\dot{y}=x-y$, it is the nullcline for the second component; there is no equilibrium point, because the first component never vanishes.

This makes isoclines useful even when no explicit solution of a differential equation is available: organize a direction field by common slopes, read where function graphs rise or fall, and identify possible turning points on nullclines.

A scalar linear differential equation of order $n\ge 1$ has the formHere the coefficients $a_j$ and the forcing term $g$ are prescribed functions of the independent variable $t$. A classical solution of a differential equation is an $n$-times differentiable function satisfying the identity at every point of its interval. The unknown function $y$ and its derivatives enter to the first power, without products between them or coefficients depending on $y$. Equivalently, the differential operator $L$ is a linear map: for constant scalars $\alpha ,\beta$,The forcing term may be zero, giving a homogeneous linear differential equation, or nonzero, giving an inhomogeneous linear differential equation. Linearity of $L$ does not mean that every function graph of a solution of a differential equation is a straight line.

To use the usual order-$n$ initial value problem theory on an interval $I$, assume that $a_n(t)\ne 0$ throughout $I$ and that all $a_j$ and $g$ are continuous functions there. Dividing by $a_n$ normalizes the highest derivative; specifying $y(t_0),y^{\prime }(t_0),\dots ,y^{(n-1)}(t_0)$ at $t_0\in I$ then selects one solution of a differential equation throughout $I$. A zero of $a_n$ needs separate analysis: one cannot divide through it and retain these conclusions. The first-order normalization appears in OpenStax's definition and standard form of linear equations; the corresponding system theorem is in MIT's existence and uniqueness notes for linear systems.

The coefficients need not be linear functions of $t$. For example, $t^2{y}^{\prime }+y=0$ is a linear differential equation on $t>0$, although $t^2$ is quadratic. Its solution of a differential equation with initial condition $y(1)=1$ is $y(t)=e^{1/t-1}$: its derivative is $y^{\prime }=-y/t^2$, so substitution gives $t^2{y}^{\prime }+y=0$, and $y(2)=e^{-1/2}\approx 0.60653$. The zero of the highest coefficient at $t=0$ explains the domain restriction. In contrast, $y^{\prime }+y^2=0$, $y{y}^{\prime }+y=0$ and $y^{\prime }+\sin y=0$ are nonlinear ordinary differential equations because of their dependence on the unknown function. Variable coefficients and nonlinear ordinary differential equations are different issues; see MIT's definitions and examples of first-order linear equations.

The practical benefit of linearity is that it separates the effect of the initial condition from the effect of the forcing term. Once one particular solution $y_p$ satisfies $L[y_p]=g$, every solution of a differential equation isHere $\ker L$ is the kernel of a linear map, consisting of solutions of a differential equation to the associated homogeneous linear differential equation. Indeed, $L[y-y_p]=g-g=0$, and conversely $L[y_p+h]=g+0=g$. Thus the family for a fixed forcing term is an affine space obtained by translating $\ker L$. If $g$ is not identically zero, the zero function is absent, so this family is not a vector space under the usual operations on functions. This short argument explains why a general solution is a particular solution plus a homogeneous solution.

The logistic differential equation is the nonlinear ordinary differential equationAs a mathematical model of population dynamics, it treats abundance $N$ as a continuous nonnegative quantity. The constant $r$ has units of inverse time, while the carrying capacity $K$ has the same units as $N$. The useful idea is to combine growth resembling exponential growth at low abundance with a slowing approach to a finite abundance. The three trajectories below already show an important distinction: starting above the carrying capacity gives a decreasing curve, whereas starting below it gives an increasing one.

For $N>0$, dividing the derivative by abundance gives the instantaneous per-capita rate:It decreases linearly with $N$: it approaches $r$ at low abundance, is $r/2$ at $N=K/2$, vanishes at the carrying capacity, and is negative above it. This per-capita rate differs from the total rate $N^{\prime }$, which also counts how many individuals contribute. It is an instantaneous rate, rather than a fixed percentage increase over a whole day. The linear per-capita assumption and the resulting qualitative picture are developed in MIT 18.03SC's logistic-model notes.

Writing $f(N)=rN(1-N/K)$, the equilibrium points satisfy $f(N)=0$, so the constant solutions of a differential equation are $N(t)=0$ and $N(t)=K$. The phase line points upward for $0<N<K$ and downward for $N>K$. Also $f^{\prime }(0)=r>0$ and $f^{\prime }(K)=-r<0$: the criterion for fixed point stability for an autonomous differential equation makes zero an unstable equilibrium and $K$ an asymptotically stable equilibrium point. Every positive initial condition approaches $K$ as $t\to \infty$; $N_0=0$ stays exactly zero. Instability of zero means that a small positive population grows away from it, not that a zero population spontaneously appears.

For every $N_0>0$, the unique solution of a differential equation for all $t\ge 0$ isThe denominator is positive on this domain, including when $N_0>K$. For $0<N_0<K$, this is a translated and scaled logistic function; its extension to all real times is a sigmoid function. For $N_0>K$ the coefficient of the exponential function is negative, and the future curve instead decreases toward $K$. No nonconstant positive solution of a differential equation crosses $K$ or reaches it at finite time. A short derivation uses the dimensionless variable $z=K/N-1$: the chain rule gives $z^{\prime }=-rz$, so exponential decay yields $z(t)=(K/N_0-1)e^{-rt}$. The substitution divides by $N$, so the zero equilibrium point must be included separately.

The total growth rate is a downward-opening quadratic function of abundance. Completing the square gives its maximum without solving the differential equation:The balance is beautiful: more individuals contribute to growth, but each contributes less, and the strongest total increase occurs halfway to the carrying capacity. This is a maximum over possible abundances. A trajectory starting above $K/2$ never passes through that abundance in its future.

The chain rule connects this maximum to the shape of the time trajectory through its second derivative:For $0<N_0<K/2$, the trajectory initially has convexity and then concavity, with its one future inflection point atFor $N_0=K/2$, this point lies at the initial time, with concavity for $t>0$. For $K/2<N_0<K$, the inflection point belongs to the past extension, so the future trajectory has concavity throughout. Above $K$, the decreasing trajectory has convexity and no inflection point. The two constant solutions of a differential equation have no change of concavity. Thus an entire logistic function has an S shape, but its observed future segment need not show both bends. See also ^[ref].

For the plotted example with $K=1000$, $r=0.4{\mathrm{day}}^{-1}$ and $N_0=100$, time $t$ measured in days givesThe initial total rate is $N^{\prime }(0)=0.4\cdot 100\cdot 0.9=36$ individuals per day, and the initial per-capita rate is $0.36{\mathrm{day}}^{-1}$. The fastest total increase is $rK/4=100$ individuals per day, reached at $t_{1/2}=\ln 9/0.4\approx 5.49306$ days. At that time the per-capita rate has already fallen to $0.2{\mathrm{day}}^{-1}$.

These parameters are illustrative choices, rather than an empirical fit. The equation specifies deterministic evolution under a phenomenological assumption: it summarizes how net growth depends on abundance without separately modelling every process that affects it. The carrying capacity is a parameter for a specified setting, not a universal constant for all populations. In his 1838 paper "Notice sur la loi que la population suit dans son accroissement", Correspondance mathématique et physique 10, pp. 113–121, Pierre-François Verhulst proposed $p^{\prime }=mp-n{p}^2$, equivalent to this equation with $r=m$ and $K=m/n$. He also noted that different assumed growth-retarding terms could describe the observations similarly while giving very different limiting populations. This is an early illustration of why a good-looking curve alone does not establish a unique long-term prediction.^[ref]

The logistic differential equation describes continuous time. The logistic map instead advances in discrete steps and can exhibit a period-doubling cascade and Devaney chaos; the monotone positive trajectories here cannot. Their connection through the forward Euler method is explained under Euler discretization of logistic growth, where a large step can introduce behaviour absent from the original differential equation.

For the homogeneous linear differential equation $\dot{Y}=K(t)Y$ with continuous coefficient matrix of period $P>0$, the monodromy matrix of a periodic linear system is the one-period map from $Y(0)$ to $Y(P)$. If $X$ is any fundamental matrix,Its eigenvalues are the Floquet multipliers. Multipliers are nonzero, since a fundamental matrix is invertible. Changing the starting phase conjugates $N$, so the Floquet multipliers and Jordan blocks are independent of that choice.^[ref]

This turns a continuous-time problem into repeated matrix multiplication:If $Nv=mv$, the associated solution of a differential equation obeys $Y(t+P)=mY(t)$ at every time, not merely at the sampled times. In particular, $Nv=v$ gives a nonzero $P$-periodic solution, and every nonzero $P$-periodic solution arises this way. If $Nv=-v$, the state changes sign after one period and repeats after two. A unit-modulus Floquet multiplier different from $1$ need not give repetition after one period, and a generalized eigenvector in a nontrivial Jordan block need not even give bounded motion.

For the scalar example under periodic solution, $N$ is the one-by-one matrix $(e^{2\pi a})$. Numerically solving a matrix problem uses the same idea: integrate one initial value problem for each column of $X(0)=I$ through one period, then compute the eigenvalues of $X(P)$. This is useful when an explicit solution of a differential equation is unavailable: one period determines the Floquet growth rate over arbitrarily many periods.

For piecewise constant coefficients, which are also covered by the one-period construction, $N$ is the time-ordered product of the segment matrix exponentials, with the earliest segment on the right. For example, take period $P=2$ and alternateSince $K_1^2=K_2^2=0$,Replacing the coefficient by its mean would instead give $e^{K_1+K_2}=\left(\begin{array}{cc}\cosh 1& \sinh 1\\ \sinh 1& \cosh 1\end{array}\right)$ and the incorrect rate $0.5$. The matrices do not commute. The scalar mean-coefficient formula therefore cannot simply be transferred to matrix systems. Nevertheless the Liouville formula for a fundamental matrix gives the correct determinant even in the noncommuting case:In the example, this is $1=m_{+}{m}_{-}$; it fixes the product of the Floquet multipliers, rather than either multiplier separately.

In Sur les équations différentielles linéaires à coefficients périodiques (1883), Gaston Floquet calls a function satisfying $f(t+P)=mf(t)$ periodic of the second kind, and distinguishes the genuinely periodic function obtained when $m=1$. This historical terminology captures precisely the difference between repetition of the coefficient and repetition of the state.^[ref]

A nonlinear ordinary differential equation depends nonlinearly on the unknown function or its derivatives: it cannot be written in the form defining a linear differential equation. For example, $y^{\prime }=y^2$ contains a square of the unknown function, and $y^{\prime }=y(1-y)$ is a logistic differential equation. A coefficient such as $t^2$ depending only on the independent variable does not make an ordinary differential equation nonlinear. This distinction concerns dependence on the unknown, rather than the shape of its function graph.

For a concrete initial value problem, consider $y^{\prime }=y^2$, $y(0)=y_0>0$. Its solution of a differential equation isTaking its derivative gives $y^{\prime }=y_0^2/(1-y_{0}t{)}^2=y^2$, and evaluating at zero checks the initial condition. The stated interval contains zero and ends at finite-time blowup: $y$ tends to positive infinity as $t$ approaches $1/y_0$ from below. Increasing the positive initial condition brings finite-time blowup closer. This is a simple mathematical example with dimensionless $t$ and $y$, rather than a calibrated physical model.

Choose $y_0=1$ and $y_0=2$, giving $y_1=1/(1-t)$ and $y_2=2/(1-2t)$. At $t=1/4$ they have values $4/3$ and $4$, so their sum $s=y_1+y_2$ has value $16/3$. On the common interval $0\le t\le 0.4$, both are valid solutions of a differential equation, but their sum fails the same ordinary differential equation:Already at $t=0$, $s=3$ and its derivative is $1+4=5$, whereas $y^{\prime }=y^2$ would require $9$. At $t=1/4$, the defect $s^{\prime }-s^2$ is $-32/3$. The cross term explains the failure of the superposition principle, even though each individual solution of a differential equation is a smooth function.

The first panel shows $y_1$, $y_2$ and their invalid sum only up to $t=0.4$, before either finite-time blowup. The second measures the defect in the ordinary differential equation, including the computed value at $t=1/4$. For a homogeneous linear differential equation, the corresponding defect of a sum is zero. Nonlinear ordinary differential equations can describe state-dependent growth such as the logistic differential equation, whose negative quadratic term limits growth instead of accelerating it without bound. Recognizing the distinction tells us when we may assemble solutions of a differential equation from simpler pieces and when we must account for their interaction. See MIT's worked counterexample to nonlinear superposition.

An ordinary differential equation relates an unknown function of one independent variable to its ordinary derivatives. Its order is the highest derivative appearing. Thus $y^{\prime }=-k(y-T_a)$ is first order, while the harmonic oscillator equation $y^{\prime \prime }+{\omega }^{2}y=0$ is second order. A first-order equation in explicit form isHere $t$ need not be time, and $y$ can be a vector, giving a system of coupled ordinary differential equations. A classical solution of a differential equation is a sufficiently differentiable function on an interval which satisfies the equation at every point. For the explicit first-order equation, if $f$ is continuous, this solution of a differential equation has a continuous derivative.

For Newton's law of cooling, the state is a single temperature; for a second-order differential equation, a first-order system also records its first derivative. This is why a starting position alone does not specify a harmonic oscillator trajectory: its starting velocity matters too.

The power of an ordinary differential equation is that one rule describes a whole family of histories. An initial value problem chooses one history when uniqueness holds. An autonomous differential equation has $f$ independent of $t$; its equilibrium points satisfy $f(y_{\ast })=0$ and give constant solutions of a differential equation. Equations with explicit time dependence can instead respond to a changing environment.

The method of characteristics replaces the first-order partial differential equation by the ordinary differential equationsAn integrating factor for the second equation gives $d(e^{-2s}y)/ds=\beta C{e}^{-s}$, so their general solution of a differential equation isHere $C,D,E$ are real constants. For $C\ne 0$, eliminating $s$ gives the characteristic projectionseach restricted to either $x>0$ or $x<0$. These are arcs of parabolas, with the straight-line exception $K=0$. Along an oriented characteristic curve, $|x|$ increases with $s$.

For $C=0$, the characteristic projections lie on $x=0$: $y=D{e}^{2s}$ describes either vertical half-line when $D\ne 0$. For $C=D=0$ the projection is the single point $(0,0)$, although $u=2s+E$ still changes. This last lifted characteristic curve cannot be contained in the function graph of a classical solution: at $(0,0)$ the original partial differential equation would require $0=2$.

Every point with $x>0$ has exactly one backward characteristic curve meeting $x=1$, at the point $(1,q(x,y))$. It meets the specified data precisely when $-1\le q\le 1$. Thus the whole region determined by these data, including the two bounding characteristic curves, isIts interior is the corresponding open domain for the partial differential equation. Uniqueness here means uniqueness in the class of continuous functions satisfying the characteristic equations for a transport equation, including the piecewise smooth function defining the solution of a differential equation above. The endpoints of the initial segment determine the two boundary values by continuity.

This region is maximal for determination by the given data: on $x>0$, arbitrary extensions of $g$ outside $[-1,1]$ give different values on the characteristic curves with $q\notin [-1,1]$. No characteristic curve from $x=1$ reaches $x\le 0$.

For $\beta =1$,The lower boundary is a straight line, the upper boundary is a parabola, and the characteristic carrying the corner is $y=x^2-x$. All three approach the excluded origin as $x\downarrow 0$.

Use the initial conditions $x(0,r)=1$, $y(0,r)=r$, $u(0,r)=g(r)$ in the characteristic equations for a transport equation. The resulting parametrized solution surface isThe Jacobian determinant of $(s,r)\mapsto (x,y)$ is $e^{3s}>0$. In fact the inverse is explicit:Consequently the explicit solution of a differential equation on the region reached from the data isTo verify it away from $q=0$, set $D=x{\partial }_x+(2y+\beta x){\partial }_y$. Direct differentiation gives $Dq=0$ and $D(x-1+2\log x)=x+2$. At $x=1$ we have $q=y$, giving precisely the prescribed boundary condition.

Across $q=0$ the formula remains continuous and locally Lipschitz, is a smooth function on each side, and solves the partial differential equation along every characteristic curve. It is also a weak solution, since its first distributional derivatives agree almost everywhere with its ordinary derivatives. It is not a classical solution throughout the region because of the inherited corner.

The second-order differential operator factors asTherefore setting $u=v_x$ turns the second-order partial differential equation into exactly the first-order partial differential equation in part (a). Conversely, given a sufficiently smooth solution of a differential equation $u$ of that equation on a coordinate rectangle, defineThe fundamental theorem of calculus gives $v_x=u$, so this solves the second-order equation, with an arbitrary twice continuously differentiable function $F$. This classical correspondence is local on smooth patches; for the cornered solution of part (b), it can instead be interpreted using distributional derivatives.

For the Weierstrass M-test proof of the Picard-Lindelöf theorem, the required bound isTo see this, use Picard iteration starting with $y_0(x)=b$ andIf $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\le M|x-a|\le Mh\le k$. Thus every Picard iterate remains in the rectangle where the hypotheses hold. Successive applications of the Lipschitz condition and integration give the factorial estimate for Picard iterationFor example, the $n=0$ estimate is $M|x-a|$, and integrating the $n$th estimate multiplied by $L$ proves the next one. The sum of the uniform bounds isfor every finite $Lh$, by the exponential series. Hence the telescoping series of Picard iterates converges uniformly by the Weierstrass M-test. The uniform limit stays in the closed rectangle, and the Lipschitz condition permits passage to the limit in the integral equation. The fundamental theorem of calculus then gives the desired solution of a differential equation on $[a-h,a+h]$.

For the direct contraction mapping theorem proof with the usual supremum norm, take the complete metric spaceand the Picard iteration map $Tz(x)=b+{\int }_a^{x}f(s,z(s))ds$. The estimates areThus this proof requiresThe first inequality makes $T$ preserve $\mathcal{B}$; the strict second inequality makes it a contraction mapping. Its unique fixed point is the desired solution of a differential equation.

This is more restrictive than the preceding Weierstrass M-test argument: a single application of $T$ must shrink every difference in the supremum norm, whereas repeated integration in the other proof produces the factorial $(n+1)!$ and convergence even when $Lh\ge 1$. If $L=0$, the additional condition is automatic. These are conditions for the two stated direct proofs; they are not necessary bounds on the actual existence interval.

Near $(1,1)$ the solution of a differential equation has $y>0$, so the separable differential equation givesEquivalently its trajectory lies on the level setOn $y>0$, the function $G(y)=y-\log (1+y)$ is strictly increasing because $G^{\prime }(y)=y/(1+y)>0$, and its range is $(0,\infty )$. Thus the positive branch exists exactly whereThe connected component containing $x=1$ isThe implicit function theorem gives a smooth solution of a differential equation at every such $x$. It remains finite at every finite $x>x_{\ast }$, because $G$ increases without bound. As $x\downarrow x_{\ast }$, we have $y\downarrow 0$, and the original right side $x+x/y$ becomes singular. Hence this is the maximal interval $(x_{\ast },\infty )$ through the prescribed initial condition.

In particular, the solution is not defined for all $x\in (0,2)$. It cannot pass through $x=x_{\ast }$ as a classical solution, since its required value there is $y=0$.

Substitute the power series and its termwise derivatives into the ordinary differential equation. The coefficient of $x^n$, for $n\ge 0$, isThe bracket factors as $(n-3)(n-4)$, so the power-series solution of a differential equation satisfiesThe odd and even recurrence relations terminate: $a_3=-a_1$, $a_5=0$, and $a_2=-6a_0$, $a_4=a_0$, $a_6=0$. Thus the full solution is $y=a_0(1-6x^2+x^4)+a_1(x-x^3)$. The two polynomials have independent initial data at zero, and the leading coefficient $1+x^2$ never vanishes on the real axis, so every real solution of a differential equation for this equation is their linear combination on the entire real axis. The given initial data select $a_0=0$, $a_1=1$:

The first characteristic polynomials of a linear multistep method isThe root condition for a multistep method requires $|\alpha |\le 1$ and excludes $\alpha =1$, where the root of a polynomial at one is double. At $\alpha =-1$, the two unit-modulus roots of a polynomial are distinct, so zero-stability holds. For $\alpha \ne 1$, the first-order consistency of a numerical method identity is ${\rho }^{\prime }(1)=1-\alpha$, equal to the coefficient of $hf$.

Using the permitted Dahlquist equivalence theorem in this multiderivative multistep method setting givesThis means convergence on every fixed finite time interval where the smooth exact solution of a differential equation exists, with consistent starting values and the nearby branch of each implicit solve. More specifically, the convergence of a zero-stable multiderivative method gives order $p$ when the starting errors are $O(h^p)$ and both $f$ and $f^{\prime }f$ are locally uniformly Lipschitz continuous on the relevant region.

The exclusions are genuine. For $y^{\prime }=0$ and $|\alpha |>1$, an arbitrarily small parasitic starting error is amplified as ${\alpha }^n$. At $\alpha =1$, take $y_0=0$, $y_1=h$, also for $y^{\prime }=0$. The exact solution is zero, but the numerical recurrence gives $y_n=nh$, an error of order one at fixed positive time. This proves failure even though the starting values converge.

Insert an exact smooth solution of a differential equation and expand about $t=t_{n+2}$. The unscaled local truncation error, with every term moved to the left, isHere the constant, first-derivative, and second-derivative terms all cancel, using the chain rule identity $y^{\prime \prime }=f^{\prime }(y)f(y)$. Hence the formal defect convention $d_h=O(h^{p+1})$ givesAt $\alpha =1/7$, the coefficient of $h^4{y}^{(4)}$ is $1/21$, so there is no further cancellation. At any other $\alpha$, the cubic coefficient is nonzero.

There is a degeneracy at $\alpha =1$: both the first-derivative coefficient and ${\rho }^{\prime }(1)$ vanish. The raw defect still has the displayed formal order two, but it cannot be normalized by $(1-\alpha )h$ into an ordinary first-order consistency of a numerical method test. Thus a convention requiring this nondegenerate normalization would leave the ODE order undefined at $\alpha =1$. In either convention, it is not a convergent approximation to arbitrary initial-value data; the next part explains why.

Use the same difference $w$ and wave energy as in part (i). The damped wave equation now gives the energy dissipation identity for a linearly damped stringBecause $E(0)=0$ and $E(t)\ge 0$, the energy remains zero. Hence $w_t=w_x=0$, and the homogeneous Dirichlet boundary conditions force $w=0$. Therefore the solution of a differential equation is unique also for $\boxed{\alpha >0}$; positive damping can only decrease this energy.

Substitute the separated solution $y=e^{-\alpha t}F(t)\sin (nx)$ into the damped wave equation. Its time derivatives givewhereas $y_{xx}=-n^2{e}^{-\alpha t}F\sin (nx)$. ThusThe last two conditions follow from $y(x,0)=0$ and $y_t(x,0)=\sin (nx)$. For $0<\alpha <n$, let $\nu =\sqrt{n^2-{\alpha }^2}>0$. The simple harmonic motion equation has solution of a differential equationThis is an underdamped normal mode: its oscillations have the decaying factor $e^{-\alpha t}$.

For $\alpha =3$, let $F_n$ denote the time factor normalized by $F_n(0)=0$, $F_n^{\prime }(0)=1$. By superposition the Fourier sine series solution isAt $t=0$, its initial velocity is ${\sum }_{n\ge 1}{b}_n\sin (nx)={\sin }^{2}x$. Multiply by $\sin (mx)$, integrate on $[0,\pi ]$, and use the permitted interchange and Fourier orthogonality. This yields the requested integral expressionUnder $x\mapsto \pi -x$, ${\sin }^{2}x$ is unchanged and $\sin (n(\pi -x))=(-1{)}^{n+1}\sin (nx)$. Hence every even-index Fourier coefficient is zero, including $b_2$.

For odd $n$, use ${\sin }^{2}x=(1-\cos 2x)/2$ and the product-to-sum formula:ThusIn particular $b_1=8/(3\pi )$ and $b_3=-8/(15\pi )$. An explicit version of the solution of a differential equation isOnly the first, overdamped normal mode, the third, critically damped normal mode, and the odd underdamped normal modes of index at least five occur. No odd-index Fourier coefficient vanishes.

Divide by $x>1$. The product rule turns the linear ordinary differential equation intoEquivalently, $\ln x$ is an integrating factor after division by $x\ln x$. An antiderivative gives $\ln xy=\frac{1}{2}(\ln x{)}^2+C$. The initial condition at $e$ gives $C=-1/2$, henceIts derivative is $y^{\prime }=[1+(\ln x{)}^{-2}]/(2x)$, so substitution gives $x\ln x{y}^{\prime }+y=\ln x$, and $y(e)=0$. Every solution of a differential equation has the integrated form above, and the initial condition fixes its only constant. This agrees with the uniqueness theorem for ordinary differential equations on the stated interval.