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An equation $a(x)y^{\prime \prime }+b(x)y^{\prime }+c(x)y=0$ can remain meaningful at a zero of $a$ even though its normalized second-order linear differential equation has singular coefficients. A global continuation problem must specify its solution regularity. Across a point where a homogeneous solution vanishes to second order, its coefficient may change in a piecewise $C^1$ continuation; requiring continuity of the second derivative can eliminate that freedom. The usual continuous-coefficient uniqueness theorem for ordinary differential equations cannot simply be applied across such a point.

For a second-order linear differential equationthe substitution $y=u\exp [-\frac{1}{2}\int p(x)dx]$ eliminates the first-derivative term and gives

For a real pair of indicial roots $r_1\ge r_2$ at a regular singular point of a second-order linear differential equation, the second local solution is classified by the gap $\Delta$ and by the recurrence relation for the smaller indicial root. Case I has $\Delta \notin \mathbb{Z}$ and two pure Frobenius solutions. Case II has $\Delta =0$ and an independent Logarithmic Frobenius solution. With a positive integer gap, Case IIIa means the exceptional coefficient equation is inconsistent, so a nonzero multiple of the first Frobenius solution times a logarithm is necessary. Case IIIb means that equation is consistent and leaves a free coefficient, so there are two pure Frobenius solutions. This IIIa/IIIb convention is the one used in the Oxford course notes.^[ref]

The free coefficient in Case IIIb adds a multiple of the first Frobenius solution, as explained by the undetermined coefficient at Frobenius resonance. Stating whether the logarithm occurs is useful alongside the case name because other courses may choose different labels.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

Work first on $x>0$. For $b=0$, the power-law ansatz $y=x^r$ gives the indicial equation $r(r-1)+r-a^2=r^2-a^2=0$. Therefore the Cauchy-Euler equation has general solutionFor $a=0$ the repeated indicial root supplies the logarithm rather than two copies of the constant solution. Alternatively, the change of variables $t=\log x$ turns the differential equation into $Y^{\prime \prime }-a^{2}Y=0$. On $x<0$ the corresponding real formulas use $|x|$ and $\log |x|$, with independent constants on the two sides of zero.

For general $b$, normalize the second-order linear differential equation asThe coefficient $P$ is singular at zero, but $xP(x)=1$ and $x^{2}Q(x)=-a^2+b{x}^{\beta }$ are analytic functions there because $\beta$ is a positive integer. The regular singular point criterion for a second-order equation therefore proves that zero is a regular singular point for every allowed $a,b,\beta$. The indicial equation remains $r^2-a^2=0$.

Rearrange as $x^{\prime \prime }-4x^{\prime }+3x=10t$. The characteristic polynomial factors as $(r-1)(r-3)$, giving the homogeneous solutions $e^t,e^{3t}$. A particular polynomial solution $At+B$ obeys $3A=10$ and $3B-4A=0$. HenceThis is the full two-constant solution of the second-order linear differential equation.

For two solutions $y_1,y_2$, define the WronskianDifferentiate and substitute the second-order linear differential equation:Integration proves the Abel identity, including the identically zero case:For $p=-2/x$, this is $W=C{x}^2$ on either interval excluding zero, since the integral uses $\log |x|$. The coefficient singularity at zero must not be ignored.

Multiply the particular equation by $x^2$ and substitute the Frobenius solution $y=x^{\lambda }{\sum }_{n\ge 0}{a}_n{x}^n$. With $a_{-1}=a_{-2}=0$, equating coefficients givesThe leading coefficient is nonzero, so the indicial equation is $\lambda (\lambda -3)=0$. Thus only $\lambda =0,3$ are possible, and the recurrences below construct both.

For $\lambda =0$, $a_1=0$, $a_2=-a_0/2$, and the $n=3$ recurrence reads $0=a_1=0$, leaving $a_3$ free. For all later indices,This is an undetermined coefficient at Frobenius resonance; choosing $a_3=0$ selects a convenient even basis solution but is not forced by the differential equation.

For $\lambda =3$, all odd-indexed coefficients vanish andThe factorial denominators show convergence for every finite $x$. Expanding the given hyperbolic functions directly yieldsConsequently the exponent-zero family is $a_0(\cosh x-x\sinh x)-3a_3(\sinh x-x\cosh x)$, and the normalized exponent-three solution is $-3a_0(\sinh x-x\cosh x)$. This verifies the coefficient formulas and supplies two independent basis functions.

Finally $y_1^{\prime }=-x\cosh x$ and $y_2^{\prime }=-x\sinh x$, soThis agrees with the Abel calculation. Its vanishing at zero does not contradict independence on regular intervals: the original equation is singular there.

The origin is an ordinary point, since the coefficients of the second-order linear differential equation are polynomials. Seek a power series $y={\sum }_{n\ge 0}{a}_n{x}^n$. Matching the coefficients of $x^0$ and $x^1$ gives $a_2=a_3=0$. Matching the coefficient of $x^{j+2}$ givesThus the four residue classes of indices modulo four evolve separately. The initial conditions for the first solution give $a_0=1$, $a_1=0$, and henceThe second set of initial conditions gives $a_0=0$, $a_1=1$, andAn empty product equals one. In either power series, the ratio of successive nonzero terms tends to zero for every fixed $x$: the coefficient ratio is of order $1/m$, while the power increases by four. The ratio test therefore proves convergence for every $x$, and termwise differentiation verifies the differential equation. These normalized solutions are linearly independent.

For their Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, differentiating and substituting the differential equation givesThus the Abel identity yieldsOn the interval containing zero where $y_1$ has no zero,Since $y_2(0)/y_1(0)=0$, integration proves the requested reduction of order formula:This is a local formula around the origin, as requested. The power series define both solutions globally; an ordinary integral through a zero of $y_1$ is not intended. Reduction of order across a zero of the known solution instead uses continuation of the actual solution.

Work first on $x>0$, where the proposed logarithm is defined. Substitution of $y=x^m$ into the Cauchy-Euler differential equation gives the indicial equationIts distinct indicial roots are $m=\pm i$. Using $x^{\pm i}=e^{\pm i\log x}$ and Euler's formula, a real basis of solutions is $\cos (\log x),\sin (\log x)$. Their Wronskian is $1/x$, so they are linearly independent and span the two-dimensional solution space of a homogeneous linear differential equation. ThereforeFor the requested change of variables, set $Y(z)=y(e^z)$. The chain rule givesThe transformed second-order linear differential equation is $Y^{\prime \prime }+Y=0$. Its characteristic roots are again $\pm i$, giving $Y(z)=A\cos z+B\sin z$. Substituting $z=\log x$ recovers precisely the previous general solution.

If solutions on $x<0$ are wanted, the same calculation uses $z=\log |x|$ and gives independent constants on that interval. Zero is a singular point of a differential equation; the logarithmic oscillations do not define a nonzero continuous solution through it.

On an interval where $p,q$ are continuous, define the WronskianUsing the two second-order linear differential equations to eliminate the second derivatives givesThe integrating factor yields the Abel identityThus the Wronskian either vanishes identically or is nonzero everywhere on the interval. If $u_1,u_2$ are linearly dependent, it is identically zero. Conversely, if $W(x_0)=0$, the two initial-data vectors are dependent, so some nontrivial linear combination has both value and first derivative zero at $x_0$. The uniqueness theorem for ordinary differential equations makes that combination identically zero. Hence the solutions are linearly independent exactly when their Wronskian is nonzero at one, and therefore every, point. Continuity and the common differential equation are essential to this converse.

For the particular equation, divide by $x^2$ on either interval $x>0$ or $x<0$. The coefficient of $y^{\prime }$ is zero, so its Wronskian is constant there. The proposed $y_1=x^2$ satisfies $x^2{y}_1^{\prime \prime }-2y_1=2x^2-2x^2=0$. To construct a second solution by reduction of order, imposeDividing by $x^4$ shows that $(y_2/x^2{)}^{\prime }=C/x^4$, givingTaking $C=-3,D=0$ producesThe nonzero Wronskian proves linear independence; hence the general solution on either nonsingular interval is $y=A{x}^2+B/x$. Zero is a singular point of a differential equation and is excluded from this construction.

Set $v=u^{\prime \prime }$. The resulting second-order linear differential equation is $v^{\prime \prime }+2v=4t^2$. A polynomial particular solution $v_p=a{t}^2+b$ requires $2a=4$ and $2a+2b=0$, hence $v_p=2t^2-2$. The homogeneous characteristic roots of a constant-coefficient differential equation are $\pm i\sqrt{2}$, soIntegrating twice and renaming constants gives the general real solutionThe four independent homogeneous terms account for the four initial data of the fourth-order linear ordinary differential equation.

For the substitution $u(t)=y(t^2)$, repeated use of the chain rule, with $x=t^2$, givesThereforeFor $x>0$, the map $t=\sqrt{x}$ is invertible, and the transformed equation is exactly the equation just solved. Hence the general real solution on $x>0$ isThere is no requirement that $u$ extend as an even function through $t=0$: only the branch $t>0$ is needed, so the sine and square-root terms must be retained.

For two solutions $y_1,y_2$, define their Wronskian byDifferentiation and substitution of the second-order linear differential equation giveThis proves the Abel identity and hence

On an interval where the known solution $y_1$ does not vanish,Integrating gives the reduction of order formulaChoosing the normalization $C=1$ makes the Wronskian nonzero, so this solution is linearly independent of $y_1$. To verify it really solves the equation, write $y_2=y_{1}h$. Its equation reduces tothe integrand gives $h^{\prime }=\exp (-\int p)/y_1^2$, which satisfies this equation on differentiation. Adding an integration constant only adds a multiple of $y_1$. The formula is local where $y_1\ne 0$; at a zero of $y_1$ with regular coefficients, the actual solution can be continued by initial-value existence and uniqueness, even if its integral representation has an apparent singularity.

Work on an interval where the change of variables is smooth and invertible, with $s=\dot{x}\ne 0$. If $v(t)=u(x(t))$, the chain rule gives $\dot{v}=s{u}_x$ and $\ddot{v}=s^2{u}_{xx}+\dot{s}{u}_x$. Eliminating $u_{xx}$ using the original second-order linear differential equation givesFor the real transformation as printed, take $s>0$. Set $v=s^{1/2}U$, which is the Liouville transformation of a second-order equation. Differentiating twice makes the coefficient of $\dot{U}$ cancel. Dividing by $s^{1/2}$ leavesThe additional term is exactly $s^{-1/2}[(\dot{s}/s)d(s^{1/2})/dt-d^2(s^{1/2})/d{t}^2]$, as required.

On an interval with $f>0$, choose $s=f^{-1/2}$ and therefore $t={\int }^x\sqrt{f(\zeta )}d\zeta$ up to a constant. Primes in the following expression mean derivatives with respect to $x$. We have $\dot{s}=-f^{\prime }/(2f^2)$ and $\ddot{s}=-f^{\prime \prime }/(2f^{5/2})+(f^{\prime }{)}^2/f^{7/2}$, giving the WKB correction termNeglecting $F$ gives $U=c_{+}{e}^t+c_{-}{e}^{-t}$. Since $u=s^{1/2}U$, the WKB approximation isChanging the lower integration limit only rescales the constants. Small $|F|$ is the stated local approximation criterion; rapid variation or a turning point can invalidate it, and small local residual alone need not imply uniform accuracy over an arbitrarily long interval.

The printed hypothesis only says $f\ne 0$. If real $f<0$ on the interval, the printed square roots require a consistent complex branch. Equivalently a real WKB approximation isThis supplies the oscillatory interpretation for the negative-$f$ case instead of silently assuming that the original nonzero hypothesis means positive. Smoothness through the derivatives used above and absence of zeros are required locally.

The characteristic polynomial of this second-order linear differential equation factors as $r^2-r-6=(r-3)(r+2)$. Its two distinct roots give the general solution $y=A{e}^{3x}+B{e}^{-2x}$. Boundedness as $x\to \infty$ forces $A=0$, and the initial condition gives $B=1$.

Divide the second-order difference equation by $h^2$ and set $x=nh$. The first quotient is a central finite difference approximation to $y^{\prime \prime }(x)$; the quotient $(y_{n+1}-y_{n-1})/(2h)$ approximates $y^{\prime }(x)$. Their Taylor series errors are $O(h^2)$ for a sufficiently smooth function. This is therefore a finite difference method for the second-order linear differential equation in part (a).

At a fixed $x\ge 0$, take $h\to 0$ with $n=x/h$ integral. Since $\log r_{-}=-2h+O(h^3)$, the exact bounded discrete solution satisfiesLikewise $(1-2h{)}^{x/h}\to e^{-2x}$. The decay selected by boundedness in the linear recurrence relation becomes exactly the decay selected by boundedness in the ordinary differential equation.

The defining Wronskian identity gives the first-order linear differential equationWork first on an interval where $y_1\ne 0$. Dividing by $y_1$ recognizes a derivative:This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant $C$ only adds a multiple of $y_1$. If $y_1$ has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

For the stated order of the two solutions, define their Wronskian byTaking a derivative cancels the two mixed products: $W^{\prime }=y_1{y}_2^{\prime \prime }-y_1^{\prime \prime }{y}_2$. Substituting the second-order linear differential equation for each second derivative cancels the $q$ terms and leavesThis is the Abel identity. On an interval with continuous coefficients, linear independence means $W(x_0)\ne 0$; hence the Wronskian never vanishes anywhere on that interval.

Classification and characteristic exponents. For a normalized second-order linear differential equation, an ordinary point at zero means $p$ and $q$ are real analytic functions there. A regular singular point is a singular point where $xp$ and $x^{2}q$ nevertheless extend as real analytic functions to zero. Write $p_0={\lim }_{x\to 0}xp(x)$ and $q_0={\lim }_{x\to 0}{x}^{2}q(x)$. The lowest-power coefficient in the Frobenius method gives the indicial equationIts roots are the characteristic exponents at a regular singular point. If their difference is not an integer, there are two linearly independent solutions $x^{{\sigma }_j}{A}_j(x)$, with $A_j$ real analytic and $A_j(0)\ne 0$. For distinct real roots ${\sigma }_1>{\sigma }_2$ with integer difference, a solution for the larger root exists, and a second has the form $x^{{\sigma }_2}B(x)+c{y}_1\log x$; the logarithmic coefficient can vanish. For a repeated root, the second solution has the form $y_1\log x+x^{{\sigma }_1}B(x)$ after normalization. This is a Logarithmic Frobenius solution. Complex conjugate roots $a\pm ib$ give real solutions whose leading behaviour is $x^a\cos (b\log x)$ and $x^a\sin (b\log x)$, with analytic-series corrections. All these descriptions are on $x>0$, or on a fixed complex branch.

For the present equation, division by $x$ gives $p=1+(1-m)/x$ and $q=(1-m)/x$. Therefore $xp=x+1-m$ and $x^{2}q=(1-m)x$ are real analytic. Zero is regular singular when $m\ne 1$; when $m=1$, it is ordinary after removing the common factor $x$. In that exceptional case the normalized equation is $y^{\prime \prime }+y^{\prime }=0$.

For the Frobenius method substitution, the lowest coefficient and the general linear recurrence relation areThese equations come respectively from the coefficients of $x^{\sigma -1}$ and $x^{n+\sigma -1}$. At $m=1$, the same formal exponents $0,1$ describe the two possible leading powers at an ordinary point, rather than singular behaviour.

Assume the usual continuous-coefficient setting for a second-order linear differential equation. The Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$ has derivativewhich proves the Abel identity. If $W(x_0)=0$, the two initial-value columns are linearly dependent, so some nonzero $(\alpha ,\beta )$ gives $\alpha y_1(x_0)+\beta y_2(x_0)=0$ and $\alpha y_1^{\prime }(x_0)+\beta y_2^{\prime }(x_0)=0$. The uniqueness theorem for ordinary differential equations makes that linear combination identically zero. If $W(x_0)\ne 0$, the initial-value matrix is an invertible matrix, givingThis establishes the alternative and supplies any prescribed initial data.

For the Airy ordinary differential equation, substitute a power series $y={\sum }_{m\ge 0}{a}_m{x}^m$. Coefficient comparison gives $a_2=0$ andChoose $(a_0,a_1)=(1,0)$ and $(0,1)$. The Airy power-series fundamental pair isEmpty products mean one. Their first terms are $y_1=1+x^3/6+x^6/180+\cdots$ and $y_2=x+x^4/12+x^7/504+\cdots$. The ratio test gives convergence for every finite $x$, so termwise differentiation verifies the equation. Their Wronskian is one at zero, and hence everywhere because $p=0$. Thus every solution is $A{y}_1+B{y}_2$.

Set $v=y^{\prime }$ to reduce the second-order linear differential equation to $x{v}^{\prime }+v=0$. Equivalently, $(xv{)}^{\prime }=0$, so $v=C_1/x$ on any interval excluding zero. The derivative condition at $x=1$ gives $C_1=2$.

Integrating on the positive interval containing the initial point yields $y=2\ln x+C_2$. The value $y(1)=-1$ then fixes $C_2=-1$, givingIts derivative is $2/x$, and direct substitution verifies the equation. The selected solution diverges as $x\downarrow 0$ and cannot extend through zero as a finite-valued function.