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For the linear partial differential equation $2y{u}_x-x{u}_y=0$, the method of characteristics gives $\dot{x}=2y$, $\dot{y}=-x$, $\dot{u}=0$. Thus $\varphi =x^2+2y^2$ is constant on each characteristic curve, whose nonzero level sets are ellipses. On a domain containing entire such ellipses, the solutions have the form $u=F(\varphi )$, subject to the desired regularity at the origin.

For the inhomogeneous equation $2y{u}_x-x{u}_y=xy(2y^2-x^2)$, a particular solution is $u_p=x^2{y}^2/2$, so the general solution is $u=x^2{y}^2/2+F(x^2+2y^2)$. Prescribing $u(x,1)=x^2$ fixes $F(s)=(s-2)/2$ only for $s\ge 2$. The inner ellipses $s<2$ never intersect that data line, so the boundary values leave their solution values undetermined. Smooth extensions must match the prescribed $F$ at $s=2$; choosing the same linear formula everywhere gives one polynomial extension.

A homogeneous linear differential equation is a linear differential equation with identically zero forcing term, $L[y]=0$. On a common interval, its solutions of a differential equation form the vector space $\ker L$, the kernel of a linear map. They include the zero function and are closed under addition and scalar multiplication. For any two solutions of a differential equation $u,v$ and constant scalars $\alpha ,\beta$, linearity gives $L[\alpha u+\beta v]=\alpha 0+\beta 0=0$: this is the superposition principle. The scalars must be constant; variable multipliers introduce extra derivatives.

For example, $y^{\prime }+2y=0$ has general solution $y=C{e}^{-2t}$, where $C$ is any constant real number. A short verification uses the product rule: $(e^{2t}y{)}^{\prime }=e^{2t}(y^{\prime }+2y)=0$, so $e^{2t}y$ is constant. For $y_1=e^{-2t}$ and $y_2=2e^{-2t}$, the sum is $3e^{-2t}$ and also solves the same homogeneous linear differential equation. At $t=1$, their values are approximately $0.13534$, $0.27067$ and $0.40601$. Every curve is a multiple of one exponential function; the solution space of a homogeneous linear differential equation has dimension one. Both $t$ and $y$ in this example and the following diagram are dimensionless mathematical variables.

Now keep the same differential operator $L[y]=y^{\prime }+2y$ but prescribe the forcing term $1$. The inhomogeneous linear differential equation $y^{\prime }+2y=1$ has particular solution $y_p=1/2$ and general solution $y=1/2+C{e}^{-2t}$. The initial conditions $y(0)=0,1,2$ select, respectively,At $t=1$, their values are approximately $0.43233$, $0.56767$ and $0.70300$. The midpoint $\frac{1}{2}(u_0+u_2)=u_1$ solves the same inhomogeneous linear differential equation, because $L[\frac{1}{2}(u_0+u_2)]=\frac{1}{2}(1+1)=1$. The sum $s=u_0+u_2=1+e^{-2t}$ instead satisfies $s^{\prime }+2s=2$, so it fails the equation with forcing term $1$. More generally, a linear combination of solutions of a differential equation for the same fixed nonzero forcing term retains that forcing term precisely when its scalar weights add to one. This is the affine space structure, rather than unrestricted superposition within that fixed family.

On the left, all three curves solve the same homogeneous linear differential equation. On the right, $u_0,u_1,u_2$ share the particular solution $1/2$, and the shaded difference $u_2-1/2$ is the homogeneous solution $\frac{3}{2}{e}^{-2t}$. The dashed sum has twice the forcing term, even though its starting value coincides with that of $u_2$. Adding the two initial conditions also adds their inputs; taking their midpoint averages both. This distinction makes the superposition principle useful for assembling responses to different inputs without mistaking that assembly for a solution of a differential equation with the original input. See MIT's superposition notes, including the particular solution one half.

An inhomogeneous linear differential equation has a nonzero forcing term. Its general solution is one particular solution plus the general solution of the associated homogeneous linear differential equation.

An integrating factor is a nonzero multiplier that makes a differential equation directly integrable. For a first-order linear differential equation, it gathers the unknown function and its derivative into one product rule expression. Suppose $a$ and $b$ are continuous functions on an interval containing $t_0$, and consider the initial value problemDefine $A(t)={\int }_{t_0}^{t}a(s)ds$ and $\mu (t)=e^{A(t)}$. The exact solution of a differential equation throughout that interval isThe definite integral fixes $\mu (t_0)=1$, so the initial condition appears without an extra constant to determine. If the equation originally has a coefficient multiplying $y^{\prime }$, first divide by that coefficient on an interval where it is nonzero; a zero of that coefficient needs separate attention. See OpenStax's first-order linear equations chapter for this normalization and the integrating factor procedure.

For a concrete balance, solve $y^{\prime }+2y=3$ with $y(0)=0$. Here $\mu =e^{2t}$, so multiplying the entire ordinary differential equation gives $(e^{2t}y{)}^{\prime }=3e^{2t}$. Accumulating from zero to $t$ givesThe computed values are $y(0.5)\approx 0.94818$, $y(1)\approx 1.29700$ and $y(2)\approx 1.47253$. Its derivative is $3e^{-2t}$, which verifies $y^{\prime }+2y=3$ exactly. Interpreting $3$ as a steady input and $2y$ as proportional removal explains the shape: initially all input increases $y$, then removal catches up. The constant particular solution $y_p=3/2$ balances input and removal, and the homogeneous solution $y_h=-\frac{3}{2}{e}^{-2t}$ is the transient response that corrects it to the initial condition. Because this autonomous differential equation has a time-independent right-hand side, $3/2$ is also an equilibrium point.

The first panel adds the particular solution and homogeneous solution; the second shows the same balance in rates, $3=y^{\prime }+2y$. This links the algebra to a physical accounting rule. Newton's law of cooling has this form with $a=k$ and $b=k{T}_a$, while an electrical circuit can have a varying input. In both cases the integrating factor systematically handles the input rather than requiring a guess for a particular solution.

Variable coefficients work just as directly. For $y^{\prime }+2ty=t$, $y(0)=0$, the integrating factor is $e^{t^2}$, andThus $y(0.5)\approx 0.11060$, $y(1)\approx 0.31606$ and $y(2)\approx 0.49084$. The derivative $y^{\prime }=t{e}^{-t^2}$ verifies $y^{\prime }+2ty=t$, including at zero, where both coefficient and input vanish. No division by $t$ is necessary. The general solution is $y=1/2+C{e}^{-t^2}$: $1/2$ is a constant particular solution, and $C{e}^{-t^2}$ solves the associated homogeneous linear differential equation. Although this special forcing term admits a constant trajectory, this is a nonautonomous differential equation: it depends explicitly on time. Merely setting its right-hand side to zero at one instant would not establish a constant solution of a differential equation; at $t=0$ every state has zero slope.

The direction field shows the selected trajectory leaving its initially horizontal tangent line. For $t\ge 0$, the difference between any two solutions of a differential equation is multiplied by $e^{-t^2}$, rather than $e^{-2t}$ in the constant-rate example. This dependence on elapsed history is clearer if the exact formula is rewritten asFor $t\ge t_0$, each contribution from the input at $s$ is weighted by the accumulated coefficient between $s$ and $t$. The homogeneous solution remembers the initial condition; the particular solution given by the second term accumulates the forcing term with zero initial value. This decomposition differs from choosing a constant particular solution, but gives the same total. Initial-data memory tends to zero when ${\int }_{t_0}^{t}a\to +\infty$; a negative accumulated coefficient instead amplifies it. The reciprocal relation between integrating factor and homogeneous solution is developed in MIT's notes on solutions to linear first-order equations.

Why does the trick work? The fundamental theorem of calculus and chain rule give ${\mu }^{\prime }=a\mu$. Consequently the product rule givesIntegrating this identity and dividing by the positive exponential function $\mu$ produces the stated formula. Any nonzero constant multiple of $\mu$ works equally well; normalization is a convenience. This is the beauty of the method: a sum involving an unknown function becomes one exact derivative, leaving only an integral of known functions. See Oregon State's product-rule derivation. The related integrating factor for a differential one-form extends the idea of making an expression exact beyond this linear setting.

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.

A matched asymptotic expansion constructs approximations in regions with different distinguished limits and fixes their free constants by requiring agreement in an overlap region. The inner expansion resolves rapid variation; the outer expansion describes the surrounding region. Matching compares their common limiting behavior, rather than imposing distant boundary conditions directly on a local approximation.^[ref]

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, the leading outer solution is $y_0=1$. The inner variable $X=x/ϵ$ gives $Y_0^{\prime \prime }+Y_0^{\prime }=0$ and $Y_0(0)=0$, hence $Y_0=A(1-e^{-X})$. In the overlap region $ϵ\ll x\ll 1$, the inner variable tends to infinity while the outer coordinate tends to zero. The asymptotic matching condition isso $A=1$. One concrete path through the overlap region is $x=\sqrt{ϵ}$: then $x\to 0$ and $X={ϵ}^{-1/2}\to \infty$. Matching selects the free constant that the left boundary condition alone could not determine.

Why choose the right boundary condition for the outer solution? If instead one selects $y_0=0$ from the left, a proposed right boundary layer with $Z=(1-x)/ϵ$ satisfies $U_0^{\prime \prime }-U_0^{\prime }=0$. Its general solution is $A+B{e}^Z$. Matching to zero as $Z\to \infty$ forces $A=B=0$, contradicting $U_0(0)=1$. The direction in which the exponential function decays determines which endpoint can carry the boundary layer; endpoint selection is part of the calculation.

For a positive eigenvalue write $\mu =\sqrt{\lambda }>0$. The general solution of the limiting constant-coefficient differential equation is $y=A\cos (\mu x)+B\sin (\mu x)$. The endpoint at zero gives $A=0$. A nonzero eigenfunction then requires $B\ne 0$, and the endpoint at $\pi$ gives $\sin (\mu \pi )=0$. ConsequentlyFor real eigenfunctions the constants may be any nonzero real amplitudes. Requiring unit L2 norm would set $|C_k|=\sqrt{2/\pi }$, but that normalization is not required by the question. The lowest eigenvalue is ${\lambda }_1=1$ and its eigenspace is spanned by $\sin x$. For completeness, zero yields only an affine solution annihilated by both endpoints, and negative eigenvalues yield hyperbolic functions whose two zero endpoint values also force the zero solution.

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 $r^2-a^2=0$.

Normalize both leading coefficients to one. For the indicial root $a=1/2$, the recurrence relation from part (b) givesRecognizing the Taylor series of the sine gives the first Frobenius solutionFor the smaller indicial root $-1/2$, $n(n-1)d_n=-d_{n-2}$. The equation at $n=1$ is $0=0$; choose the free coefficient $d_1=0$ to avoid adding the first Frobenius solution. The remaining coefficients areThus the Taylor series of the cosine givesTheir leading terms are $x^{1/2}$ and $x^{-1/2}$, as required for Case IIIb. A direct check uses $y=x^{-1/2}u$: the differential equation becomes $x^{3/2}(u^{\prime \prime }+u)=0$, so these two functions solve it on $x>0$. Their Wronskian is $-1/x\ne 0$, proving linear independence. The general solution is $(C_1\sin x+C_2\cos x)/\sqrt{x}$.

The power-law ansatz $y=x^r$ givesSince $ϵ>0$, the equation roots $1$ and $-1/ϵ$ are distinct. The Cauchy-Euler equation therefore has general solutionFor $C_1=C_2=1$, the first derivative and second derivative areThe strictly convex function diverges to $+\infty$ as $x\to 0^{+}$, passes through $(1,2)$, and has a unique lowest point atAs $x\to \infty$, $y-x\to 0$, so the line $y=x$ is an asymptote. When $0<ϵ<1$, the lowest point is to the right of one; at $ϵ=1$ it is at one, and for $ϵ>1$ it is to the left of one.

The ratio of the two terms isThus $x^{-1/ϵ}$ is larger for $0<x<1$, the terms are equal at $x=1$, and $x$ is larger for $x>1$. In the stronger asymptotic sense as $ϵ\to 0^{+}$, $x^{-1/ϵ}$ dominates at each fixed $x<1$, and $x$ dominates at each fixed $x>1$. Their crossover has $|\log x|=O(ϵ)$, equivalently $x=1+O(ϵ)$, as described by the power-law crossover in a singular perturbation. In that region $x=1+ϵX$ with bounded $X$ gives $y\sim 1+e^{-X}$.

Substitute the proposed regular perturbation $y=y_0+ϵy_1+\cdots$ into the differential equation. The leading and subsequent equations areConsequently $y_0=Cx$, and in fact each successive coefficient has the form $y_j=C_{j}x$, because the preceding second derivative is zero. The leading approximation is thereforeIt has only one adjustable constant and generally cannot impose two independent boundary conditions. Those assumed in the question require a nonzero $x^{-1/ϵ}$ coefficient, which a power series in $ϵ$ cannot recover: at fixed $x>1$ it is exponentially small, and at fixed $x<1$ it grows without bound as $ϵ\to 0^{+}$. Thus the regular perturbation describes the specified solution only in the region to the right of the crossover, away from $x=1$; it is not a uniform approximation on $(0,\infty )$.

The reduced equation and the left boundary condition give the leading outer solutionIntroduce the right inner variable $X=(2-x)/ϵ\ge 0$ and write $Y(X)=y(2-ϵX)$. The chain rule gives $y^{\prime }=-Y_X/ϵ$ and $y^{\prime \prime }=Y_{XX}/{ϵ}^2$. Multiplying the rescaled differential equation by $ϵ$ gives the exact equationAt leading order, $Y_{0,XX}+2Y_{0,X}=0$, with general solution $Y_0=A+B{e}^{-2X}$. The right boundary condition gives $Y_0(0)=0$. The asymptotic matching condition in the overlap region $ϵ\ll 2-x\ll 1$ givesThus $A=e$ and $B=-e$, yielding the leading inner expansionEquivalently, the inner variable $\xi =2(2-x)/ϵ$ gives $Y_0=e(1-e^{-\xi })$. Both normalizations describe the same boundary layer.

When $n=0$, the forcing has nonzero integral ${\int }_{-1}^{1}1dx=2$, violating the Fredholm alternative condition. There is no bounded solution. Directly, integration gives $(1-x^2)y^{\prime }=x+B$; cancellation of its logarithmic divergence at $x=-1$ requires $B=1$, while cancellation at $x=1$ requires $B=-1$.

For $n\ge 1$, the constant $1/[n(n+1)]$ is a particular solution. Subtract it from any solution. The remainder satisfies the Legendre differential equation, whose general solution is a combination of $P_n$ and the logarithmically unbounded Legendre function of the second kind. Boundedness removes the latter coefficient, givingThe constant forcing is orthogonal to $P_n$ for $n\ge 1$, consistently with Orthogonality of Legendre polynomials and the Fredholm alternative.

For the positive coefficient of $y^{\prime \prime }$, setting $ϵ=0$ gives the reduced equation $y_0^{\prime }+x{y}_0=0$, with general solution $y_0=C{e}^{-x^2/2}$. A fast decaying homogeneous solution of $ϵy^{\prime \prime }+y^{\prime }=0$ is $e^{-x/ϵ}$, placing the boundary layer at the left endpoint. Thus use the right boundary condition for the outer solution: $y_0(1)=0$ impliesThe dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives width $\delta =ϵ$. With the inner variable $X=x/ϵ$ and $Y(X)=y(ϵX)$, the exact rescaled equation isAt leading order, $Y_{0,XX}+Y_{0,X}=0$, so $Y_0=A+B{e}^{-X}$. The asymptotic matching condition $Y_0(\infty )=y_{\mathrm{out},0}(0)=0$ sets $A=0$, and $Y_0(0)=1$ sets $B=1$. ThereforeThe additive composite expansion has zero overlap to subtract. Its graph starts at one, decreases steeply over $x=O(ϵ)$, and then stays close to zero; its initial slope is $-1/ϵ$. The right endpoint discrepancy $e^{-1/ϵ}$ is exponentially small. The left panel of the figure below sketches this profile. A right-end inner expansion would instead contain a growing $e^X$ term and could not repair the left endpoint mismatch.

Separate variables and integrate on $|y|<1$:Hence the general solution in this region isEvery such solution increases from values near $-1$ towards $1$, remains strictly between the two equilibria, and is defined for all real $x$.

With $h=\delta t>0$, the central finite difference converts the decay equation toThe continuous ordinary differential equation has the general solution $x(t)=C{e}^{-Kt}$. The discrete linear recurrence is second order and requires two initial values, such as $x_0,x_1$. Trying $x_n={\rho }^n$ gives the characteristic equation of a linear recurrence ${\rho }^2+2Kh\rho -1=0$. Its roots aresoThere are no repeated roots, and the same formula extends to negative integer $n$ if required.

To demonstrate the continuum limit, put $a_h=\mathrm{arsinh}(Kh)$. Then ${\rho }_{+}=e^{-a_h}$ and ${\rho }_{-}=-e^{a_h}$, with $a_h/h\to K$. For $t_n=nh\in [0,T]$,uniformly, since $0\le t_n\le T$. More precisely $a_h/h=K-K^3{h}^2/6+O(h^4)$, giving an $O(h^2)$ error on a fixed interval. Thus $B=0$, $A=C$ selects the physical discrete mode and corresponds to $x_1={\rho }_{+}{x}_0$.

The other root is a parasitic amplification root:For fixed $B\ne 0$, even and odd mesh points approach values differing by $2B{e}^{Kt}$. Equivalently successive mesh values have a nonvanishing alternating jump despite their time separation tending to zero. This mode cannot converge uniformly to a continuous ODE solution; its limiting envelope also grows rather than decays.

For step-dependent initial data the precise result is the centered two-step discretization of exponential decay criterion:Sufficiency follows from the two uniformly bounded exponential envelopes on a fixed interval. Necessity follows from convergence at the adjacent points $0,h$: $x_0\to C$, $x_1-x_0\to 0$, and the coefficient formulas force $B_h\to 0$, $A_h\to C$. In particular a nonzero but vanishing starting error is not excluded. Starting with the exact value $x_1=C{e}^{-Kh}$ gives $B_h=C{K}^3{h}^3/12+O(h^4)$ and still converges on each fixed interval. The assertion about “remaining” solutions therefore concerns a fixed nonzero parasitic amplitude, not every nonzero step-dependent amplitude. The method is nevertheless not damping-stable for arbitrarily long times because $|{\rho }_{-}|>1$.

An integrating factor for a differential one-form makes the multiplied form exact. If $dF=\mu gdx+\mu fdy$, then $F_x=\mu g$ and $F_y=\mu f$. Equality of the mixed partial derivatives yieldsThis is a local compatibility condition under the usual smoothness assumptions.

For the trigonometric equation, where its coefficients are defined and division is legitimate, the slope is $y^{\prime }=\tan x\tan y$. In the first quadrant it is positive, small near either axis and large near the upper-right corner. The direction field below uses normalized line segments, so steep directions remain visible.

Multiplication by $\sin x\cos y$ givesThus the multiplying function is an integrating factor, and the implicit general solution isIn the larger square, $\cos x\ge 0$. The positive-$C$ curves have positive $y$ and are symmetric about the $y$-axis; the negative-$C$ curves are their reflections. For $0<|C|<1$, a graph representation iswith endpoints tending to $y=\mathrm{sgn}(C)\pi /2$. The $C=0$ interior curve is $y=0$. The domain boundaries include singular coefficients or zeros of the integrating factor, so these endpoint limits and smooth continuations through $x=0$ describe the regularized relation; they must not be mistaken for additional solutions of the literal original equation at undefined points.

For the homogeneous solution, expand in the independent eigenvectors of $A$. Each coefficient obeys $c_j^{\prime }={\lambda }_j{c}_j$, so the complementary function isIf ${\mathbf{x}}_p$ is any particular integral, subtracting it from any other solution gives a homogeneous solution. Hence $\mathbf{x}={\mathbf{x}}_p+{\mathbf{x}}_c$ is the general solution. Complex-conjugate modes can be combined to produce real solutions when the forcing is real.

For a real two-by-two matrix, the characteristic polynomial is ${\lambda }^2-(\mathrm{tr}A)\lambda +\det A$. Under the stated distinct-eigenvalue assumption, all nonzero homogeneous modes are purely oscillatory exactly when the eigenvalues are $\pm i\omega$ with $\omega >0$. Their sum and product imply $\mathrm{tr}A=0$, $\det A={\omega }^2>0$. Conversely these trace and determinant conditions force those eigenvalues, so there is no exponential growth, decay or Jordan-block secular growth.

For the final specified initial-value problem, direct multiplication gives $A^2=-9I$. The forcing vector $\mathbf{b}=(2,3i-1{)}^T$ satisfies $A\mathbf{b}=3i\mathbf{b}$, so the forcing is resonant and $t\mathbf{b}{e}^{3it}$ is a particular integral. The matrix exponential isSince the resonant particular integral vanishes initially, the initial-value solution isThis complex solution is appropriate to the complex forcing as written. Taking real parts gives the corresponding solution for the real part of that forcing.

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.

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 $B=1$.