Question bank — Second-order linear ODEs — superposition principle, general theory

False — linearity is about how $y$ enters, not $x$. Here $y$ appears only to the first power, so it is linear with a variable coefficient $q(x)=x^2$.

False — the product $y\cdot y^{\prime }$ multiplies two things involving the unknown, which is exactly what linearity forbids.

True — superposition says $c_1{y}_1+c_2{y}_2$ is again a solution for any constants, and $y\equiv 0$ solves it, so the solutions are closed under scaling and addition. See Linear algebra — vector spaces and bases.

False — the zero function does not solve it ($L[0]=0\ne g$), and adding two solutions gives $L[y_1+y_2]=2g\ne g$. It is an affine shift of the homogeneous space, not a vector space.

True — the highest derivative is the second, so the general homogeneous solution has two free constants, pinned down by two conditions such as $y(x_0),y^{\prime }(x_0)$.

True — a single nonzero Wronskian value forces the coefficient matrix to be nonsingular there, so no nontrivial $(c_1,c_2)$ can kill the combination.

True — Abel's theorem gives $W(x)=W(x_0)\exp (-\int p)$, and the exponential is never zero, so $W$ is all-or-nothing.

False — $L[y_1+y_2]=(-2)+(-2)=-4\ne -2$. Free adding/scaling is exclusively a homogeneous property.

True — $L[y_1-y_2]=g-g=0$. This is the seed of the structure theorem $y=y_h+y_p$.

True — the existence–uniqueness theorem for linear ODEs guarantees the solution extends across the whole interval of continuity, unlike the nonlinear case. See Existence and uniqueness theorems for ODEs.

The two functions are dependent ($y_2=3y_1$), so $c_1{e}^x+c_2\cdot 3e^x=(c_1+3c_2)e^x$ has really one free constant — not a fundamental set.

A vanishing Wronskian is only conclusive for solutions of the same linear ODE. These two functions are genuinely independent despite $W\equiv 0$; the converse of the Wronskian test fails without that context.

You add one particular solution to the whole homogeneous family: $y=c_1{y}_1+c_2{y}_2+y_p$. Distributing $y_p$ across pieces is meaningless.

The forcing is a constant, so try a constant $y_p=A$. Since $L$ maps a constant to a constant, $-A=-2\Rightarrow A=2$. See Method of undetermined coefficients.

$L$ is linear over functions with the linearity identity $L[c_1{y}_1+c_2{y}_2]=c_{1}L[y_1]+c_{2}L[y_2]$; adding a constant $3$ gives $L[y]+L[3]=L[y]+3q$, not $L[y]+3$. Constants are not fixed by $L$ unless $q=0$.

The fundamental-set conclusion requires both to solve the same homogeneous equation. Independent solutions of different ODEs say nothing about spanning one ODE's solution space.

Because $L$ is linear: $L[c_1{y}_1+c_2{y}_2]=c_{1}L[y_1]+c_{2}L[y_2]$. When both outputs are $0$ the result is $0$ for any scalars; when both are $g$ the result is $(c_1+c_2)g$, which equals $g$ only if $c_1+c_2=1$.

Because $e^{rx}$ reproduces itself under differentiation, so substituting collapses the ODE into the algebraic characteristic equation in $r$. See Characteristic equation — constant coefficient ODEs.

Dependence forces both $c_1{y}_1+c_2{y}_2=0$ and its derivative $c_1{y}_1^{\prime }+c_2{y}_2^{\prime }=0$; a nontrivial $(c_1,c_2)$ exists iff this $2\times 2$ system is singular, i.e. its determinant $y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$ is zero.

The formula $W(x)=W(x_0)e^{-\int p}$ makes $W$ a nonzero-exponential multiple of its value at $x_0$, so its sign and nonvanishing are fixed by that single value — all-or-nothing.

It certifies that the two-parameter family $c_1{y}_1+c_2{y}_2+y_p$ captures every solution; without it, exotic solutions might hide outside the family. See Existence and uniqueness theorems for ODEs.

The monic form makes the theory (Wronskian, Abel, existence) read cleanly with a single well-defined $p(x)$; the leading-coefficient clutter would otherwise infect every formula.

Each integration introduces one arbitrary constant; second order 'integrates twice'. Contrast with First-order linear ODEs, which carry a single constant.

Undetermined coefficients only guesses forms for special forcings (polynomials, exponentials, sines); Variation of parameters builds $y_p$ from the fundamental set for any continuous $g$.

Yes — $L[0]=0$ always, which is why the homogeneous solution set is a genuine vector space containing the zero vector.

The existence–uniqueness guarantee only holds on an open interval where $p,q,g$ are all continuous; across a discontinuity you may lose uniqueness or global extension, and $W$'s all-or-nothing rule can break at that point.

Yes, they may both hit zero at one point; independence only requires $W\ne 0$ somewhere, and $W(x_1)$ can still be nonzero even if $y_1(x_1)=y_2(x_1)=0$ (their derivatives differ).

Only where $g\equiv 0$; if $g$ is nonzero on any sub-interval the ODE is non-homogeneous there, and you must patch a particular solution to that region.

No — you still need two independent solutions; when the root repeats, the second is $x{e}^{rx}$, preserving a two-dimensional solution space. See Characteristic equation — constant coefficient ODEs.

No — then that $y_p$ is absorbable into $y_h$, so you must find a genuinely non-homogeneous particular solution (often by multiplying the guess by $x$) to keep $y_h$ and $y_p$ distinct.

No — replacing $y_2$ by $5y_1$ collapses independence; but $y_1$ and $y_1+y_2$ is still a valid fundamental set, since it is an invertible mix of independent functions.