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With the Einstein summation convention, the magnetic force has components $F_i=q{ϵ}_{ijk}{v}_j{B}_k$. Thus its matrix acting on velocity isThe Levi-Civita symbol changes sign on exchanging $i,j$, so $M^T=-M$; the diagonal vanishes and the matrix has zero trace. For nonzero $\mathbf{B}$, $M\mathbf{B}=q\mathbf{B}\times \mathbf{B}=0$, exhibiting a nontrivial kernel and proving noninvertibility. If $\mathbf{B}=0$ or $q=0$, $M$ is the zero matrix and is again noninvertible.

Equivalently, an odd-dimensional real antisymmetric matrix has $\det M=\det M^T=\det (-M)=-\det M$, hence zero determinant. The same antisymmetry gives ${\mathbf{v}}^{T}M\mathbf{v}=0$, expressing that the magnetic force does no work.

For a unit vector, the radial coordinate is $r=1$ and the spherical coordinates giveThe Pauli matrices therefore giveThe diagonal entries are real and the off-diagonal entries are complex conjugates, so $H^{\ast }=H$. Its trace is zero and its determinant is $-1$, giving eigenvalues $+1$ and $-1$. A convenient orthonormal eigenbasis isThe half-angle identities verify $H{v}_{\pm }=\pm v_{\pm }$, $v_{\pm }^{\ast }{v}_{\pm }=1$ and $v_{+}^{\ast }{v}_{-}=0$. Multiplication of either vector by any unit-modulus phase gives an equivalent normalized eigenvector. At a pole, the azimuth is undefined geometrically, but the remaining phase freedom makes the chosen expressions consistent with the same one-dimensional eigenspaces.

An integrating factor for $u^{\prime }+P(x)u=Q(x)$ is a nonzero function $\mu (x)=\exp (\int P(x)dx)$, chosen so that $(\mu u{)}^{\prime }=\mu Q$. Integrating this total derivative reduces the first-order linear ordinary differential equation to one quadrature.

For the given nonlinear equation, let $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. Multiplication by $2y$ gives $u^{\prime }+u=x^2+2x$. Using the integrating factor $e^x$, or recognizing the particular solution $u=x^2$, yields $u=x^2+C{e}^{-x}$. The initial value gives $C=1$ and selects the positive branch:The radicand is strictly positive for every real $x$. This solution never reaches the singular denominator $y=0$ of the original equation, so multiplication by $y$ has introduced no forbidden crossing. The negative branch would violate $y(0)=1$.