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Here the dependent variable is $x$ and the independent variable is $y$. Multiplying the equation by $2x$ and setting $q=x^2$ gives a linear equation:An integrating factor is $y^{-4}$, soThus $x^2=C{y}^4-y^2$. The initial point $(x,y)=(1,1)$ fixes $C=2$, and the positive branch through that point isEquivalently, inverting this branch givesThe source equation contains $xy$ in the denominator. The endpoint with $x=0$ is therefore excluded even though the inverse expression has a finite limit there. A negative-$x$ branch cannot be joined through that endpoint to produce a solution $x(y)$ of the original equation. The interval shown is the maximal real positive branch containing the initial point.

The original basis matrix has columns ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ and determinant $D=3\cdot 3-4\cdot 4=-7$. Evaluating the defining cross products givesFor instance, ${\mathbf{e}}_2\times {\mathbf{e}}_3=(3,-4,0)$, which must be divided by the negative denominator $-7$. These vectors satisfy ${\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}_j={\delta }_{ij}$, the defining reciprocal basis relations.

Dotting the desired expansion with the reciprocal vectors yieldsIndeed $\mathbf{a}={\mathbf{e}}_1+2{\mathbf{e}}_3$. The original basis is not orthonormal, so taking dot products with the original vectors would not give these expansion coefficients.

Let $A\mathbf{v}=\lambda \mathbf{v}$ with $\mathbf{v}\ne 0$ and $A^{†}=-A$. Taking the conjugate of the scalar ${\mathbf{v}}^{†}A\mathbf{v}$ givesSince ${\mathbf{v}}^{†}\mathbf{v}>0$, it follows thatThus every nonzero eigenvalue of an anti-Hermitian matrix is purely imaginary; zero is allowed as well.

For $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, compute the same mixed scalar in two ways:Using ${\lambda }^{\ast }=-\lambda$ gives $(\mu -\lambda ){\mathbf{u}}^{†}\mathbf{v}=0$. ThereforeThe relevant orthogonality uses the complex inner product, not the unconjugated bilinear dot product. Repeated eigenvalues do not force every possible pair of eigenvectors within that one eigenspace to be orthogonal.

Write all vectors in Cartesian components. The first component of the nested cross product isThe missing $a_x$ terms cancel as $-b_x{a}_x{c}_x+c_x{a}_x{b}_x=0$. Computing the $y$ and $z$ components by cyclic permutation gives the same form, proving the vector triple product identityApply this expression to the three cyclic terms. Their sum isEach coefficient cancels because the dot product is symmetric. This is the Jacobi identity for the cross product.

Let $w$ satisfy $\sin w=z$ and put $t=e^{iw}\ne 0$. The exponential definition of the complex sine givesThe quadratic formula therefore gives $t=iz\pm \sqrt{1-z^2}$. Taking the complex logarithm yields the multivalued complex inverse sine formulaThe two quadratic roots have product $-1$, so neither is zero and their logarithms exist as multivalued quantities. If $w_0$ is one local inverse value, the complete set of values is $w_0+2\pi k$ and $\pi -w_0+2\pi k$, with $k\in \mathbb{Z}$.

Branch qualification: the source's $\pm$ formula describes the branches of the complex inverse sine, not two interchangeable expressions for a single chosen inverse branch. At $z=0$, the plus sign with the usual local choices gives $0$, whereas the minus sign with $\log (-1)=i\pi$ gives $\pi$. Thus both signs cannot simultaneously equal the principal inverse at zero. A principal complex inverse sine requires compatible square-root and logarithm branches, normalized to $w_0=0$ at zero and continued on the chosen cut domain.

Rationalizing the denominator givesHence its real part, imaginary part, complex modulus, and principal argument are respectivelyThe quadrant matters: the real part is positive and the imaginary part negative. For the exponential expression, Euler's formula gives $e^{2+i\pi /4}=e^2(1+i)/\sqrt{2}$, so the same four quantities areGeneral complex arguments are obtained from each stated principal argument by adding $2\pi k$, $k\in \mathbb{Z}$.

The constant term belongs on the forcing side: the equation is $y^{\prime \prime }-2y^{\prime }=x+1$. Set $v=y^{\prime }$ to obtain the first-order linear equation $v^{\prime }-2v=x+1$. A particular solution is $v_p=-x/2-3/4$, while $v_h=A{e}^{2x}$. Imposing $v(0)=0$ fixes $A=3/4$. Integrating and then using $y(0)=0$ givesDirect differentiation gives $y^{\prime }=3e^{2x}/4-x/2-3/4$ and $y^{\prime \prime }=3e^{2x}/2-1/2$, so $y^{\prime \prime }-2y^{\prime }-1=x$. Both initial values vanish. The nonzero exponential term is required by the initial derivative, even though the forcing is polynomial.