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Using the hyperbolic functions identities $\sin (ix)=i\sinh x$ and $\cos (ix)=\cosh x$, divide by $i$ and put $u=e^x\ne 0$. The equation becomes $-u/2-5/(2u)=3$, henceTaking all branches of the complex logarithm givesIf $x$ is required to be real, there are no real solutions, since $e^x$ cannot be negative. The complex branches are essential to the unrestricted answer.

The complex exponential identity $\sin \theta =(e^{i\theta }-e^{-i\theta })/(2i)$ givesThus the locus is a circle of radius $1/2$ centered at $i/2$, traversed counterclockwise twice as $\theta$ runs through the specified interval. Its period as a parametrized curve is $\pi$. Near $\theta =0$, the Taylor expansion gives $\Re z=\theta +O({\theta }^3)$ and $\Im z={\theta }^2+O({\theta }^4)$, so locally $\Im z=(\Re z{)}^2+O((\Re z{)}^4)$. There is a smooth horizontal tangent, not a cusp: $dz/d\theta =e^{2i\theta }$ is nonzero at each passage through the origin.

An exact differential $Mdx+Ndy$ is $d\Phi$ for a scalar potential function $\Phi$, so $M=\partial \Phi /\partial x$ and $N=\partial \Phi /\partial y$. Equality $\partial M/\partial y=\partial N/\partial x$ is necessary, and is sufficient on a simply connected space when these partial derivatives are continuous.

Here, after multiplying the ordinary differential equation by $dx$,On either half-plane $y>0$ or $y<0$, integrate $M$ with respect to $x$:Comparison of $\partial \Phi /\partial y$ with $N$ gives $h^{\prime }(y)=6y$, hence $h(y)=3y^2$ up to a constant. The solution is therefore the family of level setsLocally these define differentiable graphs $y(x)$ wherever $N\ne 0$, by the implicit function theorem. When $N=0$, $M\ne 0$ on this domain, so a finite derivative cannot satisfy $M+N{y}^{\prime }=0$ there; such points are vertical tangents of the implicit curves, rather than additional classical graph solutions.

An invertible matrix is a square matrix $X$ admitting a matrix inverse $X^{-1}$ with $X{X}^{-1}=X^{-1}X=I$, where $I$ is the identity matrix. Equivalently, its associated linear map is bijective.

Let $X\mathbf{v}=\lambda \mathbf{v}$ with eigenvector $\mathbf{v}\ne 0$. The eigenvalue cannot be zero: otherwise applying $X^{-1}$ gives $\mathbf{v}=0$. Apply $X^{-1}$ to the eigenvector equation:Thus $1/\lambda$ is an eigenvalue of $X^{-1}$ with the same eigenvector.

The characteristic polynomial isThus the eigenvalues are $i$ and $-i$. There is no real diagonalization, because a real diagonal matrix would have real eigenvalues. Over the complex numbers, the two distinct eigenvalues give linearly independent eigenvectors, so the matrix is diagonalizable.

For $\lambda =i$, the first row gives $(1-i)x+2y=0$. Choose $x=2$, giving $y=-1+i$. Its standard complex norm is $\sqrt{|2{|}^2+|-1+i{|}^2}=\sqrt{6}$; the conjugate eigenvector belongs to $-i$. Hence normalized eigenvectors and a suitable change of basis areDirectly, $AP=P\mathrm{diag}(i,-i)$ and $\det P=-2i/3\ne 0$, soNormalization does not assert that these eigenvectors are orthogonal.