Do not let the rigour of mathematics impede your intuition.

Posted on October 2, 2012 by

Let D be an operator defined by D=\displaystyle\frac{d}{dx}(.)

Note: An operator maps a function to another function. (or vector space to another vector space) ie. \displaystyle D:(\mathbb{R}\mapsto\mathbb{R})\mapsto(\mathbb{R}\mapsto\mathbb{R})

Consider the following first order linear differential equation.

\displaystyle\frac{dy}{dx}+ky=Q(x)

\displaystyle Dy+kIy=Q(x)

\displaystyle (D+kI)y=Q(x)

\displaystyle y=(D+kI)^{-1}Q(x)

\displaystyle =e^{-kx}\int e^{kx}Q(x)dx (defined to be so)

Consider the second order linear differential equation.

\displaystyle\frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=Q(x)

\displaystyle(D^2+aD+bI)y=Q(x)

\displaystyle y=\frac{1}{D^2+aD+bI}Q(x)

\displaystyle y=(\frac{m_1}{D+\alpha}+\frac{m_2}{D+\beta})Q(x) (apply partial fractions ridiculously with complex numbers)

\displaystyle =m_1e^{-\alpha x}\int e^{\alpha x}Q(x)dx+m_2e^{-\beta x}\int e^{\beta x}Q(x)dx.

Does one need to understand your digestive system before you take your dinner? – Heaviside

The point of rigour is not to destroy all intuition; instead, it should be used to destroy bad intuition while clarifying and elevating good intuition. It is only with a combination of both rigorous formalism and good intuition that one can tackle complex mathematical problems; one needs the former to correctly deal with the fine details, and the latter to correctly deal with the big picture.

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