# Basic Concepts

An nth-order linear differential equation has the form ${b}_{{n}}{\left({x}\right)}{{y}}^{{{\left({n}\right)}}}+{b}_{{{n}-{1}}}{\left({x}\right)}{{y}}^{{{\left({n}-{1}\right)}}}+\ldots+{b}_{{2}}{\left({x}\right)}{y}''+{b}_{{1}}{\left({x}\right)}{y}'+{b}_{{0}}{\left({x}\right)}{y}={g{{\left({x}\right)}}}$, where ${g{{\left({x}\right)}}}$ and all coefficients ${b}_{{j}}{\left({x}\right)},j={\overline{{{0}..{n}}}}$ depend solely on the variable ${x}$. In other words, they do not depend on ${y}$ or on any derivative of ${y}$.

If ${g{{\left({x}\right)}}}\equiv{0}$, the differential equation is called homogeneous; otherwise, it is called non-homogeneous.

For example, ${{x}}^{{3}}{y}'''+\frac{{1}}{{x}}{y}''+{2}{y}'={0}$ is homogeneous, and ${y}''+{x}{y}'={\cos{{\left({x}\right)}}}$ is non-homogeneous.

A linear differential equation has constant coefficients, if all the coefficients ${b}_{{j}}{\left({x}\right)}$ are constants; if one or more of these coefficients are not constant, the differential equation has variable coefficients.

For example, ${{y}}^{{{\left({4}\right)}}}+{{x}}^{{2}}{y}'+{x}{y}={6}$ has variable coefficients, and ${3}{y}''+{y}'={{e}}^{{x}}$ has constant coefficients.

Theorem. Consider an initial value problem given by a linear differential equation and ${n}$ initial conditions ${y}{\left({x}_{{0}}\right)}={y}_{{0}}$, ${y}'{\left({x}_{{0}}\right)}={y}_{{0}}'$, ${y}''{\left({x}_{{0}}\right)}={y}_{{0}}''$, ..., ${{y}}^{{{\left({n}-{1}\right)}}}{\left({x}_{{0}}\right)}={{y}_{{0}}^{{{n}-{1}}}}$. If ${g{{\left({x}\right)}}}$ and ${b}_{{j}}{\left({x}\right)},j={\overline{{{0}..{n}}}}$ are continuous in some interval ${I}$ containing ${x}_{{0}}$ and if ${b}_{{n}}{\left({x}\right)}\ne{0}$ in ${I}$, the given differential equation together with the initial conditions has a unique (only one) solution defined throughout ${I}$.

When the conditions on ${b}_{{n}}{\left({x}\right)}$ in the theorem hold, we can divide the differential equation by ${b}_{{n}}{\left({x}\right)}$ to obtain ${{y}}^{{{\left({n}\right)}}}+{a}_{{{n}-{1}}}{\left({x}\right)}{{y}}^{{{\left({n}-{1}\right)}}}+\ldots+{a}_{{2}}{\left({x}\right)}{y}''+{a}_{{1}}{\left({x}\right)}{y}'+{a}_{{0}}{\left({x}\right)}{y}=\phi{\left({x}\right)}$, where ${a}_{{j}}{\left({x}\right)}=\frac{{{a}_{{j}}{\left({x}\right)}}}{{{b}_{{n}}{\left({x}\right)}}},j={\overline{{{0}..{n}-{1}}}}$ and $\phi{\left({x}\right)}=\frac{{{g{{\left({x}\right)}}}}}{{\phi{\left({x}\right)}}}$.

Now, let's define the differential operator ${L}{\left({y}\right)}$ by ${L}{\left({y}\right)}={{y}}^{{{\left({n}\right)}}}+{a}_{{{n}-{1}}}{\left({x}\right)}{{y}}^{{{\left({n}-{1}\right)}}}+\ldots+{a}_{{2}}{\left({x}\right)}{y}''+{a}_{{1}}{\left({x}\right)}{y}'+{a}_{{0}}{\left({x}\right)}{y}$; then, the differential equation can be rewritten as ${L}{\left({y}\right)}=\phi{\left({x}\right)}$, and, in particular, the linear homogeneous differential equation can be expressed as ${L}{\left({y}\right)}={0}$.

The differential operator has two properties:

If ${L}{\left({y}\right)}={0}$, we have that ${L}{\left({c}{y}\right)}={0}$ for any constant ${c}$.

If ${L}{\left({y}_{{1}}\right)}={0}$ and ${L}{\left({y}_{{2}}\right)}={0}$, we have that ${L}{\left({y}_{{1}}+{y}_{{2}}\right)}={0}$.

These two properties can be combined into one property: if ${L}{\left({y}_{{1}}\right)}={0}$, ${L}{\left({y}_{{2}}\right)}={0}$, ..., ${L}{\left({y}_{{n}}\right)}={0}$, we have that ${L}{\left({c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}+\ldots+{c}_{{n}}{y}_{{n}}\right)}={0}$ for any constants ${c}_{{i}},i={\overline{{{1}..{n}}}}$.

What does it give us? It gives us the following fact: if we have ${n}$ solutions ${y}_{{1}}$, ${y}_{{2}}$, ..., ${y}_{{n}}$ that satisfy the given homogeneous differential equation, their linear combination will also satisfy this homogeneous differential equation. The only question is what form ${y}_{{1}}$, ${y}_{{2}}$, ..., ${y}_{{n}}$ should have for their linear combination to provide the general solution of the homogeneous differential equation.

Example. ${y}_{{1}}={\cos{{\left({t}\right)}}}$ and ${y}_{{2}}={\sin{{\left({t}\right)}}}$ are solutions of the differential equation ${y}''+{y}={0}$. So, ${y}_{{{g}}}={c}_{{1}}{y}_{{1}}+{c}_{{2}}{y}_{{2}}={c}_{{1}}{\cos{{\left({t}\right)}}}+{c}_{{2}}{\sin{{\left({t}\right)}}}$ is also the solution for any constants ${c}_{{1}}$ and ${c}_{{2}}$.