characteristic equation (calculus)

Starting with a linear homogeneous differential equation with constant coefficients {{math|a_n, a_n − 1, ..., a₁, a₀}},

:$a\_n\; y^\{(n)\}\; +\; a\_\{n-1\}y^\{(n-1)\}\; +\; \backslash cdots\; +\; a\_1\; y^\backslash prime\; +\; a\_0\; y\; =\; 0,$

it can be seen that if {{math|y(x) {{=}} e^rx}}, each term would be a constant multiple of {{math|e^rx}}. This results from the fact that the derivative of the exponential function {{math|e^rx}} is a multiple of itself. Therefore, {{math|y′ {{=}} re^rx}}, {{math|1=y″ = r²e^rx}}, and {{math|1=y⁽ⁿ⁾ = rⁿe^rx}} are all multiples. This suggests that certain values of {{mvar|r}} will allow multiples of {{math|e^rx}} to sum to zero, thus solving the homogeneous differential equation. In order to solve for {{mvar|r}}, one can substitute {{math|1=y = e^rx}} and its derivatives into the differential equation to get

:$a\_n\; r^n\; e^\{rx\}\; +\; a\_\{n-1\}r^\{n-1\}e^\{rx\}\; +\; \backslash cdots\; +\; a\_1\; re^\{rx\}\; +\; a\_0\; e^\{rx\}\; =\; 0$

Since {{math|e^rx}} can never equal zero, it can be divided out, giving the characteristic equation

:$a\_n\; r^n\; +\; a\_\{n-1\}r^\{n-1\}\; +\; \backslash cdots\; +\; a\_1\; r\; +\; a\_0\; =\; 0.$

By solving for the roots, {{mvar|r}}, in this characteristic equation, one can find the general solution to the differential equation. For example, if {{mvar|r}} has roots equal to 3, 11, and 40, then the general solution will be $y(x)\; =\; c\_1\; e^\{3\; x\}\; +\; c\_2\; e^\{11\; x\}\; +\; c\_3\; e^\{40\; x\}$, where $c\_1$, $c\_2$, and $c\_3$ are arbitrary constants which need to be determined by the boundary and/or initial conditions.

Solving the characteristic equation for its roots, {{math|r₁, ..., r_n}}, allows one to find the general solution of the differential equation. The roots may be real or complex, as well as distinct or repeated. If a characteristic equation has parts with distinct real roots, {{mvar|h}} repeated roots, or {{mvar|k}} complex roots corresponding to general solutions of {{math|y_D(x)}}, {{math|y_R1(x), ..., y_Rh(x)}}, and {{math|y_C1(x), ..., y_Ck(x)}}, respectively, then the general solution to the differential equation is

:$y(x)\; =\; y\_\backslash mathrm\{D\}(x)\; +\; y\_\{\backslash mathrm\{R\}\_1\}(x)\; +\; \backslash cdots\; +\; y\_\{\backslash mathrm\{R\}\_h\}(x)\; +\; y\_\{\backslash mathrm\{C\}\_1\}(x)\; +\; \backslash cdots\; +\; y\_\{\backslash mathrm\{C\}\_k\}(x)$

The linear homogeneous differential equation with constant coefficients

:$y^\{(5)\}\; +\; y^\{(4)\}\; -\; 4y^\{(3)\}\; -\; 16y\text{'}\text{'}\; -20y\text{'}\; -\; 12y\; =\; 0$

has the characteristic equation

:$r^5\; +\; r^4\; -\; 4r^3\; -\; 16r^2\; -20r\; -\; 12\; =\; 0$

By factoring the characteristic equation into{{explain|reason=How is this factorization found?|date=August 2024}}

:$(r\; -\; 3)(r^2\; +\; 2r\; +\; 2)^2\; =\; 0$

one can see that the solutions for {{mvar|r}} are the distinct single root {{math|r₁ {{=}} 3}} and the double complex roots {{math|r_2,3,4,5 {{=}} 1 ± i}}. This corresponds to the real-valued general solution

:$y(x)\; =\; c\_1\; e^\{3x\}\; +\; e^x(c\_2\; \backslash cos\; x\; +\; c\_3\; \backslash sin\; x)\; +\; xe^x(c\_4\; \backslash cos\; x\; +\; c\_5\; \backslash sin\; x)$

with constants {{math|c₁, ..., c₅}}.

The superposition principle for linear homogeneous says that if {{math|u₁, ..., u_n}} are {{mvar|n}} linearly independent solutions to a particular differential equation, then {{math|c₁u₁ + ⋯ + c_nu_n}} is also a solution for all values {{math|c₁, ..., c_n}}.{{cite web|url=http://tutorial.math.lamar.edu/Classes/DE/PDETerminology.aspx|title=Differential Equation Terminology|last=Dawkins|first=Paul|work=Paul's Online Math Notes|accessdate=2 March 2011}} Therefore, if the characteristic equation has distinct real roots {{math|r₁, ..., r_n}}, then a general solution will be of the form

:$y\_\backslash mathrm\{D\}(x)\; =\; c\_1\; e^\{r\_1\; x\}\; +\; c\_2\; e^\{r\_2\; x\}\; +\; \backslash cdots\; +\; c\_n\; e^\{r\_n\; x\}$

If the characteristic equation has a root {{math|r₁}} that is repeated {{mvar|k}} times, then it is clear that {{math|1=y_p(x) = c₁e^r₁x}} is at least one solution. However, this solution lacks linearly independent solutions from the other {{math|k − 1}} roots. Since {{math|r₁}} has multiplicity {{mvar|k}}, the differential equation can be factored into

:$\backslash left\; (\; \backslash frac\{d\}\{dx\}\; -\; r\_1\; \backslash right\; )^k\; y\; =\; 0\; .$

The fact that {{math|1=y_p(x) = c₁e^r₁x}} is one solution allows one to presume that the general solution may be of the form {{math|1=y(x) = u(x)e^r₁x}}, where {{math|u(x)}} is a function to be determined. Substituting {{math|ue^r₁x}} gives

:$\backslash left(\; \backslash frac\{d\}\{dx\}\; -\; r\_1\; \backslash right)\backslash !\; ue^\{r\_1\; x\}\; =\; \backslash frac\{d\}\{dx\}\backslash left(ue^\{r\_1\; x\}\backslash right)\; -\; r\_1\; ue^\{r\_1\; x\}\; =\; \backslash frac\{d\}\{dx\}(u)e^\{r\_1\; x\}\; +\; r\_1\; ue^\{r\_1\; x\}-\; r\_1\; ue^\{r\_1\; x\}\; =\; \backslash frac\{d\}\{dx\}(u)e^\{r\_1\; x\}$

when {{math|1=k = 1}}. By applying this fact {{mvar|k}} times, it follows that

:$\backslash left(\; \backslash frac\{d\}\{dx\}\; -\; r\_1\; \backslash right)^k\; ue^\{r\_1\; x\}\; =\; \backslash frac\{d^k\}\{dx^k\}(u)e^\{r\_1\; x\}\; =\; 0.$

By dividing out {{math|e^r₁x}}, it can be seen that

:$\backslash frac\{d^k\}\{dx^k\}(u)\; =\; u^\{(k)\}\; =\; 0.$

Therefore, the general case for {{math|u(x)}} is a polynomial of degree {{math|k − 1}}, so that {{math|1=u(x) = c₁ + c₂x + c₃x² + ⋯ + c_kx^k −1}}. Since {{math|1=y(x) = ue^r₁x}}, the part of the general solution corresponding to {{math|r₁}} is

:$y\_\backslash mathrm\{R\}(x)\; =\; e^\{r\_1\; x\}\backslash !\backslash left(c\_1\; +\; c\_2\; x\; +\; \backslash cdots\; +\; c\_k\; x^\{k-1\}\backslash right).$

If a second-order differential equation has a characteristic equation with complex conjugate roots of the form {{math|r₁ {{=}} a + bi}} and {{math|r₂ {{=}} a − bi}}, then the general solution is accordingly {{math|y(x) {{=}} c₁e^{(a + bi )x} + c₂e^{(a − bi )x}}}. By Euler's formula, which states that {{math|e^iθ {{=}} cos θ + i sin θ}}, this solution can be rewritten as follows:

:$\backslash begin\{align\}$

y(x) &= c_{1}e^{(a + bi)x} + c_{2}e^{(a - bi)x}\\

&= c_{1}e^{ax}(\cos bx + i \sin bx) + c_{2}e^{ax}( \cos bx - i \sin bx ) \\

&= \left(c_{1} + c_{2}\right)e^{ax} \cos bx + i(c_{1} - c_{2})e^{ax} \sin bx

\end{align}

where {{math|c₁}} and {{math|c₂}} are constants that can be non-real and which depend on the initial conditions. (Indeed, since {{math|y(x)}} is real, {{math|c₁ − c₂}} must be imaginary or zero and {{math|c₁ + c₂}} must be real, in order for both terms after the last equals sign to be real.)

For example, if {{math|c₁ {{=}} c₂ {{=}} {{sfrac|1|2}}}}, then the particular solution {{math|y₁(x) {{=}} e^ax cos bx}} is formed. Similarly, if {{math|c₁ {{=}} {{sfrac|1|2i}}}} and {{math|c₂ {{=}} −{{sfrac|1|2i}}}}, then the independent solution formed is {{math|y₂(x) {{=}} e^ax sin bx}}. Thus by the superposition principle for linear homogeneous differential equations, a second-order differential equation having complex roots {{math|r {{=}} a ± bi}} will result in the following general solution:

:$y\_\backslash mathrm\{C\}(x)\; =\; e^\{ax\}(C\_1\; \backslash cos\; bx\; +\; C\_2\; \backslash sin\; bx)$

This analysis also applies to the parts of the solutions of a higher-order differential equation whose characteristic equation involves non-real complex conjugate roots.

• Characteristic polynomial

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Category:Ordinary differential equations