Inmathematics, theexponential response formula (ERF), also known asexponential response and complex replacement, is a method used to find a particular solution of anon-homogeneous linear ordinary differential equation of any order.^[1][2] The exponential response formula is applicable to non-homogeneous linear ordinary differential equations with constant coefficients if the function ispolynomial,sinusoidal,exponential or the combination of the three.^[2] The general solution of a non-homogeneous linearordinary differential equation is a superposition of the general solution of the associated homogeneous ODE and a particular solution to the non-homogeneous ODE.^[1] Alternative methods for solving ordinary differential equations of higher order aremethod of undetermined coefficients and method ofvariation of parameters.

The ERF method of finding a particular solution of a non-homogeneous differential equation is applicable if the non-homogeneous equation is or could be transformed to form${\displaystyle f(t)=B_1{e}^{{\gamma }_{1}t}+B_2{e}^{{\gamma }_{2}t}+\cdots +B_n{e}^{{\gamma }_{n}t}}$; where${\displaystyle B,\gamma }$ arereal orcomplex numbers and${\displaystyle f(t)}$ is homogeneous linear differential equation of any order. Then, the exponential response formula can be applied to each term of the right side of such equation. Due to linearity, the exponential response formula can be applied as long as the right side has terms, which are added together by thesuperposition principle.

Complex replacement is a method of converting a non-homogeneous term of equation into a complex exponential function, which makes a given differential equation a complex exponential.

Consider differential equation${\displaystyle y^{″}+y=\cos (t)}$.

To make complex replacement,Euler's formula can be used;

Therefore, given differential equation changes to${\displaystyle z^{″}+z=e^{it}}$. The solution of the complex differential equation can be found as${\displaystyle z(t)}$, from which the real part is the solution of the original equation.

Complex replacement is used for solving differential equations when the non-homogeneous term is expressed in terms of a sinusoidal function or an exponential function, which can be converted into a complex exponential function differentiation and integration. Such complex exponential function is easier to manipulate than the original function.

When the non-homogeneous term is expressed as an exponential function, the ERF method or theundetermined coefficients method can be used to find aparticular solution. If non-homogeneous terms can not be transformed to complex exponential function, then the Lagrange method ofvariation of parameters can be used to find solutions.

Thedifferential equations are important in simulating natural phenomena. In particular, there are numerous phenomena described ashigh order linear differential equations, for example the spring vibration,LRC circuit,beam deflection,signal processing,control theory andLTI systems with feedback loops.^[1][3]

Mathematically, the system istime-invariant if whenever the input${\displaystyle f(t)}$ has response${\displaystyle x(t)}$ then for any constant "a", the input${\displaystyle f(t-a)}$ has response${\displaystyle x(t-a)}$. Physically, time invariance means system’s response does not depend on what time the input begins. For example, if a spring-mass system is atequilibrium, it will respond to a given force in the same way, no matter when the force was applied.

When the time-invariant system is also linear, it is called a linear time-invariant system (LTI system). Most of these LTI systems are derived from linear differential equations, where the non-homogeneous term is called the input signal and solution of the non-homogeneous equations is called the response signal. If the input signal is given exponentially, the corresponding response signal also changes exponentially.

Considering the following${\displaystyle n}$th order linear differential equation

${\displaystyle a_n\frac{d^{n}y}{d{t}^n}+a_{n-1}\frac{d^{n-1}y}{d{t}^{n-1}}+\cdots +a_1\frac{dy}{dt}+a_{0}y=f(t)(1)}$

and denoting

${\displaystyle L=a_n{D}^n+a_{n-1}{D}^{n-1}+\cdots +a_1{D}^1+a_{0}I,}$

${\displaystyle D^k=\frac{d^k}{d{t}^k}(k=1,2,\dots ,n),}$

where${\displaystyle a_0,\dots ,a_n}$ are the constant coefficients, produces differential operator${\displaystyle L}$, which is linear and time-invariant and known as theLTI operator. The operator,${\displaystyle L}$ is obtained from itscharacteristic polynomial;

${\displaystyle P(s)=a_n{s}^n+a_{n-1}{s}^{n-1}+\cdots +a_0}$

by formally replacing the indeterminate s here with thedifferentiation operator${\displaystyle D}$

${\displaystyle L=P(D)}$

${\displaystyle P(D)=a_n{D}^n+a_{n-1}{D}^{n-1}+\cdots +a_{0}I}$

Therefore, the equation (1) can be written as

${\displaystyle P(D)y=f(t)(2)}$

Considering LTI differential equation above, with exponential input${\displaystyle f(t)=B{e}^{\gamma t}}$, where${\displaystyle B}$ and${\displaystyle \gamma }$ are given numbers. Then, a particular solution is

provide only that${\displaystyle P(\gamma )\ne 0}$.

Proof: Due tolinearity of operator${\displaystyle P(D)}$, the equation can be written as

${\displaystyle P(D)(y_p)=P(D)\left(\frac{B{e}^{\gamma t}}{P(\gamma )}\right)=\frac{B}{P(\gamma )}P(D)(e^{\gamma t})(3)}$

On the other hand, since

${\displaystyle P(D)\left(e^{\gamma t}\right)=P(\gamma )e^{\gamma t},}$

substituting this into equation (3), produces

${\displaystyle P(D)(y_p)=P(D)\left(\frac{B{e}^{\gamma t}}{P(\gamma )}\right)=\frac{B}{P(\gamma )}P(D)\left(e^{\gamma t}\right)=\frac{B}{P(\gamma )}P(\gamma )e^{\gamma t}=B{e}^{\gamma t}.}$

Therefore,${\displaystyle y_p}$ is a particular solution to the non-homogeneous differential equation.

Thus, the above equation for a particular response${\displaystyle y_p}$ is called the exponential response formula (ERF) for the given exponential input.

In particular, in case of${\displaystyle P(\gamma )=0}$, a solution to equation (2) is given by

${\displaystyle y_p=\frac{Bt{e}^{\gamma t}}{P^{\prime }(\gamma )},P^{\prime }(\gamma )\ne 0}$

and is called theresonant response formula.

Let's find the particular solution to 2nd order linear non-homogeneous ODE;

${\displaystyle 2x^{″}+x^{\prime }+x=1+2e^t+e^{-t}\cos (t).}$

The characteristic polynomial is${\displaystyle P(s)=2s^2+s+1}$. Also, the non-homogeneous term,${\displaystyle f(t)=1+2e^t+e^{-t}\cos (t)}$ can be written as follows

${\displaystyle f(t)=f_1(t)+f_2(t)+f_3(t),f_1(t)=1,f_2(t)=2e^t,f_3(t)=e^{-t}\cos (t).}$

Then, the particular solutions corresponding to${\displaystyle f_1(t),f_2(t)}$ and${\displaystyle f_3(t)}$, are found, respectively.

First, considering non-homogeneous term,${\displaystyle f_1(t)=1}$. In this case, since${\displaystyle f_1(t)=1=e^{0\cdot t},\gamma =0}$ and${\displaystyle P(\gamma )=P(0)=1\ne 0}$.

from the ERF, a particular solution corresponding to${\displaystyle f_1(t)}$ can be found.

${\displaystyle x_{1p}=\frac{f_1(t)}{P(0)}=\frac{1}{1}=1}$.

Similarly, a particular solution can be found corresponding to${\displaystyle f_2(t)}$.

${\displaystyle x_{2p}=\frac{f_2(t)}{P(1)}=\frac{2e^t}{4}=\frac{e^t}{2}.}$

Let's find a particular solution to DE corresponding to 3rd term;

${\displaystyle 2x^{″}+x^{\prime }+x=e^{-t}\cos (t).}$

In order to do this, equation must be replaced by complex-valued equation, of which it is the real part:

${\displaystyle 2z^{″}+z^{\prime }+z=e^{(-1+i)t}.}$

Applying the exponential response formula (ERF), produces

and the real part is

${\displaystyle x_{3p}=-\frac{1}{3}{e}^{-t}\sin (t).}$

Therefore, the particular solution of given equation,${\displaystyle x_p}$ is

${\displaystyle x_p=x_{1p}+x_{2p}+x_{3p}=1+\frac{e^t}{2}-\frac{1}{3}{e}^{-t}\sin (t).}$

Theundetermined coefficients method is a method of appropriately selecting a solution type according to the form of the non-homogeneous term and determining the undetermined constant, so that it satisfies the non-homogeneous equation.^[4] On the other hand, the ERF method obtains a special solution based on differential operator.^[2] Similarity for both methods is that special solutions of non-homogeneous linear differential equations with constant coefficients are obtained, while form of the equation in consideration is the same in both methods.

For example, finding a particular solution of${\displaystyle y^{″}+y=e^t}$ with the method of undetermined coefficients requires solving the characteristic equation${\displaystyle {\lambda }^2+1=0,\lambda =\pm i}$. The non-homogeneous term${\displaystyle f(t)=B{e}^{\gamma t},B=1,\gamma =1}$ is then considered and since${\displaystyle \gamma =1}$ is not acharacteristic root, it puts a particular solution in form of${\displaystyle y_p(t)=A{e}^{\gamma t}}$, where${\displaystyle A}$ is undetermined constant. Substituting into the equation to determine the tentative constant yields

${\displaystyle {\lambda }^{2}A{e}^{\lambda t}+A{e}^{\lambda t}=e^{\lambda t}}$

therefore

${\displaystyle A=\frac{1}{{\lambda }^2+1}.}$

The particular solution can be found in form:^[5]

${\displaystyle y_p(t)=A{e}^{\lambda t}=\frac{e^{\lambda t}}{{\lambda }^2+1}.}$

On the other hand, the exponential response formula method requires characteristic polynomial${\displaystyle P(s)=s^2+1}$ to be found, after which the non-homogeneous terms${\displaystyle f(t)=B{e}^{\gamma t},B=1,\gamma =1}$ is complex replaced. The particular solution is then found using formula

${\displaystyle y_p(t)=\frac{e^{\lambda t}}{P(\lambda )}=\frac{e^{\lambda t}}{{\lambda }^2+1}.}$

The exponential response formula method was discussed in case of${\displaystyle P(\gamma )\ne 0}$. In the case of${\displaystyle P(\gamma )=0,P^{\prime }(\gamma )\ne 0}$, theresonant response formula is also considered.

In the case of${\displaystyle P(\gamma )=P^{\prime }(\gamma )=\cdots =P^{(k-1)}(\gamma )=0,P^k(\gamma )\ne 0}$, we will discuss how the ERF method will be described in this section.

Let${\displaystyle P(D)}$ be a polynomial operator with constant coefficients, and${\displaystyle P^{(m)}}$ its${\displaystyle m}$-th derivative. Then ODE

${\displaystyle P(D)y=B{e}^{\gamma t}}$, where${\displaystyle \gamma }$ is real or complex.

has the particular solution as following.

• ${\displaystyle P(\gamma )\ne 0}$. In this case, a particular solution will be given by${\displaystyle y_p(t)=\frac{B{e}^{\gamma t}}{P(\gamma )}}$.(exponent response formula)

• ${\displaystyle P(\gamma )=0}$ but${\displaystyle P^{\prime }(\gamma )\ne 0}$. In this case, a particular solution will be given by${\displaystyle y_p(t)=\frac{Bt{e}^{\gamma t}}{P^{\prime }(\gamma )}}$.(resonant response formula)

• ${\displaystyle P(\gamma )=P^{\prime }(\gamma )=\cdots =P^{(k-1)}(\gamma )=0}$ but${\displaystyle P^k(\gamma )\ne 0}$. In this case, a particular solution will be given by

Above equation is calledgeneralized exponential response formula.

To find a particular solution of the following ODE;

${\displaystyle y^{‴}-3y^{\prime }+2y=6e^t.}$

the characteristic polynomial is${\displaystyle P(s)=s^3-3s+2}$.

By the calculating, we get the following:

${\displaystyle P(1)=0,P^{\prime }(1)=0,P^{″}(1)=6\ne 0.}$

Original exponential response formula is not applicable to this case due to division by zero. Therefore, using the generalized exponential response formula and calculated constants, particular solution is

${\displaystyle y_p(t)=\frac{6t^2{e}^t}{P^{″}(1)}=\frac{6t^2{e}^t}{6}=t^2{e}^t.}$

Object hanging from a spring with displacement${\displaystyle d}$. The force acting is gravity, spring force, air resistance, and any other external forces.

FromHooke’s law, the motion equation of object is expressed as follows;^[6][4]

${\displaystyle m\frac{d^{2}x}{d{t}^2}+r\frac{dx}{dt}+kx=F(t),}$

where${\displaystyle F(t)}$ is external force.

Now, assumingdrag is neglected and${\displaystyle F(t)=F_0\cos (\omega t)}$, where${\displaystyle \omega =\sqrt{\frac{k}{m}}}$ (the external force frequency coincides with the natural frequency). Therefore, theharmonic oscillator with sinusoidal forcing term is expressed as following:

${\displaystyle m\frac{d^{2}x}{d{t}^2}+kx=F(t).}$

Then, a particular solution is

${\displaystyle x_p=\frac{F_0}{2\sqrt{km}}t\sin (\omega t).}$

Applying complex replacement and the ERF: if${\displaystyle z_p}$ is a solution to the complex DE

${\displaystyle m\frac{d^{2}z}{d{t}^2}+kz=F_0{e}^{i\omega t},}$

then${\displaystyle x_p=\mathrm{Re}(z_p)}$ will be a solution to the given DE.

The characteristic polynomial is${\displaystyle P(s)=m{s}^2+k}$, and${\displaystyle \gamma =i\omega }$, so that${\displaystyle P(\gamma )=0}$. However, since${\displaystyle P^{\prime }(s)=2ms}$, then${\displaystyle P^{\prime }(\gamma )=P^{\prime }(i\omega )=2m\omega i\ne 0}$. Thus, the resonant case of the ERF gives

Considering the electric current flowing through an electric circuit, consisting of a resistance (${\displaystyle R}$), a capacitor (${\displaystyle C}$), a coil wires (${\displaystyle L}$), and a battery (${\displaystyle E}$), connected in series.^[3][6]

This system is described by an integral-differential equation found by Kirchhoff calledKirchhoff’s voltage law, relating the resistor${\displaystyle R}$, capacitor${\displaystyle C}$, inductor${\displaystyle L}$, battery${\displaystyle E}$, and the current${\displaystyle I}$ in a circuit as follows,

${\displaystyle L{I}^{\prime }(t)+RI(t)+\frac{1}{C}{\int }_{t_0}^{t}I(s)ds={\int }_{t_0}^{t}E(s)ds}$

Differentiating both sides of the above equation, produces the following ODE.

${\displaystyle L{I}^{″}(t)+R{I}^{\prime }(t)+\frac{1}{C}I(t)=E(t)}$

Now, assuming${\displaystyle E(t)=E_0\sin ({\omega }_{0}t)}$, where${\displaystyle {\omega }_0=\sqrt{\frac{1}{LC}}}$. (${\displaystyle {\omega }_0}$ is calledresonance frequency inLRC circuit). Under above assumption, the output (particular solution) corresponding to input${\displaystyle E(t)}$ can be found. In order to do it, given input can be converted in complex form:

${\displaystyle E(t)=E_0\sin ({\omega }_{0}t)=\mathrm{Im}(E_0{e}^{i{\omega }_{0}t})}$

The characteristic polynomial is${\displaystyle P(s)=L{s}^2+Rs+\frac{1}{s}}$, where${\displaystyle P(i{\omega }_0)=i{\omega }_{0}R\ne 0}$. Therefore, from the ERF, a particular solution can be obtained as follows;

Considering the general LTI system

${\displaystyle P(D)x=Q(D)f(t)}$

where${\displaystyle f(t)}$ is the input and${\displaystyle P(D),Q(D)}$ are given polynomial operators, while assuming that${\displaystyle P(s)\ne 0}$.In case that${\displaystyle f(r)=F_0\cos (\omega t)}$, a particular solution to given equation is

${\displaystyle x_p(t)=\mathrm{Re}\left(F_0\frac{Q(i\omega )}{P(i\omega )}{e}^{i\omega t}\right).}$

Considering the following concepts used in physics and signal processing mainly.

• The amplitude of the input is${\displaystyle F_0}$. This has the same units as the input quantity.
• The angular frequency of the input is${\displaystyle \omega }$. It has units of radians/time. Often it will be referred to it as frequency, even though technically frequency should have units of cycles/time.
• The amplitude of the response is${\displaystyle A=F_0|Q(i\omega )/P(i\omega )|}$. This has the same units as the response quantity.
• The gain is${\displaystyle g(\omega )=|Q(i\omega )/P(i\omega )|}$. The gain is the factor that the input amplitude is multiplied by to get the amplitude of the response. It has the units needed to convert input units to output units.
• The phase lag is${\displaystyle \varphi =-\mathrm{Arg}(Q(i\omega )/P(i\omega ))}$. The phase lag has units of radians, i.e. it’s dimensionless.
• The time lag is${\displaystyle \varphi /\omega }$. This has units of time. It is the time that peak of the output lags behind that of the input.
• The complex gain is${\displaystyle Q(i\omega )/P(i\omega )}$. This is the factor that the complex input is multiplied by to get the complex output.

• Operators and the exponential response formula
• Generalized exponential response formula

• Ordinary differential equations

• CS1: long volume value