Ordinary linear differential equations with constant coefficients are extensively used in solving engineering problems. This experiment helps in building a clear understanding of the mathematical concepts of linear algebra used in the standard method for finding solutions of these equations. This experiment is not focused on learning the standard method of solution.

1. Notation:

(i) F: collection of functions from R to R differentiable twice
(ii) D: d/dx
(iii) D²: d/dx(d/dx), i.e. d²/dx²

2. Second order linear differential equation:

The equation
y’’+Py’+Qy=f(x) …(i)
is a second order linear differential equation with constant coefficients, where P, Q∈R and f:R→R is a continuous function. If f(x)=0, then this equation is called a homogeneous second order linear differential equation with constant coefficients. It is well known that linear ordinary differential equations with constant coefficients are used in solving engineering problems.

3. Standard method of solution:

To find solutions of equation (i), consider the equation am²+bm+c=0, which is called the auxiliary equation. Then y(x) = c₁e^m₁^x+c₂e^m₂^x, where c₁, c₂ are constants and m₁, m₂ are the roots of the auxiliary equation is known as complementary function (C.F.) and is denoted by y_c. A specific function g(x) which satisfies equation (i) is called the particular integral (P.I.) and is denoted by y_p. Any solution of equation (i) is of the form y(x)≡y_c+y_p=c₁e^m₁^x+c₂e^m₂^x+g(x). This is called the complete integral (C.I.).

In particular, if f(x)=0, then the complementary function gives all solutions of equation (i), so that C.I.=C.F, that is solutions are given by y(x) = c₁e^m₁^x+c₂e^m₂^x

Note. There are typical approaches for calculating particular integral, but here we will consider hit and trial approach only.

4. Examples and explanation matrix:
4.1. Examples:

(i) Let y’’-y=0 be a homogeneous second order linear differential equation. Then y’’=D²(y), which implies that (D²-I)y=0. Clearly auxiliary equation is m²-1=0, thus m=1, -1. Then the complementary function and hence the complete integral of this equation is c₁e^x+c₂e^-x, where c₁, c₂ are constants.
(ii) Let y’’-y=2sinx be the second order linear differential equation. Let y’’=D²(y). Then (D²-1)y=2sinx, y=-sinx satisfies this equation. It can be seen that -sinx is a particular integral of this equation. Thus complete integral is C₁e^x+C₂e^-x -sinx, where c₁, c₂ are constants.

4.2 Summary:

Examples given above are summarized in the explanation matrix given below.

5. Understanding:
5.1. Collection of differentiable functions as a vector space:

Collection F of twice differentiable functions from R to R is a vector space over R with addition and scalar multiplication of functions.

5.2. Derivative as an operator:

Notion of derivative is well known. It operates on functions. Thus d/dx is a function from the collection of twice differentiable functions to collection of functions. It is denoted by D and is called the differential operator.

5.3. Operator defined by differential equation:

Let y’’+py’+qy=f(x), where p, q∈R be a given linear differential equation of second order. Then (D²+pD+qI)(y)=r, where D²(y)= y’’, D(y)=y’ and I is the identity operator. Thus the L.H.S. of the given equation provides the operator D²+pD+qI which is called the operator defined by the given differential equation.

5.4. Linear map equation defined by a given differential equation:

Let y’’+py’+qy=f(x), where p, q∈R be a given linear differential equation of second order. Then this can be written as T(y)=r, where T:F→F is the operator D²+pD+qI defined in Section 5.3. This is the desired linear map equation obtained from the given differential equation.

5.5. Solution of differential equation:

Recall from Experiment 7, that the solution set of the linear map equation T(y)=r described in Section 5.4, is y_o+kerT, where y_o is its particular solution and kerT is the collection of all solutions of T(y)=0. Notice that the linear map equation is just another form of the given differential equation. Therefore its solutions are same as the solutions of the given differential equation. As given in Section 3, a solution of linear differential equation is given by y_c+y_p, where y_c is a solution of the homogenous counterpart of the given differential equation and y_p is a particular solution of the given differential equation. Here it may be noted that y_p corresponds to y_o and y_c corresponds to an element of kerT.

5.6. Example:

Let y’’-5y’+6y=2e^x be the given second order linear differential equation.
Standard method for solution:
(a) The auxiliary equation is m²-5m+6=0, i.e. (m-2)(m-3)=0. Thus m=2, 3. Then the complementary function of the given differential equation is c₁e^3x+c₂e^2x, wherec₁, c₂ are constants.
(b) Clearly y=e^x satisfies the given differential equation. Thus e^x is a particular integral.
(c) Thus complete integral y(x)=c₁e^3x+c₂e^2x+e^x, where c₁, c₂ are constants.
Verification:
Substituting the value of y(x) obtained above, in the given differential equation, L.H.S. comes out to be equal to 2e^x which is the R.H.S.
Understanding of complete integral:
The linear map arising out of the given differential equation is T≡D²-5D+6I and the corresponding linear map equation is T(y)=2e^x. The complete integral obtained above is the solution set of this linear map equation, as solutions of the differential equation and induced linear map equation are same. This provides an illustration of the understanding provided in Section 5, i.e understanding as to why the standard method as given in Section 3 provides all solutions of the given differential equation.

6. Example (Engineering problem):

In an RLC circuit, resistance R=40 ohm, inductance L=2/5 henry and capacitance C=10⁻² farad. At t=0, charge Q= 0 and current I = 2 ampere. It is assumed here that voltage E(t)=0, for t>0. Aim is to determine the current I≡dQ/dt, where Q is the charge flowing in the circuit at t>0.

The given problem is mathematically represented by an initial value problem: the differential equation for the charge Q across the circuit using Kirchhoff’s law is 2/5Q′′+40Q′+100Q=E(t) and the initial conditions are Q(0)=1, Q′(0)=2. To find its solution and the required current, we proceed as follows.
The given differential equation reduces to 2Q′′+200Q′+500Q=0. The auxiliary equation of the given differential equation is 2m²+200m+500=0. Thus m=(−50+15 $\sqrt{10}$​ ) or (−50-15 $\sqrt{10}$ ). Therefore the complete integral of the given differential equation is

Q=c₁e $^{(−50+15 \sqrt{10})<i>t</i>}$​ +c₂e $^{(−50-15 \sqrt{10}​)<i>t</i>},$ … (i)
where c₁, c₂ are constants. Now differentiating equation (i) we have
Q′=(−50+15 $\sqrt{10}$​ )c₁e $^{(−50+15 \sqrt{10})<i>t</i>​}$+(−50-15 $\sqrt{10}$​ t)c₂e$^{(−50-15 \sqrt{10})<i>t</i>}$​ .
Using given initial conditions, we obtain c₁+c₂=Q(0)=1 and (−50+15 $\sqrt{10}$​ )c₁+(−50-15 $\sqrt{10}$​ )c₂=Q′(0)=2. Therefore c₂=0.285 and c₁=0.715. Thus
Q=0.715e$^{(−50+15 \sqrt{10})<i>t</i> }$ +0.285e$^{(−50-15 \sqrt{10}​)<i>t</i> }$ and the required current
I≡dQ/dt=(−50+15 $\sqrt{10}$​ ).(0.715)e $^{(−50+15 \sqrt{10})<i>t</i>}$ +(−50-15$\sqrt{10}$).(0.285) e$^{(−50-15 \sqrt{10})<i>t</i> }.$​