3
Linear Differential Equations with Constant Coefficients
3.1 INTRODUCTION
In the previous chapter, we have studied the linear differential equation of the first order
It is called a homogeneous (or reduced) equation if Q(x) ≡ 0 and a nonhomogeneous equation if Q(x) ≢ 0.
If we substitute y = u + v in Eq. (3.1), we have
which is satisfied if
This shows that y = u + v is the general solution (or complete solution) of Eq. (3.1) if u and v satisfy Eqs. (3.2) and (3.3), respectively, i.e., if
The function u(x) which contains an arbitrary constant c is called the complementary function (C.F.) and v(x) which contains no such arbitrary constant is called the particular integral (P. I.) of Eq. (3.1). There is one arbitrary constant in the C.F., because the order of the differential equation is 1.
Example 3.1.1 Solve
Solution Clearly u(x) = ce^{−x} satisfies the homogeneous equation derived from Eq. (3.6).
The general (complete) solution of Eq. (3.6) is
Thus, the general solution y of a linear equation comprises of two parts: y = y_{c} + y_{p} where y_{c}, which satisfies the homogeneous equation and contains as many arbitrary constants as the order of the differential equation, and y_{p}, which satisfies the nonhomogeneous equation and contains no arbitrary constants.
3.1.1 Linear Differential Equations of the Second Order
Second order linear differential equations have many applications in physics and engineering such as in motion of mechanical systems and fundamental electric circuits. They also lead to the study of higher functions such as Bessel's, Legendre's, and hypergeometric functions, which have wide applications.
A second order differential equation is called linear if it can be written in the standard form
where are the differential operators. When applied on a differentiable function, ‘D’ does the operation of differentiation and so it must always be written on the lefthand side of the function upon which it is operating.
 ‘D’ is a linear operator:
D(c_{1}y_{1} (x) + c_{2}y_{2} (x) = c_{1}Dy_{1} (x) + c_{2}Dy_{2} (x)
 ‘D’ satisfies index laws:
Note 3.1.2 Equation in operator notation, is
This gives a meaning to the operator D + P.
A differential equation that cannot be written in the form Eq. (3.7) is called nonlinear (see Examples 3.75 and 3.76)
If R(x) ≡ 0 then Eq. (3.7) becomes
and is called a homogeneous (or reduced) equation (illustrative Examples (2) (3) and (5) below) and if R(x) ≢ 0 then Eq. (3.7) is called a nonhomogeneous equation (Illustrative examples (1), (4) and (6) below.)
Here P, Q, R are continuous functions of x on some interval I, which is to be understood even if we do not mention it in each case, hereafter.
 y + 4y = e^{−x} sin x.
 (1 − x^{2})y″ − 2xy′ + 6y = 0.
 y″ + y′^{2}
 y″ − y = 0.
 y″ + y′ = tan x + sec^{2} x.
A solution of Eq. (3.7) is a twice differential function y on some open interval I, which satisfies Eq. (3.7) identically.
As in the case of the first order linear differential equation, the general solution of Eq. (3.7) consists of two parts y_{c} and y_{p}.
If u(x) and v(x) satisfy Eqs. (3.8) and (3.7), respectively,
then their sum y = u(x) + v(x) satisfies Eq. (3.7). u(x) is called the complementary function (C.F.) and v(x) is called the particular integral (P. I.) of Eq. (3.7).
Example 3.1.3 Solve
Solution Clearly e^{x} is a solution of the complete Eq. (3.11). Now consider the reduced equation
1 and e^{−x} is also a solution of Eq. (3.12). A linear combination of these solutions namely y = c_{1} + c_{2}e^{−x} is also a solution of Eq. (3.12). The general solution of Eq. (3.12) is
3.1.2 Homogeneous Equations—Superposition or Linearity Principle
We begin our discussion of solution of a second order linear equation with the homogeneous equation. Let us consider an example.
Example 3.1.4 Solve
Solution y_{1} = e^{x} and y_{2} = e^{−x} are solutions of the homogeneous linear differential equation for all x since
In fact, a linear combination of y_{1} and y_{2}, i.e., y = c_{1}y_{1} + c_{2}y_{2} is also a solution of Eq. (3.14) since
This is known as the superposition or linearity principle.
3.1.3 Fundamental Theorem for the Homogeneous Equation
Theorem 3.1.5 For a homogeneous linear differential Eq. (3.8), any linear combination of two solutions is again a solution.
Proof Let y_{1} and y_{2} be solutions of Eq. (3.8). Then substituting
Note 3.1.6 The above principle does not hold for nonhomogeneous equations and nonlinear equations.
Nonhomogeneous linear differential equation
Consider the nonhomogeneous differential equation
are solutions of Eq. (3.16) since
But their sum y_{1} + y_{2} = (e^{x} − 1) + (e^{−x} − 1) = e^{x} + e^{−x} − 2 is not a solution.
Nonlinear differential equation
y_{1} = x^{2} , y_{2} = 1 are solutions of the nonlinear differential equation
but (−1)y_{2} = −x^{2} and their sum y_{1} + y_{2} = x^{2} + 1 are not solutions of Eq. (3.17).
3.1.4 Initial Value Problem (IVP)
For a firstorder differential equation, a general solution contains one arbitrary constant c. If an initial condition y(x_{0}) = y_{0} is given, then a particular solution in which c will have a definite value is obtained.
Example 3.1.7 Solve the initial value problem
Solution This is a linear equation of the first order with
Multiplying Eq. (3.18) by the integrating factor ‘x’
Integrating and applying the initial condition
The required solution is
For a secondorder homogeneous linear Eq. (3.8), a general solution is of the form
involving two arbitrary constants c_{1} and c_{2}. An initial value problem now consists of Eq. (3.8) and two initial conditions
We have to find the particular solution satisfying these conditions as illustrated below.
Example 3.1.8 Solve the initial value problem
Solution cos x, sin x are solutions. We take y = c_{1} cos x + c_{2} sin x
Differentiating with respect to x
Now the required particular solution is
3.1.5 Linear Dependence and Linear Independence of Solutions
Two solutions y_{1}(x) and y_{2}(x) defined on some interval I are said to be linearly independent (L.I.) on I if
They are said to be linearly dependent (L.D.) if Eq. (3.22) holds, for some nonzero constants k_{1} and k_{2}.
y_{1} and y_{2} are linearly independent if one is not proportional to the other.
Criterion for linear independence of two functions y_{1} and y_{2} on I
Two functions y_{1} and y_{2} defined on some interval I are linearly independent on I if the Wronskian (determinant)
Example 3.1.9 Show that y_{1} = cos x and y_{2} = sin x are linearly independent
Example 3.1.10 Show that are linearly dependent
3.1.6 General Solution, Basis and Particular Solution
A general solution of a secondorder linear homogeneous (L.H.) equation
on an open interval I is of the form y = c_{1}y_{1} + c_{2}y_{2} where y_{1} and y_{2} are linearly independent on I (they are not proportional to each other) and c_{1} and c_{2} are arbitrary constants. Then, y_{1} and y_{2} are called the basis (fundamental system) of solution for the L.H. equation.
A particular solution of L.H. equation is obtained by giving specific values to c_{1} and c_{2} in the general solution.
3.1.7 Second Order Linear Homogeneous Equations with Constant Coefficients
We will now discuss the method of solution of a homogeneous linear differential equation of the form
To solve Eq. (3.24), we recall that a firstorder linear equation
has an exponential function as solution: y = e^{ax}. This gives us the idea to try as a solution of Eq. (3.24), the function
We have Dy = me^{mx}, D^{2}y = m^{2}e^{mx}.
Substituting these in Eq. (3.24), we have
Hence Eq. (3.25) is a solution of Eq. (3.26) if m is a solution of the algebraic Eq. (3.26) which is called the auxiliary equation or characteristic equation of Eq. (3.24).
It has, in general, two roots
Note 3.1.11 Auxiliary equation f(m) = am^{2} + bm + c = 0 is obtained from the homogeneous differential Eq. (3.24) by replacing ‘D’ by m. The general solution of Eq. (3.24) is obtained by combining two linearly independent solutions of Eq. (3.24) on I, which constitute a basis of solutions of Eq. (3.24) on I.
Case (i): Two distinct real roots α, β. In this case, y_{1} = e^{αx}, y_{2} = e^{βx} constitute a basis of solutions of Eq. (3.24) on any interval I. Since these are linearly independent, the general solution is
Case (ii): Two real repeated (double) roots, α, α Let β = α. The solution is
So, we have to find another independent solution for basis. We proceed as follows:
Method 3.1.12 Let β = α + h. The solution is
expanding e^{hx} by exponential theorem.
assuming that terms with h^{2} and higher powers of h tend to zero.
We can choose c_{2} to be sufficiently large so as to make hc_{2} finite as h → 0 and c_{1} large with a sign opposite to that of c_{2} so that c_{1} + c_{2} is finite. If c_{1} + c_{2} = A and hc_{2} = B, the general solution corresponding to two equal roots is y = (A + Bx)e^{α}.
Method 3.1.13 If the auxiliary equation am^{2} + bm + c = 0 has roots equal to α
The differential equation becomes (D − α)^{2}y = 0
which is a linear equation
Case (iii): Since a, b, c are real, complex roots occur in conjugate pairs if the discriminant of the auxiliary equation
The general solution in this case is
Example 3.1.14 Solve .
Solution The given differential equation is (D^{2} + 3D + 2)y = 0
The auxiliary equation is
The general solution is y = c_{1}e^{−x} + c_{2}e^{−2x}.
Example 3.1.15 Solve .
Solution The given differential equation is (2D^{2} + 3D + 1)y = 0.
The auxiliary equation is
The general solution is y = c_{1}e^{−x/2} + c_{2}e^{−x}.
Example 3.1.16 Solve (9D^{2} + 6D + 1)y = 0.
Solution The auxiliary equation is
The general solution is y = (c_{1} + c_{2}x)e^{−x/3}.
Example 3.1.17 Solve .
Solution The given differential equation is
The auxiliary equation is
The general solution is
Example 3.1.18 Solve
Solution The auxiliary equation is
The general solution is
EXERCISE 3.1
Solve the following:
 y″ + y′ − 6y = 0.
Ans: y = c_{1}e^{2x} + c_{3}e^{−x}
 8y″ − 2y′ − y = 0.
Ans: y = c_{1}e^{x/2} + c_{2}e^{−x/4}
 (9D^{2} + 12D + 4)y = 0.
Ans: y = (c_{1} + c_{2}x)e^{−2x/3}
 y″ + 4y′ + y = 0.
Ans: y = (c_{1} + c_{2}x)e^{−x/2}
 y″ + π^{2}y = 0.
Ans: y = c_{1} cosπx + c_{2}sinπx
 y″ + 2ky′ + (k^{2} + 4)y = 0.
Ans: y = e^{−kx} (c_{1}cos2x + c_{2}sin2x)

Ans: y = c_{1}e^{−ax} + c_{2}e^{−bx}

Ans: y = (c_{1} + c_{2}x)e^{−4x}
 (D^{2} − 3D + 4)y = 0. [JNTU 2003]
 (D^{2} − 5D + 6)y = 0.
Ans: y = c_{1}e^{2x} + c_{2}e^{3x}
 (D^{2} − 6D + 13)y = 0.
Ans:
 (D^{2} − 13D + 12)y = 0.
Ans: y = c_{1}e^{x} + c_{2}e^{3x}
 (D^{2} + 4)y = 0.
Ans: y = c_{1}cos2x + c_{2}sin2x
 (D^{2} − D − 2)y = 0.
Ans: y = c_{1}e^{2x} + c_{2}e^{−x}
 (3D^{2} + D − 14)y = 0.
Ans: y = c_{1}e^{2x} + c_{2}e^{−7x/3}
 (D^{2} − 4D + 1)y = 0.
Ans:
3.1.8 Higher Order Linear Equations
The method of solution of second order linear equations can be extended to higher order equations.
The differential equation
is called an nth order differential equation with variable coefficients if at least one of the coefficients P_{i}(i = 0,1,...,n) is a function of x and Eq. (3.29) is called an equation with constant coefficients if all P_{i}(i = 0,1,...,n) are constants. The righthand side member X in Eq. (3.29) is a function of x. Equation (3.29) is homogeneous if X ≡ 0 and nonhomogeneous if X ≢ 0 (Exercises 1, 3 and 4 below). In each term of Eq. (3.29), y and its derivatives occur in the first degree without being multiplied together.
Table 3.1 Classification of Linear Differential Equations
As in the case of linear equations of the first order, the general solution (or complete solution) of Eq. (3.29) comprises of two parts: y = y_{c} + y_{p} where y_{p}, called the particular integral (P. I.) satisfi es the complete nonhomogeneous (N.H.) Eq. (3.29) and contains no arbitrary constants, while y_{c}, called the complementary function (C.F.) (a linear combination of n linearly independent solutions containing n arbitrary constants) satisfies the homogeneous equation.
3.1.9 Linearly Independent (L.I.) Solutions
We now extend the definition of the two linearly independent functions given above to a set of linearly independent solutions of an nth order linear differential equation.
A set of solutions {y_{i}  i = 1,2,...,n} defined on some interval I is said to be linearly independent (L.I.) on I if
It is linearly dependent (L.D.) on I if it is not linearly independent on I.
The set of solutions {y_{i}  i = 1,2,...n} is linearly independent on I if the Wronskian determinant
The differential operator D
We can conveniently denote
as Dy, D^{2}y,..., D^{n}y respectively, and write Eq. (3.29) in the symbolic form
where is a polynomial in the symbol D.
When applied on a differentiable function, D does the operation of differentiation; so it must be always on the lefthand side of the function upon which we are applying it. It is a linear operator.
It satisfies index laws
3.1.10 Exponential Shift
Theorem 3.1.19 If f(D) = D^{n} + a_{1}D^{n−1}a_{2}D^{n−2} + … + a_{n} where a_{i} are real constants, then
where y is a function of x.
Proof
Apply (D − α) again
Repeating this r times, we have
This relation shows that the effect of shifting an exponential factor from the righthand side of the operator to its left side is to replace D by (D + α).
Note 3.1.20 By the theorem on the exponential shift, we have
where V is a function of x. Changing D to D + α in Eq. (3.32) we get
Corollary 3.1.21
Corollary 3.1.22
Rules
Higher order
Linear differential equation with constant coefficients
We consider from now on, that the coefficients in Eq. (3.29) are all constants, i.e., P_{i} = a_{i} constants.
Method of finding the complementary function
Consider the homogeneous equation
If we put y = e^{mx} as a trial solution we have the auxiliary equation
Thus, the solution of the homogeneous differential equation now reduces to that of an algebraic equation called the auxiliary equation obtained by replacing D with m. Solving the auxiliary equation we get n roots m_{i}(i = 1,2...n), each giving rise to a solution.
The general solution containing n arbitrary constants is obtained depending on the nature of roots in Table 3.2.
Table 3.2
Example 3.1.23 Solve 4y′″ + 4y″ + y′ = 0. [JNTU 2003]
Solution The differential equation is
The auxiliary equation is
The general solution is
Example 3.1.24 Solve (D^{3} + 27)y = 0.
Solution The auxiliary equation is
The general equation is
Example 3.1.25 Solve y′″ − y″ + 100y′ − 100y = 0.
Solution The auxiliary equation is m^{3} − m^{2} + 100m − 100 = 0. Clearly m = 1 is a root. By synthetic division
The general solution is
Example 3.1.26 Solve .
Solution The auxiliary equation is
The general solution is
Example 3.1.27 Solve (4D^{3} + 16D^{2} + 21D + 9)y = 0.
Solution The auxiliary equation is
The general solution is
By synthetic division
The general solution is
Example 3.1.28 Solve (D^{2} + 3)^{2} y = 0.
Solution The auxiliary equation is
The general solution is
EXERCISE 3.2
Solve:
 (D^{3} + 3D^{2} − 10D)y = 0.
Ans: y = c_{1} + c_{2}e^{2x} + c_{3}e^{−5x}
 (D^{2} − 4)^{2}y = 0.
Ans: y = (c_{1} + c_{2}x)e^{2x} + (c_{3} + c_{4}x)e^{−2x}]

Ans: x(t) = c_{1} + c_{2}e^{3t} + c_{3}e^{−t}]
 (D^{4} − 2D^{3} − 13D^{2} + 38D − 24)y = 0.
Ans: y = c_{1}e^{x} + c_{2}e^{2x} + c_{3}e^{3x} + c_{4}e^{−4x}]
 (D^{3} + 3D^{2} − 4)y = 0.
Ans: y = c_{1}e^{x} + (c_{2} + c_{3}x)e^{−2x}]
 (2D^{4} − 3D^{3} − 2D^{2})y = 0.
Ans: y = c_{1} + c_{2}x + c_{3}e^{2x} + c_{4}e^{−x/2}]
 (D^{2} + D + 1)y = 0.
Ans:
 (D^{4}+ 8D^{2} + 16)y = 0.
Ans: y = (c_{1} + c_{2}x)cos2x + (c_{3} + c_{4}x)sin2x
 (D^{3} + D^{2} + 4D + 4)y = 0 given that y = 0,y′ = −1,y″ = 5 when x = 0.
Ans: y = e−x − cos2x
 (D^{3} + 6D^{2} + 3D − 10)y = 0.
Ans: y = c_{1}e^{x} + c_{2}−2x + c_{3}e^{−5x}
 (D^{3} − 4D^{2}+ 5D − 2)y = 0.
Ans: y = (c_{1} + c_{2}x)e^{x} + c_{3}e^{2x}
 (D^{4} + 2D^{3} − 3D^{2} − 4D + 4)y = 0.
Ans: y = (c_{1} + c_{2}x)e^{x} + (c_{3} + c_{4}x)e^{−2x}
 (D^{4} − D^{3} − 9D^{2} − 11D − 4)y = 0.
Ans: y = (c_{1} + c_{2}x + c_{3}x^{2})e^{−x} + c_{4}e^{4x}
 (D^{4} − k^{4})y = 0.
Ans: y = c_{1}e^{kx} + c_{2}e^{−kx} + c_{3}coskx + c_{4}sinkx
 (D^{4} + 8D^{2} + 16)y = 0.
Ans: y = (c_{1} + c_{2}x)cos2x + (c_{3} + c4x)sin2x
 (D^{4} + 3D^{2} − 4)y = 0.
3.1.11 Inverse Operator D^{−1} or
Method of finding the particular integral
If Q is a differentiable function of x on an interval I and is a differential operator such that is then called the inverse operator such that .
If Q is a continuous function on an interval I, then
is a function of x containing no arbitrary constant such that
Theorem 3.1.29 If Q is a continuous function of x on I and a is a real or complex constant, then a particular value of
Proof
Operating (D − α) on both sides, we get
which is a linear equation of the first order, whose solution is
Since we are interested here in a particular solution , we take c = 0. Hence
If and are two inverse operators where α, β are given real/complex constants, then
This can be extended to n factors. The following method of putting into partial fractions is of practical importance.
3.1.12 General Method for Finding the P. I.
Theorem 3.1.30 If f (D) = (D − α_{1}) (D − α_{2})...(D − α_{n}) where α_{r} are real or complex constants, then particular integral
where A_{r} is the determinable constant corresponding to the partial fraction .
Example 3.1.31 Find particular values of
Solution
Example 3.1.32 Find the particular values of
Solution
Example 3.1.33 Find the particular value of
Solution
Alternative method
EXERCISE 3.3
Find the particular value of each of the following:
3.2 GENERAL SOLUTION OF LINEAR EQUATION f(D)y = Q(x)
We have seen that if y = y_{p} is a particular solution (containing no arbitrary constants) of the linear equation f(D)y = Q(x) and y = y_{c} is the general solution of the homogenous equation f(D)y = 0 with as many arbitrary constants as the order of the equation then y = y_{c} + y_{p} is the general solution (G.S.) of the nonhomogenous linear equation:
As we have studied methods of finding y_{c} and the general method of finding y_{p} earlier, we will work out some examples and give some exercises.
The method of finding y_{p} discussed above is particularly useful when Q(x) = tab tan ax, cot ax, sec ax or csc ax and also when we cannot use special short methods (discussed in section 3.2.1) when Q(x) = e^{ax}, sin ax, cos ax, x^{m}, e^{ax}V and xV.
Example 3.2.1 Solve (D^{2} − 7D + 6)y = e^{2x}(1 + x). [JNTU 2003]
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.2 Solve (D^{2} + 2D^{2} − 3D)y = xe^{3x}.
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
Partial fractions
The general solution is
Example 3.2.3 Solve (D^{2} + a^{2})y = sec ax. [JNTU 2000]
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
Replacing i by −i in the above result, we have
The general solution is
Example 3.2.4 Solve .
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
Replacing i by −i in the above result, we have
The general solution is
Example 3.2.5 Solve (D^{2} − 1)y = 2e^{x} + 3x. [JNTU 1996S]
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.6 Solve (D^{2} + 1)y = cosec x.
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
Replacing i by −i in the above result, we have
The general solution is
EXERCISE 3.4
Solve:
 (D^{2} − 3D + 2)y = e^{3x}.
Ans:
 (D^{2} − 1)y = 3 + 7x.
Ans: y = c_{1} cosh x + c_{2} sinh x − 3 − 7x
 (D^{2} + 9)y = sec 3x.
Ans:
 (D^{2} + a^{2})y = cosbx.
Ans:
 (D^{2} + 3D + 2)y = x^{2}.
3.2.1 Short Methods for Finding the Particular Integrals in Special Cases
be a nonhomogenous linear equation, where a_{r} are real constants.
We will now consider short methods for finding the P.Is in special cases when Q(x) is of the form (1) e^{aX} (2) sin ax or cos ax (3) x^{m} (4) e^{aX}V and (5) xV, where V is a function of x.
(1) Q(x) = e^{ax} where ‘a’ is a constant
Case (i) f(a) ≠ 0
Since De^{ax} = ae^{ax}, D^{2}e^{ax} = a^{2}e^{ax}, and generally, D^{r}e^{ax} = a^{r}e^{ax}
Applying on both sides of Eq. (3.35)
Rule Replace D in f(D) with a if f(a) ≠ 0, i.e., (D − a) is not a factor of f(D).
Case (ii) f(a) = 0, f′(a) ≠ 0, i.e., (D − a) is a factor of f(D) but (D − a)^{2} is not a factor of f(D).
Let f(D) = (D − a)ϕ(D), where (D − a) is not a factor of ϕ(D).
We have f′(D) = ϕ(D) + (D − a)ϕ′(D) ⇒ f′(a) = ϕ(a)
Case (iii) f(a) = 0, f′(a) = 0,...f^{(r)}(a) = 0, f^{(r+1)}(a) ≠ 0, i.e.,(D − a)^{r} is a factor of f(D) but (D − a)^{r+1} is not a factor of f(D).
Let f(D) = (D − a)^{r}ϕ(D)
where (D − a) is not a factor of ϕ(D).
We have f^{(r)}(D) = r!ϕ(D) = terms containing (D − a) as a factor
⇒f^{(r)}(a) = r!ϕ(a)
Example 3.2.7 Evaluate .
Solution We have to evaluate
∵ 4 is not a factor of (D^{2} − 5).
Example 3.2.8 Evaluate .
Solution Here f(D) = D^{2} − 4 = (D − 2)(D + 2)
Example 3.2.9 Evaluate .
Solution
Alternative method
Example 3.2.10 Find the particular integral y_{p} of .
Solution
Alternative method
Example 3.2.11 Solve y″ − y′ − 2y = 3e^{2x} given that
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Differentiating Eq. (3.44) with respect to x
Adding Eqs. (3.45) and (3.46)
Substituting these values in Eq. (3.44), the required solution is obtained as
Example 3.2.12 Solve
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.13 Solve
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.14 Solve
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.15 Solve
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
EXERCISE 3.5
Solve the following:
 (D^{2} + 1)y = cosh x.
Ans:
 (D^{2} − 1)y = sinh x.
Ans:
 (D + 1)^{3}y = e^{−x}.
Ans:
 (D^{3} − 5D^{2} + 7D − 3)y = e^{2x} cosh x.
Ans:
 (D^{3} + 1)y = 3 + 5e^{x}. [JNTU 2002]
Ans:
 (2D + 1)^{2}y = 4e^{−x/2}.
Ans:
 (D^{2} − D − 6)y = e^{x} cosh 2x.
Ans:
 (D^{2} − 2pD + p^{2})y = e^{x}.
Ans:
 (D^{4} − 1)y = e^{x}.
Ans:
 (4D^{2} + 4D − 3)y = e^{2x}.
Ans:
 (D^{2} − 3D + 2)y = e^{5x}.
Ans:
 (D^{2} + D + 1)y = e^{x}.
Ans:
 (D^{2} − 5D + 6)y = 4e^{x} + 5.
Ans:
 (D^{2} − 2D + 1)y = (1 + e^{−x})^{2}.
Ans:
 (D^{2} − 4)y = 3e^{2x} − 4e^{−2x}.
Ans:
 (D^{2} − 1)y = cosh x.
Ans:
(2) Q(x) = sin ax cos ax where ‘a’ is a constant
Case (i) f(−a^{2}) ≠ 0
We have D(sin ax) = a cos ax
and generally,
for any integer ‘r’.
Operating on both sides by , we have
if f(−a^{2}) ≠ 0.
Rule Replace D^{2} by −a^{2} if f(−a^{2}) ≠ 0
The same rule applies if Q(x) = cos ax
if f(−a^{2}) ≠ 0.
This rule holds if sin ax and cos ax are replaced by sin (ax + b) and cos (ax + b), where b is a constant.
Case (ii) f(−a^{2}) = 0,
i.e.,
The above rule fails if f(−a^{2}) = 0.In this case, (D^{2} + a^{2})f(D^{2}) so that we can write
Let us evaluate
Equating the real and imaginary parts,
We have, if ϕ(−a^{2}) ≠ 0,
General case: If f(D) contains old powers of D, then we can arrange the even and odd powers of D so that
Let
Then
Similarly,
Example 3.2.16 Evaluate .
Solution
Example 3.2.17 Evaluate .
Solution
Example 3.2.18 Solve
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
To get D^{2} in the denominator
The general solution is
Example 3.2.19 Solve (D^{2} − 1)y = sin x cos x.
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is,
Example 3.2.20 Solve (D^{2} − 4D)y = e^{x} + sin 3x cos 2x.
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.21 Solve (D^{3} + 2D)y = e^{2x} + cos (3x + 7).
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
EXERCISE 3.6
Solve the following:
 y″ + 4y′ + 47 = 4cos x + 3sin x; y(0) = 1, y′(0) = 0.
Ans: y = (1 + x)e^{−2x} + sin x
(D^{3} − 1)y = e^{x} + sin 3x + 2. [JNTU 2004]
Ans:
 (D^{2} − 4)y = 2 cos^{2} x.
Ans:
 (D^{2} + 1)y = sin x sin 2x. [JNTU 2003]
Ans:
 (D^{2} + 9)y = cos 3x + sin 2x.
 (D^{2} − 1)y = sin (x + 5).
Ans:
 (D^{2} + 4)y = e^{x} + sin 2x + cos 2x.
Ans:
 (D^{2} − 4D + 3)y = sin 3x + cos 2x.
Ans:
(3) Q(x) = x^{m} where m is a positive integer
To evaluate , we expand in increasing powers of D up to D^{m} and apply on x^{m}.
Example 3.2.22 Evaluate .
Solution
Example 3.2.23 Evaluate (D^{2} + 3D)^{−1}x.
Solution
Example 3.2.24 Solve (D^{3} − 1)y = x^{3}.
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.25 Solve (D^{3} − D^{2})y = 1 + x^{3}.
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.26 Solve (2D^{2} + D)y = x^{2}.
Solution To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
EXERCISE 3.7
Solve the following:
 y′″ − 2y″ − y′ − 2y = 1 − 4x^{3}. [JNTU 2004(3)]
Ans: (5 + cx^{2})x^{3}y^{5} = 2
(D^{2} + D + 1)y = x^{3}. [JNTU 2004]
Ans:
 (D^{2} − 3D − 2)y = x^{2}.
 (D^{3} − D^{2} − D + 1)y = 1 + x^{2}. [JNTU 1997S]
Ans: y = c_{1}e^{−x} + (c_{2} + c_{3}x)e^{x} + x ^{2} + 2x + 5
(D^{3} + 2D^{2} + D)y = e^{2x} + x^{2} + x + sin 2x. [JNTU 2003S(1)]
Ans:
 (D^{2} + 5D + 4)y = x^{2}. [JNTU 2003S(2)]
Ans:
(4) Q(x) = Q^{ax}V(x) where ‘a’ is a constant and V(x) is a function of x
By the theorem on exponential shift, we have
where V_{1} is a function of x.
Applying on both sides, we have
Let f(D + a)V_{1} = V
From Eqs. (3.75) and (3.76), we get
The effect of shifting the exponential factor e^{ax} from the right side of the operator to the left side is to replace D by (D + a) in f.
Example 3.2.27 Evaluate .
Solution
Example 3.2.28 Evaluate .
Solution
Example 3.2.29 Evaluate .
Solution
Example 3.2.30 Solve (D^{4} − 1)y = e^{x} cos x.
Solution To find the complementary function, we have to solve (D^{4} − 1)y = 0
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.31 Solve (D^{2} + 4D + 4)y = e^{−2x} cos x.
Solution To find the complementary function, we have to solve (D^{2} + 4D + 4)y = 0
The auxiliary equation is
To find the particular integral,
The general solution is
EXERCISE 3.8
Solve the following:
 (D^{2} − 7D + 6)y = e^{2x}(1 + x). [JNTU 2003]
Ans:
 (D^{2} − 2D + 1)y = xe^{x}.
Ans: y =(c_{1} + c_{2}x)e^{x} − e^{x}(x sin x + 2 cos x)
(D^{3} − 3D^{2} + 3D − 1)y = x^{2}e^{x}. [JNTU 2002S]
 (D^{2} + 4D + 3)y = e^{−x} sin x + x. [JNTU 1996, 1999]
Ans:
 (D^{2} − 4)y = x sinh x. [JNTU 2000]
Ans:
 (D^{3} − 7D^{2} + 14D − 8)y = e^{x} cos 2x. [JNTU 2003S(3)]
Ans:
 (D^{2} − 4D + 1)y = e^{2x} cos3x.
Ans:
(5) Q (x) = xV(x) where V(x) is a function of x
Theorem 3.2.32 Prove that
where V_{1} is a function of x.
Proof For n = 1
Assume that Eq. (3.77) is true for n = m
Applying D on both sides,
If Eq. (3.77) is true for n = m, then it is true for n = m + 1. Hence by the principle of mathematical induction, Eq. (3.77) is true for all n ɛ N.
Take in the above result.
Example 3.2.33 Solve (D + 1)^{2}y = x cos x.
Solution To find the complementary function, we have to solve (D + 1)^{2}y = 0.
The auxiliary equation is
To find the particular integral,
Here f(D) = (D + 1)^{2} f′(D) = 2(D + 1)
The general solution is
Example 3.2.34 Solve (D^{2} + 9)y = x sin 2x.
Solution To find the complementary function, we have to solve (D^{2} + 9)y = 0.
The auxiliary equation is
To find the particular integral,
Here f(D) = D^{2} + 9 f′(D) = 2D
The general solution is
Example 3.2.35 Solve (D − 2)^{2}y = x^{2} sin x + e^{2x} + 3. [JNTU 2001S]
Solution y_{c} = (c_{1} + c_{2}x)e^{2x}
The general solution is
Alternatively, P.I_{1} can also be obtained as follows:
EXERCISE 3.9
Solve the following:
 (D + 1)y = x cos x. [JNTU 2003]
Ans:
 (D^{2} + 4)y = x sin x. [JNTU 2003]
Ans:
 (D^{2} − 1)y = x sin 3x + cos x. [JNTU 1997]
 (D^{2} + 1)y = x^{2} sin x.
 (D − 1)^{2}y = xe^{x} sin x. [JNTU 2002]
Ans: y = (c_{1} + c_{2}x)e^{x} − e^{x} (2 cos x + x sin x)
(D^{2} − 1)y = x sin x + x^{2}e^{x}.
Ans:
 (D^{2} + 1)y = x^{2} sin x.
 (D^{2} − 1)y = x sin x + e^{x}(1 + x^{2}). [JNTU 199S]
Ans:
 (D^{2} + 3D + 2)y = xe^{x} sin x.
Ans:
 (D^{2} + 1)y = e^{−x} + x^{3} + e^{x} sin x. [JNTU 1998]
Ans:
 .
Ans: y = (c_{1} + c_{2}x)e^{−x} − e^{−x} log x
 y″ + 4y′ + 20y = 23 sin t − 15 cos; y(0) = 0, y′(0) = 1. [JNTU 2004(4)]
 (D^{3} − 1)y = e^{x} + sin 3x + 2. [JNTU 2004(1)]
 (D^{2} + 1)x = t cos 2t x = x′ = 0 at t = 0. [JNTU 2002]
Ans:
 (D^{2} − 4D + 4)y = 8x^{2}e^{2x} sin 2x. [JNTU 2001S]
Ans: y = (c_{1} + c_{2}x)e^{2x} − 2e^{2x} · [2x cos 2x + (x^{2} − 3/2)sin 2x]
3.2.2 Linear Equations with Variable Coefficients—Euler–Cauchy Equations (Equidimensional Equations)
An equation of the form
where a_{i} are all constants and Q(x) is a function of x is called an Euler–Cauchy equation. Some authors call it a homogeneous equation. Since the phrase is used for an equation with Q(x) ≡ 0, it leads to confusion if we use it for Eq. (3.80). Equation (3.80) is rightly called an equidimensional equation since the dimension of each term with respect to the independent variable is same.
Change the independent variable to z by putting
where and
Substituting these into Eq. (3.80) the equation reduces to a linear differential equation with constant coeffi cients, which can be solved by the methods discussed above.
Example 3.2.36
Solution
To equation reduces to
To find the complementary function, we solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.37
Solution
where
which is a linear equation with constant coefficients.
To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.38
Solution
where
The given equation becomes
which is a linear equation with constant coefficients.
To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.39
Solution Put
The given equation becomes
which is a linear equation with constant coefficients.
To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.40
Solution
so that xD = θ, x^{2}D^{2} = θ(θ − 1)
The given equation becomes
which is a linear equation with constant coefficients.
To find the complementary function, we have to solve
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.41
Solution Put x = e^{x} or z = log x
so that
The given equation becomes
which is a linear equation with constant coefficients.
To find the complementary function, we have to solve
The auxiliary equation is
By synthetic division
To find the particular integral,
The general solution is
EXERCISE 3.10
Solve the following:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans:

Ans: y = (c_{1} + c_{2} log x)x + log x + 4

Ans:

Ans:
3.2.3 Legendre's Linear Equation
An equation of the form
where a_{r}(r = 0,1,...,n), a, b are constants and Q(x) is a function of x is called a Legendre's linear equation.
Put a + bx = e^{z} or z = log(a + bx) so that
By induction,
Equation (3.69) now reduces to a linear equation with constant coefficients which can be solved by the methods discussed above.
Example 3.2.42 Solve
Solution This is Legendre's linear equation.
Put 3x + 2 = e^{z} or z = log(3x + 2)
The given equation becomes
To find the complementary function, we have to solve (θ^{2} − 4)y = 0.
The auxiliary equation is
To find the particular integral,
The general solution is
Example 3.2.43
Solution This is Legendre's linear equation
Put (1 + 2x) = e^{2} or z = log(1 + 2x) so that
The given equation becomes
To find the complementary function, we have to solve (θ − 2)^{2}y = 0.
The auxiliary equation is
To find the particular integral,
The general solution is
EXERCISE 3.11
Solve the following:
 [(5 + 2x)^{2}D^{2} − 6(5 + 2x)D + 8]y = 0.
Ans:
 [(2x + 1)^{2}D^{2} − 2(2x + 1)D − 12]y = 6x.
Ans:
 [(x + 2)^{2}D^{2} − (x + 2)D + 1]y = 3x + 4.
Ans:
 [(1 + x)^{2}D^{2} + (1 + x)D = 1]y = 4 cos[1 + x)].
Ans: y = c_{1} cos log(1 + x) + c_{2} sin log (1 + x) + 2 log(1 + x)sin log(1 + x)
 [(1 + x)^{2}D^{2} + (1 + x)D + 1]y = sin 2[log(1 + x)]. [JNTU 1996]
Ans:
3.2.4 Method of Variation of Parameters
We will now consider the method of variation of parameters for finding the particular integral from the complementary function and hence the general solution of a linear nonhomogeneous equation
where P, Q, R are functions of x.
Let y = C_{1}u(x) + C_{2}v(x) (3.129)
be the company function. Since u and v are solutions of the homogeneous equation
We have
We assume that a particular integral y_{p} of Eq. (3.128) is given by
where A and B are now considered as functions of x to be determined.
Differentiating Eq. (3.133) with respect to x
Now we choose A and B such that
Differentiating Eq. (3.135) with respect to x again
Since y_{p} satisfies Eq. (3.128) we have from Eqs. (3.130), (3.131) and (3.132)
by Eqs. (3.131) and (3.132).
Since u(x) and v(x) are linearly independent solutions over the interval I we have
Hence, solving Eqs. (3.134) and (3.135) we have,
Substituting these values in Eq. (3.133) we get the particular integral y_{p}. From Eqs. (3.129) and (3.133) we obtain the general solution or complete solution.
Note 3.2.44 Since the arbitrary constants C_{1} and C_{2} are replaced by A and B in the complementary function and are considered as functions of x for finding the particular integral this method is called “variation of parameters.”
Note 3.2.45 The above method is applicable when the coefficients are functions of x or constants.
Example 3.2.46 Solve the following equations by the method of variation of parameters.
 (D^{2} + 1)y = x sin x;
 (D^{2} + a^{2})y = tan ax; [JNTU 2003]
Solution (1) The given equation is
To find the complementary function, we have to solve (D^{2} + 1)y = 0
To auxiliary equation is
We take y = A cos x + B sin x, where A, B are functions of x and
The complete solution is
(2) The given equation is
To find the complementary function, we have to solve
The auxiliary equation is
We take y = A cos x + B sin x, where A, B are functions of x and
The complete solution is
(3) The given equation is
To find the complementary function, we have to solve (D^{2} + a^{2})y = 0
To auxiliary equation is
Let y = A cos ax + B sin ax where A, B are functions of x and
The complete solution is
Example 3.2.47 Solve the equation
[JNTU 1995, 1996, 1998, 1999]
Solution The given equation is
The complementary function is
Let y = A cos x + B sin x, where A, B are functions of x and
The complete solution is
Example 3.2.48 Solve the equation (D^{2} − 2D)y = e^{x} sin x.
Solution To find the complementary function, we have to solve (D^{2} − 2D)y = 0.
To auxiliary equation is
Let y = A + Be^{2x} where A, B are functions of x.
The complete solution is
Example 3.2.49 Solve the equation (D^{2} + 3D + 2)y = e^{x} + x^{2}.
Solution Here P = 3, Q = 2, R = e^{x} + x^{2}
The auxiliary equation is
The complementary function is
Let y = Ae^{−x} + Be^{−2x} where A, B are functions of x.
The complete solution is
EXERCISE 3.12
Solve the equations by the method of variation of parameters:
 (D^{2} + 1)y = cosec x cot x.
Ans: y = c_{1} cos x + c_{2} sin x − cos x log(sin x) − x sin x
 (D + 1)^{2} y = e^{−x} log x.
Ans:
 (D^{2} + a^{2})y = a^{2} sec ax.
Ans: y = c_{1} cos ax + c_{2} sin ax + ax sin ax + cos ax log(cos ax)
 (D^{2} − 3D + 2)y = xe^{x} + 2x.
Ans:
 (D^{2} + 4)y = tan 2x. [JNTU 2003, 2002, 1997, 1994]
Ans:
 ((x − 1)D^{2} − xD + 1)y=(x − 1)^{2}.
[Hint: e^{x} ,x are basis solutions.]

Ans: y = c_{1}e^{3x} + c_{2}xe^{3x} −e^{3x} logx
 (D^{3} + D)y = cosec x.
Ans: y=c_{1} + c_{2} cos x + c_{3} sin x − log(cosec x + cot x)
−cos x logsin x − x sin x
 (D^{2} − 2D + 2)y = e^{x} tan x.
Ans: y=e^{x}(c_{1} cos x + c_{2} sin x) − e^{x} cos x log(sec x + tan x)
3.2.5 Systems of Simultaneous Linear Differential Equations with Constant Coefficients
In many applied mathematical problems, we encounter systems of simultaneous linear differential equations involving two (or more) dependent variables x and y and one independent variable t, as follows:
Where f_{1},f_{2},g_{1},g_{2} are rational integral functions with constant coefficients and f, g are given functions of t.
The system Eq. (3.160) is solved by elimination and we get
where Δ = f_{1}(D)g_{2}(D) − f_{2}(D)g_{1}(D)≠ 0.
Example 3.2.50 Solve
Solution In operator notation, the equations are
Multiplying Eq. (3.161) by D + 3 and Eq. (3.162) by −2, we get
On subtraction, (D^{2}+1)x=0
The auxiliary equation is
Substituting in Eq. (3.161)
Example 3.2.51 Solve
x=0, y=0 when t = 0
Solution The given system of equations is
To obtain Δ_{x},Δ_{y} respectively replace c_{1},c_{2} in Δ by
The auxiliary equation is
The complimentary function is
The particular integral is
Substituting in Eq. (3.164)
The required solution is
Example 3.2.52 Solve
Solution The given system of equation is
Operate Eq. (3.171) by (D −1) and add to Eq. (3.172)
From Eq. (1)
Example 3.2.53
Solution Applying D^{2} on Eq. (3.176)
The roots of the auxiliary equations are
Example 3.2.54
Solution
EXERCISE 3.13
Solve the following pairs of simultaneous linear equations:

Ans:

Ans: x=c_{1} cosat − c_{2} sinat
 (D+2)x+(D+1)y=t; 5x+(D+3)y=t^{2}.
Ans:

with x(0)=6, y(0)−2.
Ans: x=4e^{t} + 2e^{−t}; y=−e^{t}−e^{−t}

Ans:
 (D+6)y−Dx=0;(3−D)x−2Dy=0; with x=2,y=3
when t = 0.
Ans: x=4e^{2t}−2e^{−3t}; y=e^{2t}+2e^{−3t}
 (D+2)x+(D−1)y=−sint;
(D−3)x(D+2)y=4cost.
Ans:
 D^{2}y=x−2,D^{2}x=y+2.
Ans: x=c_{1} sint + c_{2} cost + c_{3}e^{t} + c_{4}e^{−t} + 2; y=−c_{1} sint − c_{2} cost + c_{3}e^{t} + c_{4}e^{−t} −2
 (D+1)x+(D−1)y=e^{t}; (D^{2}+D+1)x+(D2−D+1)y=t^{2}.
Ans: