3. Linear Differential Equations with Constant Coefficients – Differential Equations

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 non-homogeneous 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) = cex satisfies the homogeneous equation derived from Eq. (3.6).

The general (complete) solution of Eq. (3.6) is

 

y = cex + ex

 

Thus, the general solution y of a linear equation comprises of two parts: y = yc + yp where yc, which satisfies the homogeneous equation and contains as many arbitrary constants as the order of the differential equation, and yp, which satisfies the non-homogeneous 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 left-hand side of the function upon which it is operating.

  1. D’ is a linear operator:

     

    D(c1y1 (x) + c2y2 (x) = c1Dy1 (x) + c2Dy2 (x)

     

  2. 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 non-linear (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 non-homogeneous 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.

  1. y + 4y = ex sin x.
  2. (1 − x2)y″ − 2xy′ + 6y = 0.
  3. y″ + y2
  4. y″ − y = 0.
  5. y″ + y′ = tan x + sec2 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 yc and yp.

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 ex is a solution of the complete Eq. (3.11). Now consider the reduced equation

 

 

1 and ex is also a solution of Eq. (3.12). A linear combination of these solutions namely y = c1 + c2ex 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   y1 = ex and y2 = ex are solutions of the homogeneous linear differential equation for all x since

 

 

In fact, a linear combination of y1 and y2, i.e., y = c1y1 + c2y2 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 y1 and y2 be solutions of Eq. (3.8). Then substituting

 

y = c1y1 + c2y2

 

 

              [∵ y1, y2 are solutions of Eq. (3.15)]

 

Note 3.1.6 The above principle does not hold for non-homogeneous equations and non-linear equations.

Non-homogeneous linear differential equation

Consider the non-homogeneous differential equation

 

 

are solutions of Eq. (3.16) since

 

 

But their sum y1 + y2 = (ex − 1) + (ex − 1) = ex + ex − 2 is not a solution.

 

Non-linear differential equation

y1 = x2 , y2 = 1 are solutions of the non-linear differential equation

 

 

but (−1)y2 = −x2 and their sum y1 + y2 = x2 + 1 are not solutions of Eq. (3.17).

3.1.4 Initial Value Problem (IVP)

For a first-order differential equation, a general solution contains one arbitrary constant c. If an initial condition y(x0) = y0 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

 

xy = log x + c
            1.0 = log 1 + cc = 0

 

The required solution is

 

xy = log x

 

For a second-order homogeneous linear Eq. (3.8), a general solution is of the form

 

y = c1y1 + c2y2

 

involving two arbitrary constants c1 and c2. An initial value problem now consists of Eq. (3.8) and two initial conditions

 

y(x0) = y0, y′(x0) = y0

 

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 = c1 cos x + c2 sin x

Differentiating with respect to x

 

 

y(0) = c1 · 1 + c2 · 0 = 2 ⇒ c1 = 2
y′(0) = −c1 · 0 + c2 · 1 = −1 ⇒ c2 = −1

 

Now the required particular solution is

 

3.1.5 Linear Dependence and Linear Independence of Solutions

Two solutions y1(x) and y2(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 non-zero constants k1 and k2.

y1 and y2 are linearly independent if one is not proportional to the other.

Criterion for linear independence of two functions y1 and y2 on I

Two functions y1 and y2 defined on some interval I are linearly independent on I if the Wronskian (determinant)

 

Example 3.1.9   Show that y1 = cos x and y2 = 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 second-order linear homogeneous (L.H.) equation

 

y″ + P(x)y′ + Q(x)y = 0

 

on an open interval I is of the form y = c1y1 + c2y2 where y1 and y2 are linearly independent on I (they are not proportional to each other) and c1 and c2 are arbitrary constants. Then, y1 and y2 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 c1 and c2 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 first-order linear equation

 

y′ − ay = (Da)y = 0

 

has an exponential function as solution: y = eax. This gives us the idea to try as a solution of Eq. (3.24), the function

 

 

We have Dy = memx, D2y = m2emx.

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) = am2 + 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, y1 = eαx, y2 = eβx 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

 

y = c1eax + c2eax = (c1 + c2)eax

 

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 ehx by exponential theorem.

 

= eax [(c1 + c2) + (c2h)x]

 

assuming that terms with h2 and higher powers of h tend to zero.

We can choose c2 to be sufficiently large so as to make hc2 finite as h → 0 and c1 large with a sign opposite to that of c2 so that c1 + c2 is finite. If c1 + c2 = A and hc2 = B, the general solution corresponding to two equal roots is y = (A + Bx)eα.

 

Method 3.1.13 If the auxiliary equation am2 + bm + c = 0 has roots equal to α

 

am2 + bm + c = a(m − α)2

 

The differential equation becomes (D − α)2y = 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

 

f(m) = am2 + bm + c = 0 is negative

 

The general solution in this case is

 

Example 3.1.14   Solve .

Solution   The given differential equation is (D2 + 3D + 2)y = 0

The auxiliary equation is

 

m2 + 3m + 2 = (m + 1)(m + 2) = 0 ⇒ m = −1, −2

 

The general solution is y = c1ex + c2e2x.

Example 3.1.15   Solve .

Solution   The given differential equation is (2D2 + 3D + 1)y = 0.

The auxiliary equation is

 

2m2 + 3m + 1 = (2m + 1)(m + 1) = 0 ⇒ m = −1/2, −1

 

The general solution is y = c1ex/2 + c2ex.

Example 3.1.16   Solve (9D2 + 6D + 1)y = 0.

Solution   The auxiliary equation is

 

9m2 + 6m + 1 = (3m + 1)2 = 0 ⇒ m = −1/3. −1/3

 

The general solution is y = (c1 + c2x)ex/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

 

y = c1 cosax + c2 sinax
EXERCISE 3.1

Solve the following:

  1. y″ + y′ − 6y = 0.

    Ans: y = c1e2x + c3ex

  2. 8y″ − 2y′ − y = 0.

    Ans: y = c1ex/2 + c2ex/4

  3. (9D2 + 12D + 4)y = 0.

    Ans: y = (c1 + c2x)e−2x/3

  4. y″ + 4y′ + y = 0.

    Ans: y = (c1 + c2x)ex/2

  5. y″ + π2y = 0.

    Ans: y = c1 cosπx + c2sinπx

  6. y″ + 2ky′ + (k2 + 4)y = 0.

    Ans: y = ekx (c1cos2x + c2sin2x)

  7. Ans: y = c1eax + c2ebx

  8. Ans: y = (c1 + c2x)e−4x

  9. (D2 − 3D + 4)y = 0.                     [JNTU 2003]

    Ans:

  10. (D2 − 5D + 6)y = 0.

    Ans: y = c1e2x + c2e3x

  11. (D2 − 6D + 13)y = 0.

    Ans:

  12. (D2 − 13D + 12)y = 0.

    Ans: y = c1ex + c2e3x

  13. (D2 + 4)y = 0.

    Ans: y = c1cos2x + c2sin2x

  14. (D2D − 2)y = 0.

    Ans: y = c1e2x + c2ex

  15. (3D2 + D − 14)y = 0.

    Ans: y = c1e2x + c2e−7x/3

  16. (D2 − 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 Pi(i = 0,1,...,n) is a function of x and Eq. (3.29) is called an equation with constant coefficients if all Pi(i = 0,1,...,n) are constants. The right-hand side member X in Eq. (3.29) is a function of x. Equation (3.29) is homogeneous if X ≡ 0 and non-homogeneous 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 = yc + yp where yp, called the particular integral (P. I.) satisfi es the complete non-homogeneous (N.H.) Eq. (3.29) and contains no arbitrary constants, while yc, 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 {yi | 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 {yi | i = 1,2,...n} is linearly independent on I if the Wronskian determinant

 

The differential operator D

We can conveniently denote

 

 

as Dy, D2y,..., Dny respectively, and write Eq. (3.29) in the symbolic form

 

f(D)y = (P0Dn + P1Dn−1 + … Pn−1D + Pn)y = X

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 left-hand side of the function upon which we are applying it. It is a linear operator.

 

D(c1y1 + c2y2) = c1(Dy1) + c2(Dy2)

 

It satisfies index laws

 

3.1.10 Exponential Shift

Theorem 3.1.19 If f(D) = Dn + a1Dn−1a2Dn−2 + … + an where ai are real constants, then

 

eax(f(D)y) = f(D − α)(eaxy)

 

where y is a function of x.

Proof

Apply (Dα) again

 

(D − α)2(eaxy) = (D − α)(eaxDy) = eaxD2y

 

Repeating this r times, we have

 

 

This relation shows that the effect of shifting an exponential factor from the right-hand 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., Pi = ai constants.

Method of finding the complementary function

Consider the homogeneous equation

 

 

If we put y = emx 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 mi(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

 

(4D3 + 4D2 + D)y = 0

 

The auxiliary equation is

 

 

The general solution is

 

y = (c1 + c2x)ex/2 + c3.

Example 3.1.24   Solve (D3 + 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 m3m2 + 100m − 100 = 0. Clearly m = 1 is a root. By synthetic division

 

 

 

The general solution is

 

y = c1ex + (c2 cos 10x + c3 sin 10x).

Example 3.1.26   Solve .

Solution   The auxiliary equation is

 

 

The general solution is

 

Example 3.1.27   Solve (4D3 + 16D2 + 21D + 9)y = 0.

Solution   The auxiliary equation is

 

 

The general solution is

 

y = c1ex + (c2 + c3x)e−3x/2

 

By synthetic division

 

 

The general solution is

 

y = c1ex + (c2 + c3x)e−3x/2

 

Example 3.1.28   Solve (D2 + 3)2 y = 0.

Solution   The auxiliary equation is

 

 

The general solution is

 

EXERCISE 3.2

Solve:

  1. (D3 + 3D2 − 10D)y = 0.

    Ans: y = c1 + c2e2x + c3e−5x

  2. (D2 − 4)2y = 0.

    Ans: y = (c1 + c2x)e2x + (c3 + c4x)e−2x]

  3. Ans: x(t) = c1 + c2e3t + c3et]

  4. (D4 − 2D3 − 13D2 + 38D − 24)y = 0.

    Ans: y = c1ex + c2e2x + c3e3x + c4e−4x]

  5. (D3 + 3D2 − 4)y = 0.

    Ans: y = c1ex + (c2 + c3x)e−2x]

  6. (2D4 − 3D3 − 2D2)y = 0.

    Ans: y = c1 + c2x + c3e2x + c4ex/2]

     

  7. (D2 + D + 1)y = 0.

    Ans:

  8. (D4+ 8D2 + 16)y = 0.

    Ans: y = (c1 + c2x)cos2x + (c3 + c4x)sin2x

  9. (D3 + D2 + 4D + 4)y = 0 given that y = 0,y′ = −1,y″ = 5 when x = 0.

    Ans: y = ex − cos2x

  10. (D3 + 6D2 + 3D − 10)y = 0.

    Ans: y = c1ex + c2−2x + c3e−5x

  11. (D3 − 4D2+ 5D − 2)y = 0.

    Ans: y = (c1 + c2x)ex + c3e2x

  12. (D4 + 2D3 − 3D2 − 4D + 4)y = 0.

    Ans: y = (c1 + c2x)ex + (c3 + c4x)e−2x

  13. (D4D3 − 9D2 − 11D − 4)y = 0.

    Ans: y = (c1 + c2x + c3x2)ex + c4e4x

  14. (D4k4)y = 0.

    Ans: y = c1ekx + c2ekx + c3coskx + c4sinkx

  15. (D4 + 8D2 + 16)y = 0.

    Ans: y = (c1 + c2x)cos2x + (c3 + c4x)sin2x

  16. (D4 + 3D2 − 4)y = 0.

    Ans: y = c1ex + c2ex + c3cos2x + c4sin2x

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 (Dα) 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) = (Dα1) (Dα2)...(Dαn) where αr are real or complex constants, then particular integral

 

 

where Ar 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 = yp is a particular solution (containing no arbitrary constants) of the linear equation f(D)y = Q(x) and y = yc 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 = yc + yp is the general solution (G.S.) of the non-homogenous linear equation:

 

f(D)y = Q(x)

 

As we have studied methods of finding yc and the general method of finding yp earlier, we will work out some examples and give some exercises.

The method of finding yp 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) = eax, sin ax, cos ax, xm, eaxV and xV.

Example 3.2.1   Solve (D2 − 7D + 6)y = e2x(1 + x).   [JNTU 2003]

Solution   To find the complementary function, we have to solve

 

(D2 − 7D + 6)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

Example 3.2.2   Solve (D2 + 2D2 − 3D)y = xe3x.

Solution   To find the complementary function, we have to solve

 

(D3 + 2D2 − 3D)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

Partial fractions

 

 

The general solution is

 

Example 3.2.3   Solve (D2 + a2)y = sec ax.    [JNTU 2000]

Solution   To find the complementary function, we have to solve

 

(D2 + a2)y = 0

 

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

 

(D2 + a2)y = 0

 

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 (D2 − 1)y = 2ex + 3x.                      [JNTU 1996S]

Solution   To find the complementary function, we have to solve

 

(D2 − 1)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

Example 3.2.6   Solve (D2 + 1)y = cosec x.

Solution   To find the complementary function, we have to solve

 

(D2 + 1)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

Replacing i by −i in the above result, we have

 

 

The general solution is

 

y = yc + yp = c1 cosx + c2 sinx + sinx log sinxx cosx.
EXERCISE 3.4

Solve:

  1. (D2 − 3D + 2)y = e3x.

    Ans:

  2. (D2 − 1)y = 3 + 7x.

    Ans: y = c1 cosh x + c2 sinh x − 3 − 7x

  3. (D2 + 9)y = sec 3x.

    Ans:

  4. (D2 + a2)y = cosbx.

    Ans:

  5. (D2 + 3D + 2)y = x2.

    Ans:

3.2.1 Short Methods for Finding the Particular Integrals in Special Cases

 

 

be a non-homogenous linear equation, where ar 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) eaX (2) sin ax or cos ax (3) xm (4) eaXV and (5) xV, where V is a function of x.

 

(1) Q(x) = eax where ‘a’ is a constant

 

Case (i)     f(a) ≠ 0

Since Deax = aeax, D2eax = a2eax, and generally, Dreax = areax

 

 

Applying on both sides of Eq. (3.35)

 

 

Rule Replace D in f(D) with a if f(a) ≠ 0, i.e., (Da) is not a factor of f(D).

 

Case (ii)   f(a) = 0, f′(a) ≠ 0, i.e., (Da) is a factor of f(D) but (Da)2 is not a factor of f(D).

Let f(D) = (Da)ϕ(D), where (Da) is not a factor of ϕ(D).

We have f′(D) = ϕ(D) + (Da)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.,(Da)r is a factor of f(D) but (Da)r+1 is not a factor of f(D).

Let f(D) = (Da)rϕ(D)

where (Da) is not a factor of ϕ(D).

We have f(r)(D) = r!ϕ(D) = terms containing (Da) 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 (D2 − 5).

Example 3.2.8   Evaluate .

Solution   Here f(D) = D2 − 4 = (D − 2)(D + 2)

 

Example 3.2.9   Evaluate .

Solution

 

Alternative method

 

Example 3.2.10   Find the particular integral yp of .

Solution

 

Alternative method

 

Example 3.2.11   Solve y″ − y′ − 2y = 3e2x 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)

 

3c1 + 1 = −2 ⇒ c1 = −1 (∵ c2 = 1)

 

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

 

(D2 + 6D + 9)y = 0

 

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:

  1. (D2 + 1)y = cosh x.

    Ans:

  2. (D2 − 1)y = sinh x.

    Ans:

  3. (D + 1)3y = ex.

    Ans:

  4. (D3 − 5D2 + 7D − 3)y = e2x cosh x.

    Ans:

  5. (D3 + 1)y = 3 + 5ex.   [JNTU 2002]

    Ans:

     

  6. (2D + 1)2y = 4ex/2.

    Ans:

  7. (D2D − 6)y = ex cosh 2x.

    Ans:

  8. (D2 − 2pD + p2)y = ex.

    Ans:

  9. (D4 − 1)y = ex.

    Ans:

  10. (4D2 + 4D − 3)y = e2x.

    Ans:

  11. (D2 − 3D + 2)y = e5x.

    Ans:

  12. (D2 + D + 1)y = ex.

    Ans:

  13. (D2 − 5D + 6)y = 4ex + 5.

    Ans:

  14. (D2 − 2D + 1)y = (1 + ex)2.

    Ans:

     

  15. (D2 − 4)y = 3e2x − 4e−2x.

    Ans:

  16. (D2 − 1)y = cosh x.

    Ans:

(2) Q(x) = sin ax cos ax where ‘a’ is a constant

 

Case (i)   f(−a2) ≠ 0

We have D(sin ax) = a cos ax

 

 

and generally,

 

 

for any integer ‘r’.

 

∴      f(D2)sin ax = f(−a2)sin ax

 

Operating on both sides by , we have

 

 

if  f(−a2) ≠ 0.

 

Rule Replace D2 by −a2 if f(−a2) ≠ 0

The same rule applies if Q(x) = cos ax

 

 

if  f(−a2) ≠ 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(−a2) = 0,

i.e.,

The above rule fails if f(−a2) = 0.In this case, (D2 + a2)f(D2) so that we can write

 

f(D2) = (D2 + a2)ϕ(D2) where ϕ(−a2) ≠ 0.

 

Let us evaluate

 

 

Equating the real and imaginary parts,

 

 

We have, if ϕ(−a2) ≠ 0,

 

 

General case: If f(D) contains old powers of D, then we can arrange the even and odd powers of D so that

 

f(D) = f1(D2) + Df2(D2)

 

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 D2 in the denominator

 

 

The general solution is

 

Example 3.2.19   Solve (D2 − 1)y = sin x cos x.

Solution   To find the complementary function, we have to solve

 

(D2 − 1)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is,

 

Example 3.2.20   Solve (D2 − 4D)y = ex + sin 3x cos 2x.

Solution   To find the complementary function, we have to solve

 

(D2 − 4D)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

Example 3.2.21   Solve (D3 + 2D)y = e2x + cos (3x + 7).

Solution   To find the complementary function, we have to solve

 

(D3 + 2D)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

EXERCISE 3.6

Solve the following:

  1. y″ + 4y′ + 47 = 4cos x + 3sin x; y(0) = 1, y′(0) = 0.

    Ans: y = (1 + x)e−2x + sin x

  2. (D3 − 1)y = ex + sin 3x + 2.        [JNTU 2004]

    Ans:

  3. (D2 − 4)y = 2 cos2 x.

    Ans:

  4. (D2 + 1)y = sin x sin 2x.      [JNTU 2003]

    Ans:

  5. (D2 + 9)y = cos 3x + sin 2x.

    Ans:

  6. (D2 − 1)y = sin (x + 5).

    Ans:

  7. (D2 + 4)y = ex + sin 2x + cos 2x.

    Ans:

  8. (D2 − 4D + 3)y = sin 3x + cos 2x.

    Ans:

(3) Q(x) = xm where m is a positive integer

To evaluate , we expand in increasing powers of D up to Dm and apply on xm.

Example 3.2.22   Evaluate .

Solution

 

Example 3.2.23   Evaluate (D2 + 3D)−1x.

Solution

 

Example 3.2.24   Solve (D3 − 1)y = x3.

Solution   To find the complementary function, we have to solve

 

(D3 − 1)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

Example 3.2.25   Solve (D3D2)y = 1 + x3.

Solution   To find the complementary function, we have to solve

 

(D3D2)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

Example 3.2.26   Solve (2D2 + D)y = x2.

Solution   To find the complementary function, we have to solve

 

(2D2 + D)y = 0

 

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

EXERCISE 3.7

Solve the following:

  1. y′″ − 2y″ − y′ − 2y = 1 − 4x3.      [JNTU 2004(3)]

    Ans: (5 + cx2)x3y5 = 2

  2. (D2 + D + 1)y = x3.        [JNTU 2004]

    Ans:

  3. (D2 − 3D − 2)y = x2.

    Ans:

  4. (D3D2D + 1)y = 1 + x2.     [JNTU 1997S]

    Ans: y = c1ex + (c2 + c3x)ex + x 2 + 2x + 5

  5. (D3 + 2D2 + D)y = e2x + x2 + x + sin 2x.      [JNTU 2003S(1)]

    Ans:

  6. (D2 + 5D + 4)y = x2.       [JNTU 2003S(2)]

    Ans:

(4) Q(x) = QaxV(x) where ‘a’ is a constant and V(x) is a function of x

By the theorem on exponential shift, we have

 

 

where V1 is a function of x.

Applying on both sides, we have

 

 

 

Let   f(D + a)V1 = V

 

 

From Eqs. (3.75) and (3.76), we get

 

 

The effect of shifting the exponential factor eax 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 (D4 − 1)y = ex cos x.

Solution   To find the complementary function, we have to solve (D4 − 1)y = 0

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

Example 3.2.31   Solve (D2 + 4D + 4)y = e−2x cos x.

Solution   To find the complementary function, we have to solve (D2 + 4D + 4)y = 0

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

y(x) = y(x) = yc + yp = (c1 + c2x)e−2xe−2x cos x.
EXERCISE 3.8

Solve the following:

  1. (D2 − 7D + 6)y = e2x(1 + x).         [JNTU 2003]

    Ans:

  2. (D2 − 2D + 1)y = xex.

    Ans: y =(c1 + c2x)exex(x sin x + 2 cos x)

  3. (D3 − 3D2 + 3D − 1)y = x2ex.          [JNTU 2002S]

    Ans:

  4. (D2 + 4D + 3)y = ex sin x + x.        [JNTU 1996, 1999]

    Ans:

  5. (D2 − 4)y = x sinh x.          [JNTU 2000]

    Ans:

  6. (D3 − 7D2 + 14D − 8)y = ex cos 2x.      [JNTU 2003S(3)]

    Ans:

  7. (D2 − 4D + 1)y = e2x cos3x.

    Ans:

(5) Q (x) = xV(x) where V(x) is a function of x

Theorem 3.2.32 Prove that

 

 

where V1 is a function of x.

Proof For n = 1

 

D1(xV1) = x(DV1) + V1 (true)

 

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)2y = x cos x.

Solution   To find the complementary function, we have to solve (D + 1)2y = 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 (D2 + 9)y = x sin 2x.

Solution   To find the complementary function, we have to solve (D2 + 9)y = 0.

The auxiliary equation is

 

m2 + 9 = 0 ⇒ m = ±3i
yc = c1 cos 3x + c2 sin 3x

 

To find the particular integral,

 

 

Here  f(D) = D2 + 9 f′(D) = 2D

 

 

The general solution is

 

Example 3.2.35   Solve (D − 2)2y = x2 sin x + e2x + 3.   [JNTU 2001S]

Solution   yc = (c1 + c2x)e2x

 

 

The general solution is

 

 

Alternatively, P.I1 can also be obtained as follows:

 

EXERCISE 3.9

Solve the following:

  1. (D + 1)y = x cos x.        [JNTU 2003]

    Ans:

  2. (D2 + 4)y = x sin x.    [JNTU 2003]

    Ans:

  3. (D2 − 1)y = x sin 3x + cos x.   [JNTU 1997]

    Ans:

  4. (D2 + 1)y = x2 sin x.

  5. (D − 1)2y = xex sin x.       [JNTU 2002]

    Ans: y = (c1 + c2x)exex (2 cos x + x sin x)

  6. (D2 − 1)y = x sin x + x2ex.

    Ans:

  7. (D2 + 1)y = x2 sin x.

  8. (D2 − 1)y = x sin x + ex(1 + x2).     [JNTU 199S]

    Ans:

  9. (D2 + 3D + 2)y = xex sin x.

    Ans:

  10. (D2 + 1)y = ex + x3 + ex sin x.        [JNTU 1998]

    Ans:

  11. .

    Ans: y = (c1 + c2x)exex log x

  12. y″ + 4y′ + 20y = 23 sin t − 15 cos; y(0) = 0, y′(0) = 1.        [JNTU 2004(4)]

    Ans:

  13. (D3 − 1)y = ex + sin 3x + 2.        [JNTU 2004(1)]

  14. (D2 + 1)x = t cos 2t x = x′ = 0 at t = 0.       [JNTU 2002]

    Ans:

  15. (D2 − 4D + 4)y = 8x2e2x sin 2x.    [JNTU 2001S]

    Ans: y = (c1 + c2x)e2x − 2e2x · [2x cos 2x + (x2 − 3/2)sin 2x]

3.2.2 Linear Equations with Variable Coefficients—Euler–Cauchy Equations (Equidimensional Equations)

An equation of the form

 

 

where ai 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

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 3.2.38

 

 

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

 

(m + 1)2 = 0   ⇒   m = −1, −1

 

 

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

 

m2 − 2m + 2 = (m − 1)2 + 1 = 0 ⇒ m − 1 = ±im = 1 ± i

 

 

 

To find the particular integral,

 

 

The general solution is

 

y = [c1 cos(log x) + c2 sin(log x)]x + x log x.

Example 3.2.40

 

 

Solution

 

 

so that xD = θ,   x2D2 = θ(θ − 1)

The given equation becomes

 

 

which is a linear equation with constant coefficients.

To find the complementary function, we have to solve

 

2 + 1)y = 0

 

The auxiliary equation is

 

m2 + 1 = 0 ⇒ m = ±i

 

 

 

To find the particular integral,

 

 

The general solution is

 

Example 3.2.41

 

 

Solution   Put x = ex     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

3 − θ2 + 2)y = 0

 

The auxiliary equation is

 

m3m2 + 2 = 0   ⇒   m = −1 is a root

 

By synthetic division

 

m2 − 2m + 2 = (m − 1)2 + 1 = 0 ⇒ m = 1 ± i

 

 

To find the particular integral,

 

 

 

The general solution is

 

EXERCISE 3.10

Solve the following:

  1. Ans:

  2. Ans:

  3. Ans:

  4. Ans: y = c1x2 + c2x3x2 log x

  5. Ans:

  6. Ans:

  7. Ans:

  8. Ans:

  9. Ans:

  10. Ans:

  11. Ans:

  12. Ans:

  13. Ans: y = x(c1 cos (log x) + c2 sin(log x)) + x log x

  14. Ans:

  15. Ans: y = (c1 + c2 log x)x + log x + 4

  16. Ans:

  17. Ans:

3.2.3 Legendre's Linear Equation

An equation of the form

 

 

where ar(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 = ez 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 = ez or z = log(3x + 2)

 

 

The given equation becomes

 

 

To find the complementary function, we have to solve (θ2 − 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) = e2 or z = log(1 + 2x) so that

 

 

The given equation becomes

 

 

To find the complementary function, we have to solve (θ − 2)2y = 0.

The auxiliary equation is

 

 

To find the particular integral,

 

 

The general solution is

 

y = [c1 + c2 log (1 + 2x)](1 + 2x)2 + (1 + 2x)2[log (1 + 2x)]2.
EXERCISE 3.11

Solve the following:

  1. [(5 + 2x)2D2 − 6(5 + 2x)D + 8]y = 0.

    Ans:

  2. [(2x + 1)2D2 − 2(2x + 1)D − 12]y = 6x.

    Ans:

  3. [(x + 2)2D2 − (x + 2)D + 1]y = 3x + 4.

    Ans:

  4. [(1 + x)2D2 + (1 + x)D = 1]y = 4 cos[1 + x)].

    Ans: y = c1 cos log(1 + x) + c2 sin log (1 + x) + 2 log(1 + x)sin log(1 + x)

  5. [(1 + x)2D2 + (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 non-homogeneous equation

 

 

where P, Q, R are functions of x.

Let        y = C1u(x) + C2v(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 yp 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 yp 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 yp. From Eqs. (3.129) and (3.133) we obtain the general solution or complete solution.

 

Note 3.2.44 Since the arbitrary constants C1 and C2 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.

  1. (D2 + 1)y = x sin x;
  2. (D2 + a2)y = tan ax;                                [JNTU 2003]

Solution   (1) The given equation is

 

 

To find the complementary function, we have to solve (D2 + 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 (D2 + a2)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

 

yc = c1 cos x + c2 sin x; P = 0, Q = 1, R = sec x

 

Let y = A cos x + B sin x, where A, B are functions of x and

 

 

The complete solution is

 

y = c1 cos x + c2 sin x + cos x log x + x sin x.

Example 3.2.48   Solve the equation (D2 − 2D)y = ex sin x.

Solution   To find the complementary function, we have to solve (D2 − 2D)y = 0.

To auxiliary equation is

 

 

Let y = A + Be2x where A, B are functions of x.

 

 

The complete solution is

 

Example 3.2.49   Solve the equation (D2 + 3D + 2)y = ex + x2.

Solution   Here P = 3, Q = 2, R = ex + x2

The auxiliary equation is

 

m2 + 3m + 2 = 0   ⇒   m = −1, −2

 

The complementary function is

 

yc = c1ex + c2e−2x; u(x) = ex, v(x) = e−2x

 

Let y = Aex + 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:

  1. (D2 + 1)y = cosec x cot x.

    Ans: y = c1 cos x + c2 sin x − cos x log(sin x) − x sin x

  2. (D + 1)2 y = ex log x.

    Ans:

  3. (D2 + a2)y = a2 sec ax.

    Ans: y = c1 cos ax + c2 sin ax + ax sin ax + cos ax log(cos ax)

  4. (D2 − 3D + 2)y = xex + 2x.

    Ans:

  5. (D2 + 4)y = tan 2x.     [JNTU 2003, 2002, 1997, 1994]

    Ans:

  6. ((x − 1)D2xD + 1)y=(x − 1)2.

    [Hint: ex ,x are basis solutions.]

    Ans: y =c1 ex + c2x −1(1+x+x2)

  7. Ans: y = c1e3x + c2xe3xe3x logx

  8. (D3 + D)y = cosec x.

    Ans: y=c1 + c2 cos x + c3 sin x − log(cosec x + cot x)

    −cos x logsin xx sin x

  9. (D2 − 2D + 2)y = ex tan x.

    Ans: y=ex(c1 cos x + c2 sin x) − ex 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 f1,f2,g1,g2 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 = f1(D)g2(D) − f2(D)g1(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

 

(D+3)(D−3)x−(D−3)2y=0−10x−(D+3)2y=0

 

On subtraction, (D2+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 Δxy respectively replace c1,c2 in Δ by

 

 

 

 

 

 

The auxiliary equation is

 

 

The complimentary function is

 

yc = (c1 + c2t)e−3t

 

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 D2 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:

  1. Ans:

  2. Ans: x=c1 cosatc2 sinat

    y=c1 sinatc2 cosat

  3. (D+2)x+(D+1)y=t; 5x+(D+3)y=t2.

    Ans:

  4. with x(0)=6, y(0)−2.

    Ans: x=4et + 2e−t; y=−ete−t

  5. Ans:

  6. (D+6)yDx=0;(3−D)x−2Dy=0; with x=2,y=3

    when t = 0.

    Ans: x=4e2t−2e−3t; y=e2t+2e−3t

  7. (D+2)x+(D−1)y=−sint;

    (D−3)x(D+2)y=4cost.

    Ans:

  8. D2y=x−2,D2x=y+2.

    Ans: x=c1 sint + c2 cost + c3et + c4et + 2; y=−c1 sintc2 cost + c3et + c4et −2

  9. (D+1)x+(D−1)y=et; (D2+D+1)x+(D2−D+1)y=t2.

    Ans: