Differential Equations- II | Mathematics for Competitive Exams PDF Download

Introduction

Definition
The general form of linear equation of order n is
    ...(1)
where Q and P₁, P₂, P₃, ...., P_n are all constants or functions of x.
If P₁, P₂, P₃, ..... P_n are all constants (Q may not be constant), then the equation is said to be a linear differential equation with constant coefficients.

Differential Operator D 

The derivatives of different orders of the dependent variable y w.r.t. the independent variable x are expressed as follows:

Clearly  have been used for the symbols D, D², D³, .........., Dⁿ respectively which are differential operators of different orders. Using these operators, the differential equation (1) can be written as follows:
(Dⁿ + P₁D^n-1 + P₂D^n-2 + .... + P_n)y = Q
or, in brief f(D)y = Q ...(2)
It can be easily seen that the above operators obey the fundamental law algebra:
(i) Law of indices : D^m.Dⁿ = D^m+n 
(ii) Commutativity : D^m + Dⁿ = Dⁿ + D^m 
D^m.Dⁿ = Dⁿ.D^m = D^m+n 
(D - α)(D - β) = (D - β)(D - α), where α, β are constants.
(iii) Distributivity : D^m (u + v) = D^m (u) + D^m (v).
D can be operated like an algebraic symbol.

For example, 
(D² + 2D + 1) sin x = D² (sin x) + 2D (sin x) + sin x
= D[D (sin x)] + 2D (sin x) + sin x
= D(cos x) + 2cos x + sin x
= - sin x + 2cos x + sin x = 2 cos x 

Inverse Operator [D^-1] 

We know the differentiation and Integration are the inverse operations. Therefore if D is used for the Derivative, then the symbol D^-1 or 1/D is used for the inverse operation i.e. integration.

For example,

Linearly Independent (LI) and Linearly Dependent (LD) 

The collection of functions y₁, y₂........y, is called LI if the only solution of the equation c₁y₁ + c₂y₂ + .... c_n.y_n = 0 is c₁ = c₂ = .... = c_n = 0 where c₁’s are constant otherwise it is called LD.
Hence by definition, it is clear that
Two functions are LD iff one function is a constant multiple of the other.
For example : If y₁ = x and y₂ = 3x then y₁ and y₂ are LD, because then non zero solutions of c₁x + c₂(3x) = 0 are c₁ = - 3 and c₂ = 1.

Wronskian Collection Definition 

The Wronskian defines the determinant of the collection of n functions W (y₁, y₂, .... , y_n) as follows:

where dash (‘) denotes the derivative.

Wronskian Test for LI 
(i) The functions y₁, y₂, .... y_n are LI if and only if W(y₁, y₂, ...., y_n) ≠ 0
(ii) The functions y₁, y₂, ..... y_n are LD if and only if W (y₁ , y₂, ....... y_n) = 0
Example : The function e^x and x e^x are LI because x e^x is not a constant multiple of e^x.

Homogeneous and Non Homogeneous Linear Differential Equations

When Q = 0, f(D)y = 0 is called homogeneous otherwise Non Homogeneous.
Properties: 
1. If y = f(x) is any solution of the equation f(D)y = 0, then y = cf(x) is also a solution of the equation, where c is any arbitrary constant.
2. If y = f₁(x) and y = f₂(x) are the two solutions of f(D)y = 0, then y = c₁ f₁(x) + c₂ f₂(x) will also a solution of the equation where c₁ and c₂ are arbitrary constants.

Complementary Function and Particular Integral 

The general form of the equation is
f(D)y = 0       ...(1)
If Q = 0, then it will be of the form
f(D)y = 0       .....(2)
If the solution of the equation (2) is taken as y = y₁, then on substituting the value of y, we will see that y = c₁y₁ will also be a solution where c₁ is an arbitrary constant.
Similarly if n LI function
y = y₁, y = y₂, ........y = y_n 
are also the solutions of the equation (1), then
f(D)y₁ = 0 = f(D)y₂ = f(D)y₃ = .... = f(D)y_n        ......(3)
Now if c₁, c₂, c₃.......c_n are arbitrary constant, then c₁y₁ + c₂y₂ + c₃y₃ + .... + c_ny_n will also be a general solution of the equation (2), because
f(D)(c₁y₁ + c₂y₂ + .... + c_n y_n)
= c_nf_n(D)y₁ + c₂f(D)y₂ + .... c_nf(D)y_n 
= c₁(0) + c₂(0) + .... + c_n(0) = 0      [by (3)]
or    f(D)u = 0
where u = c₁y₁ + c₂y₂ + .... c_ny_{n      .....}(4)
Clearly u is a general solution of the equation (2) because the number of arbitrary constants is equal to its (u) order.
Again if PS of equation (2) is v, then
f(D)v = Q         .......(5)
then f(D)(u + v) = f(D)u + f(D)v
= 0 + Q        [from (4) and (5)]
= Q
Therefore the general form of CS of equation (1) is y = (c₁y₁ + c₂y₂ + ..... + c_ny_n) + v
or y = u + v
In this solution u and v are called the Complementary Function (CF) and Particular Integral (PI) of the solution respectively.
Therefore the GS of any equation is of the following form :
y = CF + PI       .....(6)
Remarks : 
1. If in equation (1), Q = 0, then its CS (complete solution) will be y = CF.
2. In CF, the number of arbitrary constants is equal to the order of equation.
3. In PI, there is no arbitrary constant.

Methods for Finding CF 

In order to find the CF of the given DE, first of all assume the function Q on the RHS of the equation to zero, i.e., write the equation in the following form

or, in the symbolic form,
(Dⁿ + P₁ D^n-1 + .... + P_n) y = 0
or f(D) y = 0         ....(1)
Let y = e^mx be the solution of (1), then substituting the value of Dy, D²y,  ....., Dⁿy
e^mx [mⁿ + P₁ m^n-1 + P₂m^n-2 + .... + P_n] = 0
or mⁿ + P₁ m^n-1 + P₂m^n-2 + .... + P_n = 0    [∵ e^mx ≠ 0]
From this it is clear that y = e^mx will be a solution of (1), if m is a solution of the equation mⁿ + P₁m^n-1 + P₂m^n-2 + .... + P_n = 0
or f(m) = 0      .....(2)
The above equation (2) is known as the Auxiliary Equation or in brief AE of the equation (1).
Case I : When all the roots of AE are different : 
Suppose m₁, m₂, .... , m_n are the roots of AE f(m) = 0 and are all different.
 are the n distinct LI solutions and
The GS will be:
y = c₁y₁ + c₂y₂ + .... + c_ny_n = C.F.
     ....(1)

Case II : When roots of AE are same (repeated) : 
Suppose AE f(m) = 0 has two equal roots (m₁ = m₂) and rest of different roots are :
m₁ = m₂ > m₃, m₄........ m_n 
In this case the solution will be :
     ...(2)
If three roots of AE are equal say m₁ = m₂ = m₃, then in this case the solution will be
 ...(3)
Similarly solution can be written in case of more equal roots.

Case III : When the roots of AE are complex : 
Suppose a pair of roots of AE are complex say α ± β, then the solution of the corresponding equation is :
 
which can be written in the convenient form as :
y = e^αx (c₁ cos βx + c₂ sin βx)    ....(4)
The above form can be converted in the following compact form :
    ....(5)
Hence the CS of the equation will be :

Case IV : When roots of AE are complex and equal (repeated) :
Let the AE has a pair of complex roots α ± β repeated once, then the solution of the corresponding equation will be :
y = (c₁ + c₂x)e^(α+iβ) + (c₂ + c₄x)e^(α+iβ)x 
which can be written in the simplest form as follows :
y = e^ax[(c₁ + c₂x)cos βx + (c₃ + c₄x)sin βx]   ...(6)
and CS will be :

When roots of AE are real and in the surd form : 
Such roots always occur in pairs. Let the roots be in the surd form a±√b, then the solution of the corresponding equation will be :
   ......(7)

Example 1:  

The given equation can be written as (D³ - 3D² + 4) y = 0
The AE of the given equation is m³ - 3m² + 4 = 0 

or (m + 1) (m² - 4m + 4) = 0 

or (m + 1) (m - 2)² = 0 ⇒ m = 2, 2, -1
Therefore the GS is y = (c₁ + c₂x)e^2x + c₃e-x 

Example 2: Solve : (D⁴ + a⁴) y = 0

The AE of the given equation is m⁴ + a⁴ = 0 

or m⁴ + 2m²a² + a⁴ - 2m²a² = 0 

or (m² + a²)² - (√2 ma)² = 0
or (m² + a² + √2 ma) (m² + a² - √2 ma) = 0
when m² + √2 ma + a² = 0 then

and when m² - √2 ma + a² = 0, then

Therefore GS is

Example 3: Solve : (D⁴ + 2D³ + 3D² + 2D + 1)y = 0 

The AE of the given equation is m⁴ + 2m³ + 3m² + 2m + 1 = 0
or (m² + m + 1)² = 0

Therefore GS is

Example 4:

when for t = 0, θ = θ₀ and dθ/dt = 0.

The given equation can be written as (ℓD² + g)θ = 0 

The AE of the given equation is ℓm² + g = 0

Therefore GS is

∵ For t = 0, θ = θ₀ 

∴ From (1), θ₀ = c₁ cos + c₂ sinθ ⇒ c₁ = θ₀.    ....(2)

Again differentiating (1),

or

Now since at t = 0, dθ / dt is also zero, therefore

⇒ c₂ = 0     ....(3)
Substituting the values of c₁ and c₂ in (1), the required solution is

Symbolic operator  
(D² - D) (x² - x) = 3 - 2x,   we have

General Method of Finding PI
First of all, we shall find the value of  where α is any constant.

∴ (D - α)y = Q
or Dy - αy = Q
or
which is a linear equation in the variable y. Multiplying both sides by the Integrating Factor IF = e^-αx and then integrating, we have
ye^-ax = ∫Q.e^-αx dx + c
or   y = e^αx ∫Q.e^-αx dx + ce^αx 
Since we want a particular solution of this equation, therefore if c can be taken to be zero, then y = e^αx∫e^-αx.Qdx
or
   ....(9)
Now we can find the value of 1/f(D).Q with the help of (9)
The value of the integral is obtained by resolving f(D) into linear factors and 1/f(D) into fractions and integrating, we obtain

Example 1: Solve (D² + 3D + 2)y = e^x 

The AE of the given equation is m² + 3m + 2 = 0
or (m + 2)(m + 1) = 0
⇒ m = - 2, - 1
Therefore
C.F. = c₁e^-2x + c₂e^-x        ......(1)



     ...(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)
y = c₁e^-2x + c₂e^-x + (1/6)e^x 

Example 2: Solve : (D² + a²) y = sec ax 

The AE of the given equation is m² + a² = 0
∴ m = ± ia
Therefore C.F. = e^0x [c₁ cos ax + c₂ sin ax]
= c₁ cos ax + c₂ sin ax    ....(1)





Therefore the GS is y = C.F. + P.l.
i.e.

Shorter Method for Finding PI in Some Special Cases 

If Q in the different equation f(D)y = Q is of the following special form, then their PI can be found by the short methods discussed below :
1. e^ax, where a is any constant.
2. sinh ax, cosh ax.
3. xⁿ, where n is a positive integer.
4. e^ax V, where V is any function of x.
5. xV, where V is any function of x. 

(1) PI for e^ax, where a is any constant and F(a) ≠ 0
PI for e^ax :
Here we have to find the value of  when f(a) ≠ 0.
From successive differentiation we know that
Dⁿ(e^ax) = aⁿ e^ax 
Now f(D) (e^ax) = [Dⁿ + P₁D^n-2 + P₂D^n-2 + ...+ P_n] (e^ax)
= Dⁿ (e^ax) + P₁D^n-1 (e^ax) +...+ P_n(e^ax)
= aⁿ e^ax + P₁a^n-1e^ax + ...+ P_ne^ax 
= (aⁿ + P₁a^n-1 + P₂d^n-2 + ...+ P_n)e^ax 
∴ f(D)(e^ax) = f(a)e^ax     .......(1)

      [by (1)]

Now if f(a) ≠ 0, then
       ......(X)

Special case: 
(i) When f(a) = 0, then 
i.e., when a is a non-repeated root of f(D) = 0
∴ (D - a) is a factor of f(D)
Let f(D) = (D - a) ϕ(D) where ϕ(a) ≠ 0.
    [∵ ϕ(a) ≠ 0]

(ii) If a is a root repeated r times, 
i.e., (D - a)^r is a factor of f(D), then let
f(D) = (D - a)r Ψ(a) where Ψ(a) ≠ 0
   [∵ Ψ(a) ≠ 0]



Continuing in this manner, we may show that

        .....(XI)
(iii) If Q = sinh(ax) or cosh(ax), then

     .....(XII)

Example 1: Solve : {(D + 2)(D - 1)³} y = e^x 

The AE of the given equation is (m + 2) (m - 1)³ = 0
⇒ m = -2, 1, 1, 1
Therefore C.F. = c₁e^-2x + (c₂ + C₃x + c₄x²) e^x           ....(1)


           ....(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 2: Solve : {(D - 1)²(D² + 1)²} y = e^x 

The AE of the given equation is (m - 1)² (m² + 1)² = 0
∴ m = 1, 1, ± i ± i
Therefore C.F. = (c₁ + c₂x) e^x + e^0x [(c₃ + c₄x) cos x + (c₅ + c₆x) sin x]
= (c₁ + c₂x) e^x + (c₃ + c₄x) cos x + (c₅ + c₆x) sin x     ....(1)


    ....(2)
Therefore the GS is

Example 3: Solve : (D³ + 1) y = (ex + 1)² 

The AE of the given equation is m³ + 1 = 0
or (m + 1) (m² - m + 1) = 0
∴  m = -1, 1/2 (1 ± √3i)



Therefore, the GS is y = C.F. + P.l. = (1) + (2)

Example 4: Solve : (D² - 1)y = cosh x 

The AE of the given equation is m² - 1  = 0
or (m - 1)(m + 1) = 0 ⇒ m = 1, -1
Therefore C.F. = c₁e^x + c₂e^-x     ....(1)


     ....(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

Working Rule

Example 5: Solve (D² - 2D + 5)y = e^-x.

Flere auxiliary equation is m² - 2m + 5 = 0, whose roots are m = -1 ± 2i
∴ C.F. = e^-x[C₁ cos2x + C₂ sin2x], where C₁ and C₂ are arbitrary constants.


∴ The required solution is y = C.F. + P.l. i.e.

Special case: If a is a root repeated r times

When a is non-repeated:

Example 6: Solve (D² - 6D + 9)y = 4x^3x.

Here the auxiliary equation is m² - 6m + 9 = 0 or m = 3, 3.
∴ C.F. = (C₁x + C₂)e^3x, where C₁, C₂ are arbitrary constants and P.l.

 
∴ The required solution is y = C.F. + P.l. or y = (C₁x + C₂)e^3x + 2x²e^3x.

(2) P.l. for sinax or cosax and F(-a²) ≠ 0
We have to find 1/f(D²) sin ax and 1/f(D²) cos ax when f(-a²) ≠ 0.
We know that D(sin ax) = a cos ax
D²(sin ax) = -a² sin ax
D³(sin ax) = -a³ cos ax

In general, (D²)ⁿ sin ax = (-a²)ⁿ sin ax
∴ f(D²) sin ax = f(-a²) sin ax



Special Case : If (D² + a²)ⁿ is one of the factor, then the above method fails
∵ f(-a²) = 0 and in such a case we proceed as follows :
We know that e^iax = cos ax + i sin ax



Equating real and imaginary parts, we have

Remark : 
1. If f(D) contains odd powers of D also, it can be put in the form
f(D) = f₁(D²) + Df₂(D²).
2. When f (-a²) = 0, it is convenient to use the procedure given above instead of using the formula directly.

Example 1: Solve : (D² - 4D + 4)y = e^2x + sin 2x 

The AE of the given equation is m² - 4m + 4 = 0
or (m - 2)² = 0 ⇒ m = 2, 2
Therefore C.F. = (c₁ + c₂x)e^2x     ....(1)



Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 2: Solve : (D² + a²)² y = sin ax 

The AE of the given equation is (m² + a²)² = 0
∴ m = ± ai, ± ai
Therefore C.F. = e^0x [(c₁ + c₂x) cos ax + (c₃ + c₄x) sin ax]
= (c₁ + c₂x)cos ax + (c₃ + c₄x) sin ax.      ....(1)


Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 3: Solve : (D³ - 3D² + 4D - 2) y = e^x + cos x 

The AE of the given equation is m³ - 3m² + 4m - 2  = 0
or (m - 1) (m² - 2m + 2) = 0

Therefore C.F. = c₁e^x + e^x(c₂ cos x + c₃ sin x)     ....(1)



Therefore, the GS is y = C.F. + P.l. = (1) + (2)



when F(-a²) = 0 
(I) Let F(D) = D² + a² and Q = sinax.

(II) Let F(D) = D² + a² and Q = cosax.

Example 4: Solve (D² + 1)y = sin2x.

Here the auxiliary equation is m² + 1 = 0 or m = ± i.
∴ C.F. = C₁ cosx + C₂ sinx, where C₁ and C₂ are arbitrary constants and

∴ The requires solution is

Example 5: Solve (D² + a²)y = cosax.

Here the auxiliary equation is m² + a² = 0 or m = ± ai
∴ C.F. = C₁cos ax + C₂sin ax, where C’ s are arbitrary const, and


∴ The requires solution is y = C.F. + P.l.

(3) PI for x^m 

Working Rule 
Take out the lowest degree term from F(D) and the remaining factor will be of the form [1 + f(D)] or [1 - f(D)] and this factor can be taken in the numerator with a negative index which can be expanded with the help of Binomial Theorem. The expansion is to be carried upto the term D^m, since D^m+1x^m = 0, D^m+2x^m = 0 and all other higher differential coefficients of x^m are zero.

Example 1: Solve (D² - 13D + 12)y = x.

Here auxiliary equation is m² - 13m + 12 = 0 or (m - 12)(m - 1) = 0 or m = 1, 12
∴ C.F. = C₁e^x + C₂e^12x, where C’s are arbitrary constants


∴  Required solution is y = C.F. = P.l.

Example 2: Solve (D² - 5D + 6)y = x.

Its C.F. = C₁e^2x + C₂e^3x, where C₁ and C₂ are arbitrary constants.


∴ The complete solution is y = C.F. + P.l.

Example 3: Solve : (D³ + 3D² + 2D) y = x² 

The AE of the given equation is m³ + 3m² + 2m = 0
or m(m + 1)(m + 2) = 0 ⇒ m = 0, - 1, - 2
Therefore C.F. = c₁e^0x + c₂e^-x + c₃e^-2x 
= c₁ + c₂e^-x + c₃e^-2x      .....(1)



        .....(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 4: Solve : (D³ + 2D² + D)y = e^2x + x² + x 

The AE of the given equation is m³ + 2m² + m = 0
or m(m² + 2m + 1) = 0
or m (m + 1)² = 0 m = 0, - 1 , - 1
Therefore CF = c₁e^0x + (c₂ + c₃x)e^-x 
= c₁ + (c₂ + c₃x)e^-x         .....(1)


     ....(2)
the GS is y = CF + PI = (1) + (2)

Example 5: Solve : (D⁵ - D) y = 12e^x + 8 sin x - 2x 

The AE of the given equation is m⁵ - m = 0
or, m(m - 1)(m + 1)(m² + 1) = 0
∴ m = 0, 1, -1 , ± i
C.F. = c₁e^0x + c₂e^x + c₃e^-x + (c₄ cos x + c₅ sin x)
Therefore = c₁ + c₂e^x + c₃e^-x + c₄ cos x + c₅ sin x    ....(1)



      [by IV]
= 3xe^x + 2xsinx + x²        .....(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)
i.e. y = c₁ + c₂e^x + c₃e^-x + c₄ cos x + c₅ sin x + 3xe^x + 2x sin x + x².

(4) PI for eax.V, where V is a function x
Here we have to find the value of
Consider any function V₁ of x to be determined later on in terms of the given function V. By successive differentiation, we have
D(e^ax V₁) = e^ax DV₁ + ae^ax V₁ = e^ax (D + a) V₁

Dⁿ(e^ax V₁) = e^ax (D + a)ⁿ V₁ 
∴ f(D) (e^ax V₁) = e^ax f(D + a) V₁    .......   (1)
 Put f(D + a) = V₁ so that
     ....(2)
Eliminating V₁ from (1) and (2), we obtain


      ....(XVII)
Now the value of   can be determined by the methods discussed earlier.

Remark : This method can also be used to find the value of  In this case V = 1. 

Example 1: Solve : (D³ - 7D - 6) y = e^2x (1 + x) 

The AE of the given equation is m³ - 7m - 6 = 0
or (m + 1) (m² - m - 6) = 0
or (m + 1) (m - 3) (m + 2) = 0
∴  m = -1, -2, 3
Therefore C.F. = c₁e^-x + c₂e^-2x + c₃e^3x    ......(1)



     ......(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 2: Solve : (D² - 2D + 5) y = e^2x sin x.

The AE of the given equation is m² - 2m + 5 = 0

Therefore C.F. = e^x(c₁ cos 2x + c₂ sin 2x)      ...(1)

     [by XVII]


   ....(2)
Therefore, the GS is y = C.F. + P.l. = (1) + (2)

Example 3: Solve : (D² - 2D + 1) y = x²e^3x 

The AE of the given equation is m² - 2m + 1 = 0
or (m - 1)² = 0 m = 1, 1
Therefore C.F. = (c₁ + c₂x)e^x     ....(1)


      ....(b)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 4: Solve : (D⁴ + D² + 1) y = ax² + be^-x sin 2x 

The AE of the given equation is m⁴ + m² + 1 = 0
or (m² + 1)² - m² = 0
or (m² + m + 1) (m² - m + 1) = 0

Therefore






Therefore the GS is y = C.F. + P.l. = (1) + (2)


When Q is of the form e^ax.V, where V is any function of x then

Example 5: Solve (D² - 2D + 1)y = e^x.x².

Here the auxiliary equation is m² - 2m + 1 = 0 or m = 1, 1.
∴  C.F. = (C₁x + C₂)e^x, where C₁ and C₂ are arbitrary constants and


∴ The required solution is y = C.F. + P.l.

or (m - 1) 2 = 0 ⇒ m = 1 , 1
Therefore C.F. = (c₁ + c₂x)e^x     ....(1)



   .....(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 6: Solve : (D² + 2D + 1) y = x cos x 

The AE of the given equation is m² + 2m + 1 = 0
or (m + 1)² = 0 ⇒ m = -1, -1
Therefore C.F. = (c₁ + c₂x)e^-x    .....(1)



   ....(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

Example 7: Solve : (D² - 4D + 4) y = 8 x² e^2x sin 2x. 

The AE of the given equation is m² - 4m + 4 = 0
or (m - 2)² = 0 ⇒ m = 2, 2
Therefore C.F. = (c₁ + c₂x)e^x     ....(1)

 [by formula XVII]



    ...(2)
Therefore the GS is y = C.F. + P.l. = (1) + (2)

when Q is of the form xV, where V is any function of x

Example 8: Solve (D² - 2D + 1) 

Flere the auxiliary equation is m² - 2m + 1 = 0 or (m - 1)² = 0 or m = 1, 1.
∴ C.F. = (C₁x + C₂)e^x, where C₁ and C₂ are arbitrary constants and





∴ The required solution is y = C.F. + P.I.

To show that  dx, where Q is a function of x
Proof :    Then (D - a) y = Q, operating with  which is linear equation in y whose integrating factor is e^-ax. Its solution is ye-ax = ∫Qe^-ax dx, (neglecting the constant of integration as P.l. is required)
or y = e^ax ∫Qa^-ax dx Hence proved.

Example 9: 

Hence proved.

Formation of linear differential equations, whose solutions are specified.
While solving linear differential equations with constant coefficients, we observed that if the auxiliary equation f(m) = 0 had a root m = α, then the operator f(D) had a factor (D - α) and a term such as Ae^ax occurred in the general solution of the equation. In the same way, we observed that Bxeax appeared in the solution only when f(D) contained a factor (D - α)², Cx²e^ax only when f(D) contained (D  - α)³ and so on. A, B, C , ..... are independent of x. We also observed that terms as Ae^ax sin βx or Be^ax cos βx, which occurred in the solution, correspond to roots m = α ± iβ of the auxiliary equation or to factor {(D - α)² + β²} in the operator f(D).
With these facts in our possession, we try to form homogeneous linear differential equations, whose solutions are specified, assuming the superposition principle. For that, let the given function be (Ae^2x + Bx), with no restrictions on A and B.
For the term Ae^2x, we know that there will be a root m = 2 of the auxiliary equation and a factor (D - 2) of the operator f(D). The term Bx will appear, if the auxiliary equation has a double root m = 0, 0 and a factor D² in f(D). Hence we see that the homogeneous linear equation.
D² (D - 2) y = 0             ...(1)
that is, (D³ - 2D²) y = 0  
has y = Ae^2x + Bx + C as its general solution and hence y = Ae^2x + Bx will be a particular solution of (1), of which A and B are constants.
Again, if the specified function be
y = 10 + 5xe^x + sin x,         ...(2)
then we see that the term 10 is associated with the root m = 0 of the auxiliary equation, the term 5xex is associated with the double root m = 1, 1 of the auxiliary equation and the term sin x is associated with the pair of imaginary roots m = 0 ± i. Hence the auxiliary equation is m (m - 1)² (m² + 1) = 0
or, m⁵ - 2m⁴ + 2m³ - 2m² + m = 0
The differential equation corresponding to this auxiliary equation is
(D⁵ - 2D⁴ + 2D³ - 2D² + D) y = 0,         ...(3)
the general solution of which is y = A + (B + Cx) e^x + E sin x + F cos x.
With appropriate choice of the constants A = 10, B = 0, C = 5, E = 1, F = 0, the given relation (2) becomes a particular solution of the differential equation (3).
If again the specified function be
y = 7xe^x cos 3x      ....(4)
and if we are to find the homogeneous linear differential equation with real constant coefficients of which this function is a particular solution, then we observe that the desired equation will have its auxiliary equation f(m) = 0, whose roots will be m = 1 ± 3i and 1 ± 3i. Hence the auxiliary equation will be
{(m - 1)² + 9}² = 0
or, m⁴ - 4m³ + 24m² - 40m + 100 = 0.
Thus the desired equation will be
(D⁴ - 4D³ + 24D² - 40D + 100) y = 0,       ......(5)
the general solution of (5) will be y = e^x (A + Bx) cos 3x + e^x (E + Fx) sin 3x.
With appropriate choice of the constants A = E = F = 0 and B = 7, the given relation (4) becomes a particular solution of the differential equation (5).

Method of undetermined coefficients.
Let us consider the linear differential equation of n-th order
      .....(1)
where P₁, P₂......... P_n are constants and X is a function of x only.
In symbolic notation, this can be written as f(D) y = X, where f(D) = Dⁿ + P₁D^n-1 + ..... + P_n.
We know that the general solution of the equation (1) may be expressed as y = y_c + y_P, where y_c is the complementary function, that is, the general solution of the corresponding homogeneous equation f(D)y = 0 and y_p is the particular integral, that is, any solution of the equation (1) containing no arbitrary constants.
The method of undetermined coefficients is one of the procedures for determining y_p, when X is an exponential, a polynomial, a sine or cosine, or some combination of such functions only.
(i) Particular integral for X = e^ax, a being a constant.
The equation here is f(D)y = e^ax.      ......(2)
Since the derivatives of e^ax are constant multiples of e^ax, it is reasonable to guess that the particular integral might also be a constant multiple of e^ax and we assume that
y_p = Ae^ax       .......(3)
might be a particular solution of the equation (2).
Here A is a constant (called the undetermined coefficient) to be determined, such that (3) actually satisfies (2).
Thus we get  A f(a) e^ax = e^ax, giving  A = 1/f(a) , provided f(a) ≠ 0.
If f(a) = 0, that is if a be a root of the auxiliary equation f(m) = 0, then (3) reduces the left hand side of (2) to zero. But the right hand side of (2) being different from zero, (3) possibly cannot satisfy (2), as it stands. In this case, we take
y_p = A x e^ax as a trial solution.
If a be a double root of the auxiliary equation, then we take y_p = Ax² e^ax as a trial solution. In general, if a be a multiple root of the auxiliary equation of multiplicity r(<n), then we take y_p = ax^r e^ax as a trial solution. 

(ii) Particular integral for X = a₀x^k + a₁x^k-1 + ...... + a_k, where k is a positive integer and a₀, a₁, ......a_K are constants.
Here the equation is
f(D) y = a₀x^k + a₁x^k-1 + ...... + a_k.     .....(4)
Since the derivative of a polynomial is also a polynomial, we assume that
y_p = A₀x^k + A₁x^k-1 + ...... + A_k     .....(5)
might be a particular solution of the equation (4), if P_n ≠ 0.
Here A₀, A₁, ..... , A_k are constants (undetermined coefficients) to be determined, such that (5) satisfies (4). Substituting (5) in (4) and equating like powers of x from both sides, we get the values of A₀, A₁,   ....., A_k.
If P_n = 0 (that is, if there be no term containing y), then putting (5) in (4), we see that the highest power of x on the left hand side of (4) is x^k-1 but that on the right hand side of (4) is x^k. So, in this case, we take y_p = x(A₀x^k + A₁x^k-1 + ..... + A_k).
If P_n = 0 and P_n-1 = 0 , then we take y_p = x² (A₀x^k + A₁x^k-1 + ..... + A_k).
In general, if the last r P’s be zero but P_n-r ≠ 0, then we take
y_p = x^r (A₀x^k + A₁x^k-1 + ..... + A_k).

(iii) Particular integral for X = sin ax or cos ax.
Let the equation be f(D)y = sin ax.      .....(6)
Since the derivatives of sin ax are constant multiples of sin ax and cos ax, we take
y_p = A sin ax + B cos ax     ......(7)
as a trial solution of the equation (6), provided (7) does not satisfy the homogeneous equation f(D) y = 0. Here A and B are constants (undetermined coefficients) and are obtained by putting (7) in (6) and equating the resulting coefficients of sin ax and cos ax from both sides.
If (7) satisfies the equation f(D)y = 0, (that is if (7) be a part of y_c), then we take y_p = x(A sin ax + B cos ax) as a trial solution.

Note. If X be a linear combination of e^ax, sin bx and ∑a_nx^k-n then the particular integral is obtained by following the superposition principle.
If X be of the form e^ax sin bx or e^ax(∑a_nx^k-n) or (∑a_nx^k-n) sin bx or eax (∑a_nx^k-n) sin bx, then y_p is modified accordingly. For example, if X = e^ax sin bx, where a is not a root of the auxiliary equation and sin bx is not a part of y_c. then we take y_p = (A₁e^ax) (B₁ sin bx + C₁ cos bx), that is y_p = e^ax (A sin bx + B cos bx).

Example 1: Obtain linear differential equation with real constant coefficients, that is satisfied by the following functions :
(i) y = x² - 8 sin 4x;  (ii) y = 4e^-x cos 3x + 12e^-x sin 3x.

(i) The term x² will appear, if the auxiliary equation has a multiple root of multiplicity 3 and m = 0, 0, 0. The term sin 4x will be given by a pair of imaginary roots 0 ± 4i. The corresponding auxiliary equation is m³ (m² + 16) = 0, that is, m⁵ + 16m³ = 0.
Hence the given function is a solution of the homogeneous linear differential equation
(D⁵ + 16D³) y = 0 ...(1)
Note. The general solution of the equation (1) is y = A + Bx + Cx² + E sin 4x + F cos 4x.
The choice of the constants A = B = F = 0, C = 1 and E = - 8 justifies that the given function is a particular solution of the equation (1).
(ii) The roots of the auxiliary equation corresponding to the terms of the given function are (-1 ± 3i).
The auxiliary equation is thus (m + 1)² + 9 = 0 or m² + 2m + 10 = 0.
The corresponding linear homogeneous differential equation with constant coefficients is (D² + 2D + 10) y = 0

Example 2: Solve, by the method of undetermined coefficients the equation (D² + 1) y = 10e^2x, with the conditions y = 0 and Dy = 0 when x = 0.

Here the auxiliary equation is m² + 1 = 0 , which gives m = ± i.
Hence the complementary function (y_c) is (C. cos x + C₂ sin x), where C₁ and C₂ are arbitrary constants.
In this case, 2 is not a root of the auxiliary equation. So we assume that the particular integral is y_p = Ae^2x, where A is a constant to be determined, such that y_p satisfies the given equation.
This gives (D² + 1) y_p = 10e^2x 
or (D² + 1) Ae^2x = 10e^2x 
or (4 + 1) Ae^2x = 10e^2x, giving A = 2.
Thus the general solution is given by
y = y_c + y_p = C₁ cos x + C₂ sin x + 2e^2x.
Therefore, Dy = - C₁ sin x + C₂ cos x + 4e^2x.
To find C₁ and C₂, we use the given initial conditions.
From the condition y = 0 when x = 0, we get C₁ = -2 and from the condition Dy = 0 when x = 0, we get C₂ = -4.
Hence the required solution of the equation is y = 2e^2x - 2cos x - 4sin x.

Example 3: Solve, by the method of undetermined coefficients, the equation (D² - 3D) y = 2x² + 1.

The auxiliary equation is m² - 3m = 0, giving m = 0, 3.
Therefore y_c is (C₁ + C₂e^3x), where C₁ and C₂ are arbitrary constants.
In this case, there is no term containing y. So we assume y_p = x(A₀x² + A₁x + A₂) where A₀, A₁, A₂ are constants to be determined, such that y_p satisfies the given equation. In other words,
(D² - 3D)y_p = 2x² + 1
or, (D² - 3D) (A₀x³ + A₁x² + A₂x) = 2x² + 1
or, 6 A₀x + 2 A₁ - 3(3 A₀x² + 2 A₁x + A₂) = 2x² + 1
or, -9 A₀x² + (6 A₀ - 6 A₁)x + (2A₂ - 3 A₂) = 2x² + 1.
Equating like powers of x from both sides, we get
-9 A₀ = 2, 6 A₀ - 6 A₁ = 0, 2 A₁ - 3A₂ = 1.

Hence the general solution is

Cauchy-Euler Equation

Cauchy’s Linear Equation 
A differential equation of the form

where a₁, a₂,..,a_n are constants and Q is either a constant or a function of x alone is called a Non-Homogeneous Linear Differential Equation.

When Q = 0 then we say it is nth order Homogeneous Linear Differential Equation 

Note : The degree of x and the order of derivative is same in each term.

First method of solving non-homogeneous DE: 

The homogeneous linear differential equation is transformed into a linear equation with constant coefficients by changing the independent variable x to z by the substitution.
z = log x or x = e^z       ...(1)
Differentiating wrt x,
dz/dx = 1/x

  .....(2)


       .....(3)

      .....(4)
Similarly  can also be determined.
Now if we take d/dz = D, then by (2), (3), (4) ...., we observe that

Therefore, the given equation becomes
[{D(D - 1 ) (D - 2)...(D - n + 1)} + a₁{D(D - 1) (D - 2)...(D - n + 2)} + ... + a_n] y = Z ...(5) where Z is a function of z into which Q is changed.
Clearly equation (1) is a linear differential equation with constant coefficients, which can be solved by the earlier methods.
The Auxiliary Equation AE of the above transformed equation (5) is {m(m - 1) (m - 2)....(m - n + 1)} + a₁{m(m - 1) (m - 2) ....(m - n + 2)} + ... + a_n = 0 which is a nth degree equation in m giving n values of m.
Let m = m₁, m₂,....m_n be the roots of this equation

       ...(6)
Particular Cases : 
(i) When two roots of AE are equal, Let m₁ = m₂, then

(ii) When three roots of AE are equal, Let m₁ = m₂ = m₃, then
 
(iii) When two roots of AE are complex, say α ± i β, then
 
(iv) When the complex roots a ± i β are repeated twice, then
 
Now the solution of the equations can be obtained by finding the Particular Integrals [PI] by the methods already discussed earlier and then GS = CF + PI

Another method of solving non-homogeneous linear equation: 

In this method, the independent variable is not changed into another variable but solved directly. We know that if z = log x, then



If we denote the operator x(d/dx) = θ, then equation (6) of First method of solving nonhomogeneous DE can be expressed as :
[{θ(θ - 1) (θ - 2)...(θ - n + 1)} + a₁{θ(θ - 1)...(θ - 2)...(θ - n + 2)} +...+ a_n] y = Q ...(1)
or f(θ) y = Q ...(2)
(a) C.F. : It is obtained by solving the equation f(θ) y = 0.
Let y = x^m be a solution of (2), then by (1)
[{m(m - 1) (m - 2)....(m - n + 1)} + a₁ {m(m - 1) (m - 2)... (m - n + 2)} +....+ a_n] x^m = 0
or [m(m - 1) (m - 2)....(m - n + 1)} + a₁ {m(m - 1) (m - 2)... (m - n + 2)} +....+ a_n] = 0
[∵ x^m ≠ 0]
or f(m) = 0         ...(3)
which is a nth degree equation in m let m₁, m₂, ...,m_n be the n roots of equation (3), the GS of (2) i.e, CF is given by

Note. AE (3) is obtained by simply writing θ = m in equation (2). 

(b) PI : Let y = u be a PS of equation (1), then f(θ) u = Q
If 1/f(θ) be the inverse operator of f(0), then



First we find the value of

which is a linear equation where I.F. = e^{∫-α/x dx} = e^{-α logx} = x^-α. Multiplying both sides by th e Integrating Factor and then integrating


      ....(1)
Now we can evaluate

To find  when f(m) ≠ 0. 


θⁿ(x^m) = mⁿx^m 
∴ f(θ) x^m = f(m) x^m 

If f(m) ≠ 0, then
    ....(1)
Particular Cases : When f(m) = 0, then
(θ - m) is a factor of f(θ)
Therefore let f(θ) = (θ - m) f(θ) where f(m) ≠ 0.


Similarly if m is a root repeated r times, so that (θ - m)^r is a factor of f(θ) i.e.t
f(θ) = (θ - m)^r ψ(θ) where ψ (m) ≠ θ.
In this case we can see that
     ....(2)

Example 1:

We first change the independent variable x to z by substitution x = e^z, that is, z = logx, so that
The equation is then reduced to
D’(D' - 1) (D’ - 2) - D’(D’ - 1) + 2D’ - 2) y = e^3z + 3e^z 
or, (D’³ - 4D’² + 5D’ - 2) y = e^3z + 3e^z 
or, (D’ - 1)² (D' - 2) y = e^3z + 3e^z.
Here the auxiliary equation (m - 1)² (m - 2) = 0 has the roots 1, 1, 2.
Thus the complementary function is (C₁ + C₂z) ez + C₃ e^2z = (C₁ + C₂ log x) x + C₃x².
The particular integral is


Hence the complete solution is

Example 2: Solve : (x² D² + 3xD + 1) y = 1/(1-x)² , where D ≡ d/dx. 

Let us put x = e^z, so that z = log x.
The given equation then reduces to


Now, the roots of the auxiliary equation are - 1, - 1.
Thus the complementary function is
(C₁ + C₂z) e^-z = (C₁ + C₂ log x) x^-1.
The particular integral is

Hence the complete solution is

Note. The particular integral may also be obtained in the following way :

Example 3: Solve : (x⁴ D⁴ + 6 x³ D³ + 9x²D² + 3xD + 1) y = (1 + log x)², where D ≡ d/dx

Let us put x = e^z, so that z = log x.
The given equation then reduces to {D’ (D’ - 1) (D’ - 2) (D’ - 3) + 6D’(D’ - 1) (D’ - 2) + 9D’ (D’ - 1) + 3D’ + 1} y = (1 + z)².
Simplifying, we get (D’² + 1)² y = (1 + z)².
Now, the roots of the auxiliary equation are ± i, ± i.
Thus, the complementary function is (C₁ + C₂z) cos z + (C₃ + C₄z) sin z
= (C₁ + C₂ log x) cos (log x) + (C₃ + C₄ log x) sin (log x).
The particular integral is

= (1 - 2D’² + ....) (1 + 2z + z²)
= 1 + 2z + z² - 4 = z² + 2z - 3 = (log x)² + 2 log x - 3.
Hence the complete solution is y = (C₁ + C₂ log x) cos (log x) + (C₃ + C₄ log x) sin (log x) + (log x)² + 2 log x - 3.

Linear Equations of Second Order With Variable Coefficients

Definition 
The Standard Form the linear equation of second order is
     ....(1)
or y₂ + Py₁ + Qy = R       ....(2)
where P, Q, R are functions of x.

Method of Solution 

Factorization of The Operator 
Let the given equation be y₂ + Py₁ + Qy = R or (D² + PD + Q)y = R,  where D ≡ d/dx.
It will be possible in certain cases to factorise the left hand side into two linear operators in D acting on y. In such cases we first reduce the given equation to one of first order and then solve the given equation.

Example 1: Solve xy₂ + (x - 1)y₁ - y = 0.

Given equation may be written as

[xD² + (x - 1)D - 1]y = 0, where D ≡ d/dx       .....(1)
Now xD² + (x - 1)D - 1 = xD² + xD - D - 1
= xD(D + 1) - (D + 1) = (xD - 1)(D + 1)         ...(2)
Note: But if we take (D + 1)(xD - 1), then
(D + 1)(xD - 1) = D(xD - 1) + (xD - 1) = D(xD) - D + xD - 1
= [xD² + D(x).D] - D + xD - 1 = [xD² + (1)D] - D + xD - 1 = xD² + xD - 1, which is different from the left hand side of the problem.
∴ With the help of (2), the given equation (1) reduces to
(xD - 1)(D + 1)y = 0        .....(3)
Let (D + 1) y = v,        .....(4)
Hence from (3) and (4),
we get (xD - 1)v = 0,          .....(5)
which is a differential equation of first order in v.
Now (5) can be rewritten
Integrating, log v - log x = log c₁, where c₁ is arbitrary const, or log (v/x) = logc₁
or v = c₁x
or (D + 1)y = c₁x, from (4) or which is a linear equation of first order in y. Its integrating factor = e^∫dx = e^x.
∴  The solution of (6) is y.e^x = c₂ + ∫c₁xe^xdx = c₂ + c₁ ∫xe^xd_x = c₂ + c₁[xe^x - ∫1.e^xdx], integrating by parts of ye^x = c₂ + c₁[xe^x - e^x] or y = c₂e^-x + c₁(x - 1) is the required solution.
Where c₁, c₂ are arbitrary constants.

One Integral Belonging to The Complementary Function 
Let the given equation be
y₂ + Py₁ + Qy = R       ...(i)
where P, Q and R are functions of x.
Let y = z be one known integral of the complementary function of (i) i.e. one integral of
y₂ + Py₁ + Qy = 0       ...(ii)
i.e. y = z is a solution of (ii) and
so z₂ + Pz₁ + Qz = 0      ...(iii)
Now let the complete solution of (i) be
y = vz,      .......(iv)
where v is a function of x to be determined.
From (iv), we have
y₁ = v₁z + vz₁ and y₂ = (v₂z + v₁z₁) + (v₁z₁ + vz₂) = v₂z + 2v₁z₁ + vz₂.
Substituting these values of y, y₁ and y₂ in (i),
we get (v₂z + 2v₁z₁ + vz₂) + P(v₁z + vz₁) + Q(vz) = R
or v₂z + v₁(2z₁ + Pz) + v(z₂ + Pz₁ + Qz) = R
or zv₂ + (2z₁ + Pz)v₁ + v(0) = R,
from (iii)

which is a linear equation of first order in v₁.
From here we can obtain v₁ i.e. dv/dx in the usual way which on integration gives the value of the unknown function v and then substituting this value of v in (iv), where z is already known, we get the complete solution of (i).

Rulse for finding the Complementary function

1. If 1 + P + Q = 0, then complim entary function y = e^x 
2. If 1 - P + Q = 0, then com plim entary function y = e^-x 
3. If m² + mP + Q = 0, then com plim entary function y = e^mx 
4. If P + Qx = 0, then complimentary function y = x
5. If 2 + 2Px + Qx² + 0, then comp lim entary function y = x² 
6. If m(m-1) + Pmx + Qx² = 0, then complimentary function y = x^m 

Working method for finding solution of linear equation of second order :

1. Put the given equation in the standard form  in which the coefficient of d²y/dx² is unity.
2. Find an integral of CF studied earlier.
   If a solution u is given in a problem, then there is no need of this step.
3. Now take y = uv and simplify to the form of equation (4).
4. Take dv/dx = P and reduce to a linear equation in p and x.
5. Now find CS of the so obtained equation.

Example 1: Solve xy₂ - 2(x + 1)y₁ + (x + 2)y = (x - 2)e^2x.

Reducing the given equation to the standard form
y₂ + Py₁ + Qy = R,
we have
       ....(i)

And so we find that
∴ From Rule 1; y = e^x is an integral belonging to the complementary function (i.e. C.F.) of (i)
∴ Let y = zex      ...(ii)
be the complete integral (solution) of (i), where z is a function of x to be determined.
From (ii), we have y₁ = + z₁e^x = (z₁ + z)e^x 
and y₂ = (z₂ + z₁)e^x + (z₁ + z)e^x = (z₂ + 2z₁ + z)e^x 
Substituting these values of y, y₁ and y₂ in the given equation, we have
x(z₂ + 2z₁) + z)e^x - 2(x + 1)(z₁ + z)e^x + (x + 2)ze^x = (x - 2)e^2x 
or xz₂ + z₁[2x - 2(x + 1)] + z[x - 2(x + 1) + (x + 2)] = (x - 2)e^x 

 which is a linear equation of first order in z₁.
Its integrating factor



integrating first integral by parts taking ex as 2nd function

or dz = (c₁x² + e^x)dx,

Substituting this value of z in (ii), the required solution is  where a = c₁/3 , b = c₂ are the arbitrary constants.

Example 2:

Dividing the given equation by x² and putting in the standard form
    .....(1)

Thus P + Qx = 0, Therefore y = x is a part of CF
Now taking y = vx and substituting

       ....(2)
Putting dv/dx = P in (2), we have
    .....(3)
(3) is a linear equation whose I.F. = e^-2x 
Therefore solution of (3) is

Example 3:

Here P + Qx = 0, Therefore y = x is a part of CF.
Now substituting y = vx,   dy/dx and d²y/dx²  etc and simplifying,
     ....(1)
Substituting dv/dx = P in (1),
     .....(2)
(2) is a linear equation whose

Therefore solution of (2) is


Again integrating,
     ...(1)
Integrating the first factor by parts and substituting x² = t in the second factor on RHS of (1), we obtain


which is the required CS.

Example 4: 

Clearly, P + Q + 1 = 0 , therefore y₁ = e^x is a solution.
Now,

Substituting these in the given equation,
    ....(2)

(3) is a linear equation whose I.F. = e^2x/sin x
Therefore solution of (3) is  


which is the required CS.

Example 5:

given y = cot x is a solution.

The given equation is
Clearly P = 0. Now putting y = v cot x,



Integrating log α = 4 ∫ cosec 2x dx + log c₁ 
= 2 log tan x + log c₁ 
or  α = dv/dx = c₁ tan²x
dv = c₁(sec² x - 1) dx
Integrating v = c₁ (tan x - x) dx + c₂ 
or y = c₁ cot (tan x - x) + c, cot x which is the required CS.

Example 6: 

Given that  is a solution.

From the given solution, let CS of the given equation be  Find dy/dx and d²y/ dx². Substituting in the equation and simplifying to


Separating the variables and integrating




Therefore the required CS is

Removal of the first derivative (Normal form): 
To express the standard form of second order linear equation
   ....(1)
in the normal form
Let CS of (1) be y = uv where u and v are functions of x.
Differentiating this twice wrt x,


Substituting these in (2),


    ....(2)
Now to remove the first derivative dv/dx, equate its coefficient to zero

Integrating,
     ....(3)
Using (3) in (2),  
    .....(4)



Using these in (4),


     [Normal Form] ...(5)
     [Formula] .....(6)

Working method:

1. Write the given equation in the standard form (1) keeping the coefficient of d²y/dx² as unity (1).
2. Choose u = e^{1/2∫P dx} to remove the first derivative.
3. Now take y = uv as CS, reduce the given equation to the normal form d²v/dx² + Iv = S, where
4. Now (a) If I = constant, then use the earlier methods.
   (b) If I = (constant / x²), then the resulting equation reduces to homogeneous form which can be solved by the earlier methods.
5. By the above method, get v so that CS is y = uv.

Example 1:

In order to put the given equation in the standard form, divide throughout by x²,
     ......(1)
   ....(2)
To remove the first derivative,

or, u = e^x e^logx = x e^x       .....(3)
As given,

Therefore the normal form of the given equation is

or, D²v = 0, where D ≡ d / dv
To solve this, the AE is D² = 0 ⇒ D = 0, 0
∴  v = (c₁ + c₂x) e^0.x = c₁ + c₂ x    ....(4)
Therefore from (3) and (4), the required CS is
y = uv = xe^x (c₁ + c₂x)

Example 2:

Simplifying the equation and writing in the standard form
  ....(1)
Now here

To remove the first derivative
     ...(2)

       ...(3)
and
Therefore the normal form  of the given equation is

∴ v = c₁ cos (x + c₂)     ...(5)
Hence by (2) and (5), CS is
y = uv = xc₁ cos (x + c₂)

Example 3:

From the equation, it is clear that
     ...(1)
The remove the first derivatives,
   .....(2)

Now taking y = uv in the given equation, by (1) and (3), its normal form will be
   ...(4)
(4) is a homogeneous equation therefore to solve this, take x = e^z, D = d / dz
D(D - 1) v - 6 v = 0 ⇒ (D² - D - 6) v = 0
AE is D² - D - 6 = 0 ⇒ (D + 2) (D - 3) = 0 ⇒ D = - 2,3
Therefore the solution is v = c₁e^-2z + c₂e^3z = c₁x^-2 + c₂x³ ....(5)
Hence by (2) and (5), the required CS is
     .....(5)

Example 4: 

To remove the first derivative
    ....(2)

Let CS of the given equation be y = uv, then its normal form will be
    ...(4)
or, (D² + 1) v = -3 sin 2x
Its AE is D² + 1 = 0 ⇒ D = ± i
Therefore CF = c₁ cos x + c₂ sin x

Therefore the solution is v = c₁ cos x + c₂ sin x + sin 2x
From (2) and (5), the required CS is y = u v

Method of changing the independent variable 
The linear equation of second order in the standard form is
      ....(1)
where P, Q and R are functions of the variable x.
Changing the independent variable x to z by the substitution z = f(x)
By differential calculus,
     .....(i)



Using (i) and (ii) in (1), we have


    ....(2)

Here P₁, Q₁ and R₁ are all functions of x but can be changed into functions of z by the relation z = f(x).

Selection of the Variable z 
Here z should be chosen in such a manner that (2) can be easily integrated to give the solution. For this, if Q₁ = constant, then P will also become constant and (2) can be integrated easily. Since z is arbitrary, therefore this can be chosen in such a manner that it satisfies the given condition.
We choose z so that the coefficient of dy/dz become zero.
Case I. If P₁ = 0, then (2) will be of the form


Therefore (2) reduces to     ....(4)
(a) Now if Q₁ = constant, then (4) will be linear equation and can be solved.
(b) If Q₁ = constant / z², then (4) will be non-homogeneous differential equation of variable coefficients and can be solved by the earlier methods.
Case II. If Q₁ = a², then

Therefore (2) will be of the form

Now if P₁ is a constant, then the equation can be easily solved to get the solution.

Remarks : 1. Remember the values of P₁, Q₁, and R₁.
Choose P₁ = 0 and Q₁ = a² only and solve.

Example 1: 

Here,
    ....(1)
Changing the independent variable x to z, the given equation becomes


Now choose z such that Q = constant = a² (say)


Therefore the transformed equation is

Its AE is D² + a² = 0 ⇒ D = ± ai
Therefore the solution is

Example 2:

Divide throughout by x, and writing the equation in the standard form
      .....(1)

Now choose z such that (dz/dx)² = 4x² 
⇒ dz = 2x dx    or z = x²      .....(2)
By the formula,  = sin x² = sin z   [by (2)]
and Q, = -1
Change the independent variable x in the given equation to z by (2),

The AE is (D² - 1) = 0 ⇒ D² = 1 ⇒ D = ± 1

Therefore the required solution is

Example 3:

Divide the given equation throughout by cos x and writing in the standard form

Here P = tan x; Q = -2 cos² x and R = 2 cos⁴ x
Now choose z such that (dz/dx)² = 2 cos² x
dz = √2 cos x dx, Integrating, z = √2 sin x    ....(2)


Hence the transformed equation is

Therefore C.F. = c₁e^z + c₂e^-z  ......(i)


 [by binomial theorem]
   .....(ii)

or, the required CS is y = c₁e^√2sinx + c₂e ^-√2sinx + sin² x 

Aliter: If z is chosen such that P₁ = 0, then

Integrating, dz/dx = cosx, Also z = sin x
Therefore Q₁ = -2 and R₁ = 2 cos² x
Then the transformed equation is


There fore required CS is

Example 4:

Here , P = 3 sin x - cot x; Q = 2 sin² x; R = e^{-cos x} sin² x   .....(1)
To change x to z, choose z such that Q₁ = a² = 2 (say)
Therefore dz/dx = sin x or z = -cosx


There fore the transformed equation is
  .....(4)
or, (D² + 3D + 2) y = e^z 
The AE is D² + 3D + 2 = 0 ⇒ D = -1, -2
Therefore C.F. = c₁e^-z + c₂e^-2z     ....(i)

The Wronskian Function 
Let y₁(x) and y₂(x) be two solutions of the second order differential equation a₀(x) y”(x) + a₁(x) y’(x) + a₂(x) y(x) = 0
Then the wronskian of y₁(x) and y₂(x) is denoted and defined as the determinant.

The Wronskian of n functions 
Let y₁(x), y₂(x)..... ,y_n(x) be n solutions of (dⁿy/dxⁿ) + P₁(d^n-1y/dx^n-1) + ...+ P_ny = 0 of. Then the Wronskian of y₁(x), y₂(x),...y_n(x) is denoted by W(y₁,y₂,...y_n) or W (x) and defined by the determinant.

Abel’s Formula
Let y₁(x) and y₂(x) be two solutions of
y" + p(x)y' + q(x)y = 0,      ...(1)
and let W (x) = y₁ (x) y₂’ (x) - y₁’(x) y₂(x),     .......(2)
be their Wronskian. We can compute the derivative W’(x) as
W’(x) = y₁’ (x) y₂’’(x) + y₁’’(x) y₂” (x) - y₁’’(x) y₂(x) - y₁’(x) y₂’(x)
= y₁(x) y₂”(x) - y₁”(x) y₂(x)       ....(3)
Since y₁ and y₂ are solutions of (1), we have
y₁” (x) = -p(x) y₁’(x) - q (x)y₁(x),
and y₂’’(x) = - p (x) y₂’(x) - q (x) y₂(x),     ...(4)
and substituting these into (3), we get
W’(x) = y₁(x) [-p(x)y₂’(x) - q(x)y₂(x)] - [-p(x)y₁’(x) - q(x)y₁(x)]y₂(x)
= -P(x)y₁(x)y₂’(x) - q(x)y₁(x)y₂(x) + p(x)y₁’(x)y₂(x) + q(x)y₁(x)y₂(x)
= -P(x)y₁(x)y₂’(x) + p(x)y₁’(x)y₂(x)    .....(5)
= -p(x)W(x).
In other words, the Wronskian satisfies the first order linear equation
W ’(x) + p (x)W (x) = 0. . . . (6)
This fact is known as Abel’s theorem. We can easily solve (6), and derive
          .....(7)
the formula known as Abel’s formula or Abel’s identity. In particular, if the coefficient p(x) is constant, then
W(x) = W(x₀)e^{(x - x₀)p}       .....(8)
An important consequence of Abel’s formula is that the Wronskian of two solutions o/(1) is either zero everywhere, or nowhere zero.

Example: We know that y₁(x) = cos x and y₂(x) = sin x are solutions to y" + y = 0. Since p = 0 in this case , in light of Abel’s formula, we see Wronskian W(x) of y₁ and y₂ must be a constant. We confirm it by explicit computations:
W(x) = cos x(sin x)’ - (cos x)’ sin x = cos² x + sin² x = 1. ...(9) 

Example: The functions y₁(x) = e^x and y₂(x) = xe^x are solutions to y” - 2y’ + y = 0. Since p = -2, we have W(x) = ce^2x for some constantc. Explicit computation gives
W(x) = e^x(xe^x)’ - (e^x)’e^x = e^x(e^x + xe^x) - xe^2x = e^2x,        ....(10)
so c = 1.

Method of Variation of Parameters

Consider the second order non-homogeneous linear equation
a₀(x)y” + a₁(x)y' + a₂(x)y = r(x), a₀(x) ≠ 0        .......(1)
We shall discuss a general method of solution, called the method of variation of parameters, which can always be used to find a particular integral whenever we have the complementary function of the equation
a₀(x)y” + a₁(x)y' + a₂(x)y = 0, a₀(x) ≠ 0     ......(2)
Using the methods given in the previous section, we can find two linearly independent solutions y₁(x) and y₂(x) of the equation (2). The complementary function is given by
y_c(x) = Ay₁(x) + By₂(x)       ......(3)
where A and B are arbitrary constants. The idea behind the method of variation of parameters is to vary the parameters A and B. That is, we assume A and B to be functions of x and determine A(x), B(x) such that
y(x) = A(x) y₁(x) + B(x) y₂(x)        .....(4)
is the general solution of Eq.(1). Now, y(x) contains two functions A(x) and B(x) which are to be determined. Therefore, we need two equations to determine them. One equation is obtained by substituting y(x) from Eq.(4) in Eq.(1). The determination of the second is at our disposal. This equation is chosen such that the determination of A(x) and B(x) is simple. Differentiating Eq.(1), we obtain
y’(x) = A’y₁ + Ay₁’ + B’y₂ + By₂’ = (A’y₁ + B’y₂) + (Ay₁’ + By₂’)      ......(5)
If we differentiate this equation again, then the equation would contain the second derivatives A” and B” of the unknown functions. In order that these derivatives are not used, we set in Eq.(5)
A’y₁ + B’y₂ = 0     ........(6)
which gives us the second equation to determine A(x) and B(x). Now, differentiating y’(x) = Ay₁’ + By₂’ we obtain
y”(x) + Ay₁’’ + A’y₁’ + By₂’’ + B’y₂’      ...(7)
Substituting the expressions for y(x), y’(x) and y"(x) in Eq.(1), we obtain
a₀(x)[Ay₁” + A’y₁’ + By₂’’ + B’y₂’] + a₁(x)[Ay₁’ + By₂’] + a₂(x)[Ay₁ + By₂] = r(x)
or a₀(x)[A’y₁’ + B’y₂'] + A[a₀(x)y₁” + a₁(x) y₁’ + a₂(x)y₁] + B[a₀(x)y₂” + a₁(x)y₂’ + a₂(x)y₂] = r(x).
Since, y₁(x) and y2(x) are the solutions of the homogeneous equation (2), we obtain

Since a₀(x) ≠ 0 on the given interval I, g(x) is continuous on I. Solving the equations
A’y₁ + B’y₂ = 0
A’y₁’ + B’y₂’ = g(x),
we obtain

We note that the Wronskian W(y₁, y₂) is

Since y₁, y₂ are the linearly independent solutions of the homogeneous equation. Hence, we can write Eqs.(9) as
    ....(10)
Integrating, we obtain

Substituting in Eq.(4), we obtain the general solution which contains two arbitrary constants. If we do not add the arbitrary constants while carrying out integrations of Eqs.(10), then we obtain the particular solution as
y_p(x) = A(x)y₁(x) + B(x)y₂(x), which does not contain any arbitrary constants.
The general solution is given by y(x) = y_c(x) + y_p(x).
The method is applicable both for constant coefficient and variable coefficient problems. The method can also be easily extended to equations of any order. At each differentiation step, we set the part containing the derivatives of the unknown functions to zero, until we arrive at the final substitution step. For example, consider the third order equation
a₀(x)y”’ + a₁(x)y" + a₂(x)y’ + a₃(x)y = r(x), a₀(x) ≠ 0       ...(12)
The complementary function is
y(x) = Ay₁(x) + By₂(x) + Cy₃(x)
where y₁, y₂, y₃ are the linearly independent solutions of the corresponding homogeneous equation and A, B, C are arbitrary constants. We assume the solution as
y(x) = A(x) y₁(x) + B(x) y₂(x) + C(x) y₃(x)   ........(13)
Following the procedure discussed earlier, we obtain the required equations for determining A(x), B(x) and C(x) as
A’(x)y₁ + B’(x)y₂ + C’(x)y₃ = 0
A’(x)y₁’ + B’(x)y₂’ + C’(x)y₃’ = 0
and A'(x)y₁” + B’(x)y₂” + C’(x)y₃” = r(x)/a₀(x) = g(x)      .....(14)
The determinant of the coefficient matrix is the Wronskian W(y₁, y₂, y₃) ≠ 0. We determine A(x), B(x), C(x) and substitute in Eq.(13) to obtain the general solution.

Example 1: Find the general solution of the equation y” + 3y’ + 2y = 2e^x, using the method of variation of parameters.

The corresponding homogeneous equation is y” + 3y’ + 2y = 0. The characteristic equation is m² + 3m + 2 = 0 and its roots are m = -1, -2. Hence, the complementary function is
y_c(x) = Ay₁(x) + By₂(x) = Ae^-x + Be^-2x 
where y₁(x) = e^-x and y₂(x) = e^-2x are two linearly independent solutions of the homogeneous equation. Assume the general solution as
y(x) = A(x)e^-x + B(x)e^-2x 
We have g(x) = r(x)/a₀(x) = 2e^x.
The Wronskian of y₁(x), y₂(x) is given by

Using Eq.(11), we obtain the solutions for A(x) and B(x) as

The general solution is y(x) = A(x)e^-x + B(x)e^-2x 

Example 2: Solve by the method of variation of parameters :

Writing the given equation (D² + a²) y = sec ax   ...(1)
The AE is (D² + a²) = 0 ⇒ D = ± ai
Therefore the CF of (1) is C.F. = c₁ cos ax + c₂ sin ax,    .....(2)
where c₁ and c₂ are arbitrary constants.
Let GS of (1) be
y = A cos ax + B sin ax   .....(3)
where A and B are to be determined. Choose A and B such that

Using (3) and (6) in (1),

  ....(7)
Multiplying (4) by a sin ax and (7) by cos ax and adding

or dB/dx  = 1/a

    ....(8)
   ....(9)
Therefore using (8) and (9) in (3), the CS is y = A cos ax + B sin ax

Example 3: Solve by the method of variation of parameters :

Dividing throughout by x² and writing it in the normal form
  ....(1)
By inspection, P + Qx = 0 therefore y = x is a part of CF    ....(2)
Again taking y = vx in (1), we have


The AE is D² - 2D = 0 ⇒ D = 0, 2  
Therefore v = c₁ + c₂e^2x     ....(3)
Hence CF is
y = vx = c₁x + c₃x e^2x     ....(4)
Now let the solution of (1) be
y = Ax + Bxe^2x     ....(5)
where A and B are functions of x and are given by
    ....(6)
    ....(7)
Solving (6) and (7),

and

which on integrating,

and

where k₁ and k₂ are arbitrary constants of integration. Hence CS is

Example 4: Solve by the method of variation of parameters

Writing the given equation in the normal form
    ...(1)
Taking RHS of (1) zero, then

By inspection, P + Qx = 0 ⇒ y = x is a part of CF.    ...(2)
Again 1 + P + Q = 0 ⇒ y = e^x i s also a part of CF     ...(3)
Therefore the solutions
y = c₁e^x + c₂x      .......(4)
Now let the solution of (1) be
y = Ae^x + Bx     ......(5)
where A and B are functions of x. Now using the earlier methods, A and B should satisfy the following conditions:

Solving (i) and (ii)
dB/dx =1,   integrating, B = x + k₁     ......(6)
and  dA/dx = -xe^-x,  integrating A = e^-x (x + 1) + k₂    .....(7)
Using (6) and (7) in (5), we have the required CS
y = [e^-x (x + 1) + k₂] e^x + (x + k₁) x
or, y = k₁x + k₂e^x + (1 + x + x²)

Example 5: 

Here 1 - P + Q = 0. Therefore the solution of AE is y = e^-x 
Now let y = ve^-x, then taking R = 0 the given equation reduces to the form


Let dv / dx = p, then

Integrating, log p = log c₁ + x + log sin x = log (c₁ sin xe^x)



which is CF of the given equation.
Now solution of (1) be
y = A (sin x - cos x) + Be^-x,     ...(2)
where A and B are functions of x whose values can be obtained from the following equations
     .....(3)




Using (5) and (6) in (2), the required CS is obtained.