Second-Order Differential Equations

### Homogeneous Linear Equations of Second Order

A second order differential equation is called linear if it can be written as

and nonlinear if it cannot be written in this form.
The above equation is linear in the unknown function y and its derivatives, whereas  p and q as well as r may be given function of x .
If r (x) = 0 , then
y" + p (x) y' q (x) y = 0 is called homogeneous.
If r (x) ≠ 0 , then
y" + p (x) y' + q (x) y = r(x) is called nonhomogeneous.

Fundamental Theorem for homogeneous Equation
For a homogeneous linear differential equation any linear combination of two solutions is a again a solution.
If y1 and y2 are solutions of homogeneous linear differential equation, then
y = c1y1 + c2y2 (Linear combination)
is also a solution of the homogeneous linear differential equation.
Note: It does not hold for nonhomogeneous linear equations or nonlinear equations.
General Solution and Initial Value Problem
For second order homogeneous linear differential equations a general solutions is of the form
y = c1y1 + c2y2   (c1 ,c2 are arbitrary constants)
An initial value problem now consists of
y" + p (x) y'+ q(x) y = 0 and two initial conditions y (x0) = K0 and y'(x0) = K1 .
Using these conditions we can find constants c1 and c2.

### Basis

A basis of solutions is a pair y1 and y2 of linearly independent solutions.
Two functions y(x) and y2 (x) are linearly independent if
k1y(x) + k2y2 (x) = 0 ⇒ k= 0,k2 = 0 .
and we call them linearly dependent if k1y1 (x) + k2 y2 (x) = 0 for some constant k1 ,k2 not both zero.

How to Obtain Basis if One Solution Is Known (Reduction of Order)
To get y2 , set y2 = uy1 . Then y'= u'y1 + uy'1 and y'2 = u''y'1 +  2u'y'+ uy''1.
Substituting y, y'2 and y''in y'' + p(x) y' + q (x) y = 0

Collecting terms in u'',u' and u, we have

Since y1 is a solution then

Put u' = U and u'' = U' then
this is the desired first-order equation.

The quotient Can not be constant (because U ≠ 0 ), so that y1 and y2 form a basis.

Example 18: If one solution of the second-order homogeneous linear equation x2y'' - xy + y = 0  is y1 = x , then find a basis solution.

Example 19: If one solution of the second-order homogeneous linear equation
y'' - 2xy' + 2αy = 0  is y1 (x) = 1 for α = 0
Then find a basis solution.

∵ y'' - 2xy' + 2αy = 0

Such an integral is easy to find when limits are specified.

### Second Order Homogeneous Equations with Constant Coefficients

Consider

y'' + ay' + by = 0   …….(1)
where coefficient a and b are constant.
Let us assume its solution is y = eλx
Substitute y = eλx , y' = λeλx and y'' = λ2eλx in equation (1), we will get
2 + aλ + b) eλx = 0
Hence y = eλx is a solution of (1) if λ is a solution of quadratic equation

This equation is called the characteristic equation (or auxiliary equation) of (1). Its roots are

This shows that y1 = eλ1x and y2 = eλ2x are solution of (1).
Let us consider three different cases

Case I: Two Distinct Real Roots λ1 and λ2
In this case,
y1 = eλ1x and y2 = eλ2x

Constitute a basis of solutions of (1). The corresponding general solution is

Example 20: Solve y'' + y' - 2y = 0 with y (0) = 4 , y' (0) = -5 .

The characteristic equation is
λ2 + λ - 2 = 0 ⇒λ = 1, -2.
Thus general solution is
y = c1ex + c2e-2x .
∵ y = c1ex + c2e -2x ⇒ y'= c1e- 2c2 e-2x
∵ y(0) = 4 ⇒ c1+ c2 = 4  and y'(0) = -5 ⇒ c- 2c2 = -5
Thus  c1 = 1 and c2 = 3 .
Thus, general solution is
y = ex + 3e -2x .

Case II: Real Double Root
If the discriminant a2 - 4b = 0, then λ = λ1 = λ2 =  Hence only one solution
y1 = e -(a/2)x .

Let us obtain a second solution y2 of y'' + ay' + by = 0,

Thus a basis of solutions of y'' + ay + by = 0 is
e -(a /2)x,            xe -(a/2)x .
The corresponding general solution is

Example 21: Solve y" - 4y' + 4y = 0 with y (0) = 3 , y' (0) = 1 .

The characteristic equation is
λ- 4λ + 4 = 0 ⇒ λ = 2, 2 .

Thus general solution is
y = (c1 + c2x)  e2x .
∵ y = c1 + c2x) e2x ⇒ y' = c2e2x + 2(c1 + c2x) e2x
∵ y (0) = 3 ⇒ c1 = 3  and ∵ y' (0) = 1 ⇒ c+ 2c1 = 1
Thus  c1 = 3 and c2 = -5 .
Thus general solution is y = (3 - 5x) e2x .

Case III: Complex Roots
If the discriminant a2 - 4b < 0 , then

Thus

Hence

The corresponding general solution is

Example 22: Solve y" + 0.2y' + 4.01y = 0 with y (0) = 0 , y' (0) = 2 .

The characteristic equation is
λ2 + 0.2λ + 4.01 = 0 ⇒ λ = 0.1 + 2i

Thus general solution is y = e -0.1x (A cos 2 x + B sin 2 x) .
∵ y = e-0.1x (A cos 2 x + B sin 2x) ⇒ y' = B (0.1e -0.1x sin 2x + 2e -0.1x cos 2 x)
∵ y (0) = 0 ⇒ A = 0 ⇒ y = Be -0.1x sin 2 x
and ∵y'(0) = 2 ⇒ 2B = 2 ⇒ B = 1
Thus general solution is y = e-0.1x sin 2x .

Example 23: Solve y'' + ω2y = 0 .

The characteristic equation is λ2 + ω2 = 0 ⇒ λ = +
Thus general solution is y = (A cos ωx + B sin ωx) ∵ a = 0 .

Summary of Cases I-III
y'' + ay' + by = 0
Substitute x = ez ,

Similarly
Thus x2y"+ axy' + by = 0 ⇒ [D(D - 1) + aD + b] y = 0 ⇒ [D+ (a - 1) D + b] y = 0

This is second order homogeneous equation with constant coefficients.

Example 24: Solve x2y'' y" - 2.5 xy' - 2.0 y = 0 .

compare with x2y'' + axy'+ by = 0 , then a = -2.5 and b = -2.0
Substitute x = ez ⇒ y'' + (a - 1)y' + by = 0 where
⇒ y'' - 3.5y' - 2.0 y = 0
The characteristic equation is
λ2 - 3.5λ - 2.0 = 0 ⇒ λ = -0.5, 4
Thus general solution is

Example 25: Solve x2 y" - 3xy' + 4y = 0 .

compare with x2y" + axy' + by = 0 , then a = -3 and b = 4
Substitute x = ez ⇒ y'' + (a - 1)y' + by = 0 where
⇒ y" - 4y' + 4y = 0
The characteristic equation is
λ2 - 4λ + 4 = 0 λ = 2, 2
Thus general solution is
y = (c1 + c2z)e2z = (c+ c2 In x) x2.

Example 26: Solve x2y" + 7 xy' + 13y = 0 .

compare with x2y" + axy' + by = 0 , then a = 7 and b = 13
Substitute x = ez ⇒ y'' + (a - 1) y' + by = 0 where
⇒ y + 6y' +13y = 0
The characteristic equation is
λ2 + 6λ + 13 = 0 ⇒ λ = -3 + 2i
Thus general solution is
y = e -3z  (A cos 2z + B sin 2z) = x-3 [A cos 2 (2 ln x) + B sin (2 ln x)] .

### Concept of Wronskian (Linear dependence and independence)

Consider homogeneous equation
y'' + p (x) y' + q(x) y = 0
and two initial conditions y(x0) = K0 and y' (x0) = K1 .
Its general solution is
y = c1y1 + c2y2
Using these conditions we can find constants c1 and c2 .
If y1 and y2 are two solutions of homogeneous equation, then Wornskian is defined by

Note:
The two solutions y1 and y2 are linearly dependent if and only if their Wornskian is zero. If Wornskian is not zero then, then y1 and yare linearly independent.
Example: Solution of y'' + ω2y = 0 is y1 = cos ωx and y2 = sin ωx . Check whether they are linearly dependent or independent.
Solution: Wornskian is defined by

If ω ≠ 0 , they  are linearly independent.

Example 27: Solution of y'' - 2y' + y = 0 is y = (c1 + c2x) ex . Check whether they are linearly dependent or independent.

Wornskian is defined by

They are linearly independent.

Finding a Second Solution
Consider homogeneous equation
y" + p(x) y' + q (x) y = 0
If y1 and y2 are two independent solutions, then Wronskian is defined by

By differentiating Wronskian, we will get

(y1 and y2 are two independent solutions)

### Nonhomogeneous Linear Equations of Second Order

Consider nonhomogeneous linear differential equation
y" + p (x) y' + q(x) y = r (x)  …….(1)
General Solution
A general solution of the nonhomogeneous equation (1) is of the form
y(x) = yh (x) + yp (x)
where yh (x) = c1y1(x) + c2y2(x) is a general solution of the homogeneous equation
y'' + p(x) y' + q(x) y = 0 and yp (x) is any solution containing no arbitrary constants.
Methods of Finding yp (x) :
(a)

Example 28: Find particular integral for

Example 29: Find particular integral for

Example 30: Find particular integral for

∵ f (D) = D2 - 6D + 9 ⇒ f (3) = 9 - 18 + 9 = 0, f' (3) = 6 - 6 = 0
⇒ f'' (D) = 2

(b)  and then operate.

Example 31: Find particular integral for (D2 + 5D + 4)y = 3 - 2 x .

(c)

Example 32: Find particular integral for (D2 + 4)y = sin 3x .

Example 33: Find particular integral for (D+ 4)y = cos 2x .

Example 34: Find particular integral for (D2 + D + 1)y = cos 2 x .

(d)

Example 35: Find particular integral for (D2 - 4D + 4)y = x3e2x.

Example 36: Find particular integral for (D2 - 5D + 6)y = ex cos 2x.

Example 37: Find particular integral for (D2 - 6D + 13)y = 2x .

Example 38: Find particular integral for

Example 39: Solve y" + 2y' + 101y = 10.4ex with y(0) = 1.1 , y'(0) = -0.9 .

The characteristic equation is
λ+ 2λ + 101 = 0 ⇒ λ = -1 + 10i .
Hence a/2 = -1 and ω = 10
Thus solution is

Thus general solution is
y = yh (x) + yp (x) = e-x (A cos10 x + B sin 10 x) + 0.1ex

∵ y (0) = 1.1 ⇒  A + 0.1 = 1.1 ⇒ A = 1
and ∵ y' (0) = -1 + 10B + 0.1 = -0.9 ⇒ B = 0
Thus Particular solution is
y = e-x cos10x + 0.1ex .

Example 40: Solve differential equation y" + 4 y = 8x2 .

The characteristic equation is
λ2 + 4 = 0 ⇒ + 2i .
Hence a/2 = 0 and ω = 2
Thus solution is

⇒ yp (x) = 2x2 - 1
Thus general solution is
y = y(x) + yp (x) = Acos2x + Bsin 2x + 2x2 + 1

Example 41: Solve differential equation y" - 3y' + 2y ⇒ ex .

The characteristic equation is
λ2 - 3λ + 2 = 0 ⇒ λ = 1,2.
Thus solution is

Thus general solution is

Example 42: Solve y" + 2y' + y = e -x with y(0) = -1 , y' (0) = 1 .

The characteristic equation is
λ2 + 2λ +1 = 0 ⇒ λ = -1, -1.
Thus solution is

Thus general solution is

and

Thus particular solution is

Example 43: Solve
y" + 2 y' + 5y = 1.25e0.5x + 40 cos 4x - 55 sin 4 x with y (0) = 0.2 , y' (0) = 60.1 .

The characteristic equation is
λ2 + 2λ + 5 = 0 ⇒ λ = -1 + 2i .
Hence,
Thus, the solution is

Hence,

Thus, general solution is
y = yh (x) + yp (x) = e-x (Acos 2x + B sin 2x) + 0.2e0.5x + 5sin 4x
∵ y(0) = A + 0.2 = 0.2 A = 0
⇒ y = Be -x sin 2 x + 0.2e0.5x + 5 sin 4 x
and y' = e -x  (-B sin 2x + 2 B cos 2 x) + 0.1e0.5x + 20 cos 4 x
∵ y' (0) = 2B + 0.1 + 20 = 60.1 ⇒ B = 20
Thus particular solution is y = 20e -x sin 2x + 0.2e0.5x + 5 sin 4 x

## Singular Points of Linear Differential Equations of Second Order

When a linear homogeneous second-order ODE is written in the form
y'' + p (x) y' + q (x) y = 0  ……..(1)
points x0 for which p (x) and q (x) are finite are termed ordinary points of the ODE.
However, if either p (x) or q (x) diverges as x → x0 , the point x0 is called a singular point. Singular points are further classified as regular or irregular (or essential singularities):

• A singular point x0 is regular if either p (x) or q (x) diverges there, but (x - x0) p(x) and (x - x0)2 q(x) remain finite.
• A singular point x0 is irregular if p (x) diverges faster than so that (x - x0) p(x) goes to infinity as x → x0 , or if q (x) diverges faster than so that (x - x0)2 q(x) goes to infinity as x → x0 .

These definitions hold for all finite values of x0 .

To analyze the behavior at x → ∞ we set x = 1/z, substitute into the differential equation, and examine the behavior in the limit z  → 0.
The equation (1) is originally in the dependent variable y (x) , will now be written in terms of  w (z) where w (z) = y (z-1) . Thus

Substituting y' and y" in equation (1), we get

From equation (4), we see that the possibility of a singularity at z = 0 depends on the behavior of

If these two expressions remain finite at z = 0 , the point x = ∞ is an ordinary point. If they diverge no more rapidly than  respectively, x = ∞ is a regular singular point; otherwise it is an irregular singular point (or essential singularity).

Example 44: Consider Bessel’s equation x 2y'' + xy' + (x2 - n2)  y = 0 .

which shows that the point x = 0 is a regular singularity.

By inspection we see that there are no other singularities in the finite range. Let us check at x → ∞ (z → 0)  for  coefficient

Since the latter expression diverges as  the point x = ∞ is an irregular or essential singularity.

Singularities of Some Important ODEs

The document Second-Order Differential Equations | Mathematical Methods - Physics is a part of the Physics Course Mathematical Methods.
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## Mathematical Methods

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## FAQs on Second-Order Differential Equations - Mathematical Methods - Physics

 1. What are singular points in linear differential equations of second order?
Ans. Singular points in linear differential equations of second order are points in the domain of the equation where the coefficients of the equation become infinite or the equation becomes undefined. These points can have a significant impact on the behavior and solutions of the differential equation.
 2. How can singular points affect the solutions of linear differential equations of second order?
Ans. Singular points can affect the solutions of linear differential equations of second order by causing the solutions to behave differently in their vicinity. The presence of a singular point can lead to the existence of special solutions called singular solutions, which may not follow the general solution obtained without considering the singular point.
 3. What are the types of singular points in linear differential equations of second order?
Ans. There are generally three types of singular points in linear differential equations of second order: - Regular singular points: These are points where the coefficient functions are analytic and can be expanded as a power series. - Irregular singular points: These are points where the coefficient functions cannot be expanded as a power series. - Essential singular points: These are points where the coefficient functions have an essential singularity.
 4. How can we determine the nature of a singular point in linear differential equations of second order?
Ans. The nature of a singular point in linear differential equations of second order can be determined by analyzing the coefficients of the equation. Regular singular points can be identified by checking if the coefficient functions can be expanded as a power series. Irregular singular points can be identified if the coefficient functions cannot be expanded as a power series. Essential singular points can be identified if the coefficient functions have an essential singularity.
 5. Can singular points affect the existence and uniqueness of solutions in linear differential equations of second order?
Ans. Yes, singular points can affect the existence and uniqueness of solutions in linear differential equations of second order. The existence and uniqueness of solutions depend on the behavior of the solutions near the singular points. If the solutions become unbounded or have certain restrictions near the singular points, the existence and uniqueness of solutions may be affected. Therefore, it is important to analyze the behavior of solutions near singular points to understand the existence and uniqueness of solutions.

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