Second-Order Differential Equations | Mathematical Methods - Physics PDF Download

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 y₁ (x) and y₂ (x) are linearly independent if
k₁y₁ (x) + k₂y₂ (x) = 0 ⇒ k₁ = 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^'₂ = u'y₁ + uy'₁ and y'2 = u''y'₁ +  2u'y'₁ + uy''1.
Substituting y₂ , y'₂
and y''₂ 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 = e^λx
Substitute y = e^λx , y' = λe^λx and y'' = λ²e^λx in equation (1), we will get
(λ² + aλ + b) e^λx = 0
Hence y = e^λx 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 = e^λ1x and y2 = e^λ2x are solution of (1).
Let us consider three different cases

Case I: Two Distinct Real Roots λ¹ and λ²
In this case,
y₁ = e^λ1x and y₂ = e^λ2x

^{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 = c₁e^x + c₂e^-2x .
∵ y = c₁e^x + c₂e -2x ⇒ y'= c₁e^x - 2c₂ e^-2x
∵ y(0) = 4 ⇒ c₁+ c₂ = 4  and _∵ y'(0) = -5 ⇒ c₁ - 2c₂ = -5
Thus  c₁ = 1 and c₂ = 3 .
Thus, general solution is
y = e^x + 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  
λ² - 4λ + 4 = 0 ⇒ λ = 2, 2 .

Thus general solution is
y = (c₁ + c₂x)  e^2x .
∵ y = c₁ + c₂x) e^2x ⇒ y' = c₂e^2x + 2(c₁ + c₂x) e^2x
∵ y (0) = 3 ⇒ c₁ = 3  and ∵ y' (0) = 1 ⇒ c₂ + 2c₁ = 1
Thus  c₁ = 3 and c₂ = -5 .
Thus general solution is y = (3 - 5x) e^2x .

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 = + iω
Thus general solution is y = (A cos ωx + B sin ωx) ∵ a = 0 .

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



Similarly
Thus x²y"+ axy' + by = 0 ⇒ [D(D - 1) + aD + b] y = 0 ⇒ [D² + (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 = e^z ⇒ 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 = e^z ⇒ 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)e^2z = (c₁ + 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 = e^z ⇒ 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 y₂ are 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) e^x . 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) = y_h (x) + y_p (x)
where y_h (x) = c₁y₁(x) + c₂y₂(x) is a general solution of the homogeneous equation
y'' + p(x) y' + q(x) y = 0 and y_p (x) is any solution containing no arbitrary constants.
Methods of Finding y_p (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 (D² + 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 = x³e^2x.

Example 36: Find particular integral for (D² - 5D + 6)y = e^x cos ^_2x.

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

Example 38: Find particular integral for

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

The characteristic equation is
λ² + 2λ + 101 = 0 ⇒ λ = -1 + 10i .
Hence a/2 = -1 and ω = 10
Thus solution is

Thus general solution is
y = y_h (x) + y_p (x) = e^-x (A cos10 x + B sin 10 x) + 0.1e^x

∵ 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 + 0.1e^x .

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

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

Example 41: Solve differential equation y" - 3y' + 2y ⇒ e^x .

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 = 1.25e^0.5x + 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) + y_p (x) = e^-x (Acos 2x + B sin 2x) + 0.2e^0.5x + 5sin 4x
∵ y(0) = A + 0.2 = 0.2 A = 0
⇒ y = Be ^-x sin 2 x + 0.2e^0.5x + 5 sin 4 x
and y' = e ^-x  (-B sin 2x + 2 B cos 2 x) + 0.1e^0.5x + 20 cos 4 x
∵ y' (0) = 2B + 0.1 + 20 = 60.1 ⇒ B = 20
Thus particular solution is y = 20e ^-x sin 2x + 0.2e^0.5x + 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