Linear Ordinary Differential Equations of First and Second Order - 4 | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Second Order Linear Equations

 The general form of these equations is

a₂(t)y′′ + a₁ (t)y′ + a₀(t)y = b(t),

where

If b(t) ≡ 0, we call it homogeneous. Otherwise, it is called non-homogeneous.

Homogeneous equations with constant coefficients

This is the simplest case: a₂, a₁ , a₀ are all constants, and g = 0. Let’s write:

We start with an example.

Example 1. Solve y′′ − y = 0, (we have here a₂ = 1, a₁ = 0, a₀ = 1).

Answer. Let’s guess an answer of the form y₁(t) = et .

Check to see if it satisfies the equation: y′′ = e^t , so y′′ − y = e^t − e^t = 0. So it is a solution.

Guess another function: y₂(t) = e−t .

Check:  y′ = −e^−t , so y′′ = e^−t, so y′′ − y = e^t − e^t = 0.

So it is also a solution.

Claim: Another function y = c₁ y₁ + c₂ y₂ for any arbitrary constants c₁ , c₂ (this is called “a linear combination of y₁, y₂ ”) is also a solution.

Check if this claim is true:

y(t) = c₁e^t + c₂ e^−t ,

then

Actually this claim is a general property. It is called the principle of superposition.

Theorem (The Principle of Superposition) Let y₁(t) and y₂(t) be solutions of

Then, y = c₁ y₁ + c₂ y₂ for any constants c₁ , c₂ is also a solution.

Proof: If y₁ solves the equation, then

                 (I)

                     (II)

Multiple (I) by c₁ and (II) by c₂, and add them up:

therefore y is also a solution to the equation.

How to find the solutions of

a₂ y′′ + a₁y′ + a₀y = 0?

We seek solutions in the form y(t) = e^rt. Find r.

Since y ≠ 0, we get

a₂r₂ + a₁r + a₀ = 0

This is called the characteristic equation.

Conclusion: If r is a root of the characteristic equation, then y = e^rt is a solution.
If there are two real and distinct roots r₁ ≠ r₂ , then the general solution is y(t) =  where c₁ , c₂ are two arbitrary constants to be determined by initial conditions (ICs).

Example 2. Consider y′′ − 5y′ + 6y = 0.

(a). Find the general solution.

(b). If ICs are given as: y(0) = −1, y′ (0) = 5, find the solution.

(c). What happens to y(t) when t → ∞?

Answer.(a). The characteristic equation is: r₂ − 5r + 6 =, so (r − 2)(r − 3) = 0, two roots: r₁ = 2, r₂ = 3. 

General solution is:

Solve these two equations for c₁, c₂ : Plug in c₂ = −1 − c₁ into the second equation, we get 2c₁ + 3(−1 − c₁ ) = 5, so c₁ = −8. Then c₂ = 7. The solution is

(c). We see that y(t) = e^2t · (−8 + 7e^t ), and both terms in the product go to infinity as t grows. So y → +∞ as t → +∞.

Example 3. Find the solution for 2y′′ + y′ − y = 0, with initial conditions y(1) = 0, y′(1) = 3.

Answer.Characteristic equation:

.General solution is:

The ICs give

(A)+(B) gives

Plug this in (A):

The solution is

and as t → ∞ we have y → ∞.

Example 4. Consider the equation y′′ − 5y = 0. (a).

Find the general solution.

(b). If the initial conditions are given as y(0) = 1 and y′(0) = a, then, for what values of a would y remain bounded as t → +∞?

 Answer.(a). Characteristic equation

General solution is

Example 5. Consider the equation 2y′′ + 3y′ = 0. The characteristic equation is

The general solutions is

As t → ∞, the first term in y vanished, and we have y → c₂ .

Example 6. Find a 2nd order equation such that c₁ e^3t + c₂e^−t is its general solution.

 Answer. From the form of the general solution, we see the two roots are r₁ = 3, r₂ = −1.

The characteristic equation could be (r − 3)(r + 1) = 0, or this equation multiplied by any non-zero constant. So r₂ − 2r − 3 = 0, which gives us the equation

y′′ − 2y′ − 3y = 0.

Solutions of Linear Homogeneous Equations; the Wronskian

 We consider some theoretical aspects of the solutions to a general 2nd order linear equations.

 Theorem . (Existence and Uniqueness Theorem) Consider the initial value problem

If p(t), q(t) and g(t) are continuous and bounded on an open interval I containing t0, then there exists exactly one solution y(t) of this equation, valid on I .

Example 1. Given the equation

Find the largest interval where solution is valid.

Answer. Rewrite the equation into the proper form:

so we have

We see that we must have t ≠ 0 and t ≠ 3. Since t₀ = 1, then the largest interval is I = (0, 3), or 0 < t < 3. See the figure below.

Definition. Given two functions f (t), g(t), the Wronskian is defined as

W (f, g)(t) =˙ f g′ − f ′g.

Remark: One way to remember this definition could be using the determinant of a 2 × 2 matrix,

Main property of the Wronskian:

• If W (f , g) ≡ 0, then f and g are linearly dependent.
• Otherwise, they are linearly independent.

Example 2. Check if the given pair of functions are linearly dependent or not.
 (a). f = e^t , g = e^−t .
 

Answer.We have

so they are linearly independent. (b). f (t) = sin t, g(t) = cos t.

Answer. We have

W (f , g) = sin t(sin t) − cos t cos t = −1 = 0

and they are linearly independent. (c). f (t) = t + 1, g(t) = 4t + 4.

Answer.We have W (f , g) = (t + 1)4 − (4t + 4) = 0

so they are linearly dependent. (In fact, we have g(t) = 4 · f (t).)

(d). f (t) = 2t, g(t) = |t|.

Answer. Note that g′ (t) = sign(t) where sign is the sign function. So

W (f , g) = 2t · sign(t) − 2|t| = 0

(we used t · sign(t) = |t|). So they are linearly dependent.

Theorem . Suppose y₁(t), y₂ (t) are two solutions of y′′ + p(t)y′ + q(t)y = 0.

Then

(I) We have either W (y₁, y₂ ) ≡ 0 or W (y₁, y₂) never zero;

II) If W (y₁, y₂) = 0, the y = c₁ y₁ + c₂y₂ is the general solution. They are also cal led to form a fundamental set of solutions. As a consequence, for any ICs y(t₀) = y₀,  there is a unique set of (c₁ , c₂ ) that gives a unique solution.

The next Theorem is probably the most important one.

Theorem (Abel’s Theorem) Let y₁, y₂ be two (linearly independent) solutions to
y′′ + p(t)y′ + q(t)y = 0
on an open interval I . Then, the Wronskian W (y₁, y₂) on I is given by

for some constant C depending on y₁, y₂, but independent on t or on I .

Example 3. Given

t² y′′ − t(t + 2)y′ + (t + 2)y = 0.

Find W (y₁, y₂) without solving the equation.

Answer.We first find the p(t)

which is valid for t ≠ 0. By Abel’s Theorem, we have

NB! The solutions are defined on either (0, ∞) or (−∞, 0), depending on t₀.

From now on, when we say two solutions y₁, y₂ of the solution, we mean two linearly independent solutions that can form a fundamental set of solutions.

Example 4. If y₁, y₂ are two solutions of

ty′′ + 2y′ + tet y = 0,

and W (y₁, y₂)(1) = 2, find W (y₁, y₂ )(5).

Answer. First we find that p(t) = 2/t. By Abel’s Theorem we have

If W (y₁, y₂)(1) = 2, then C = 2. So we have

Example 5. If W (f , g) = 3e^4t , and f = e^2t , find g.

Answer.By definition of the Wronskian, we have

W (f , g) = f g′ − f ′g = e^2t g′ − 2e^2t g = 3e^4t ,

which gives a 1st order equation for g:

g′ − 2g = 3e^2t .

Solve it for g, by method of integrating factor :

We can choose c = 0, and get g(t) = 3te^2t .

Next example shows how Abel’s Theorem can be used to solve 2nd order differential equations.

Example 6. Consider the equation y′′ + 2y′ + y = 0. Find the general solution.
 Answer.The characteristic equation is r² + 2r + 1 = 0, which given double roots r₁ = r₂ = −1. So we know that y₁ = e^−t is a solutions. How can we find another solution y₂ that’s linearly independent?

By Abel’s Theorem, we have

and we can choose C = 1 and get W (y₁, y₂) = e^−2t . By the definition of the Wronskian, we have

These two computation must have the same answer, so

This is a 1st order equation for y₂. Solve it:

Choosing c = 0, we get y₂ = te^t . The general solution is

This is called the method of reduction of order. 

Complex Roots 

Example : Consider the equation y′′ + y = 0, find the general solution.

Answer. By inspection, we need to find a function such that y′′ = −y. We see that y₁ = cos t and y₂ = sin t both work. By the Wronskian W (y₁, y₂ ) = −2 ≠ 0, we see that these two solutions are linearly independent.

Therefore, the general solution is
y(t) = c₁ y₁ + c₂ y₂ = c₁ cos t + c₂ sin t.

Let’ try to connect this with the characteristics equation:
r₂ + 1 = 0, r₂ = −1, r₁ = i, r₂ = −i.

The roots are complex. In fact, they are acomplex conjugate pair. We see that the imaginary part seems to give sin and cos functions.

In general, the roots of the characteristic equation can be complex numbers. Consider the equation

The two roots are

If b² − 4ac < 0, the root are complex, i.e., a pair of complex conjugate numbers. We will write r_1,2 = λ ± iµ. There are two solutions:

To deal with exponential function with pure imaginary exponent, we need the Euler’s Formula:

Back to y₁, y₂ , we have

But these solutions are complex-valued. We want real-valued solutions! To achieve this, we use the Principle of Superposition. If y₁, y₂ are two solutions, then c₁ y₁ + c₂ y₂ is also a solution for any constants c₁ , c₂ . In particular, the functions  are also solutions.
Write

We need to make sure that they are linearly independent. We can check the Wronskian,

 (home work problem).

So y₁, y₂ are linearly independent, and we have the general solution

Example 1. (Perfect Oscillation: Simple harmonic motion.) Solve the initial value problem

Answer.The characteristic equation is

The general solution is

y(t) = c₁ cos 2t + c₂ sin 2t.

Find c₁ , c₂ by initial conditions: since y′ = −2c₁ sin 2t + 2c₂ cos 2t, we have

Solve these two equations, we get  So the solution is

which is a periodic oscillation. This is also called perfect oscillation or simple harmonic motion.

Example 2. (Decaying oscillation.) Find the solution to the IVP (Initial Value Problem)

y′′ + 2y′ + 101y = 0, y(0) = 1, y′(0) = 0.

Answer.The characteristic equation is

r² + 2r + 101 = 0, ⇒ r_1,2 = −1 ± 10i, ⇒ λ = −1, µ = 10.

So the general solution is

y(t) = e^−t(c₁ cos 10t + c₂ sin 10t),

so

Fit in the ICs:

Solution is

y(t) = e^−t (cos t + 0.1 sin t).

The graph is given below:

We see it is a decaying oscillation. The sin and cos part gives the oscillation, and the e^−t part gives the decaying amplitude. As t → ∞, we have y → 0.

Example 3. (Growing oscillation) Find the general solution of y′′ − y′ + 81.25y = 0.

Answer.  r² − r + 81.25 = 0, ⇒ r = 0.5 ± 9i, ⇒ λ = 0.5, µ = 2.

The general solution is y(t) = e^0.5t (c₁ cos 9t + c₂ sin 9t).

A typical graph of the solution looks like:

We see that y oscillate with growing amplitude as t grows. In the limit when t → ∞, y oscillates between −∞ and +∞.

Conclusion: Sign of λ, the real part of the complex roots, decides the type of oscillation:

•  λ = 0: perfect oscillation;
• λ < 0: decaying oscillation;
• λ > 0: growing oscillation.

We note that since λ = \(-b \over 2a\) , so the sign of λ follows the sign of −b/a.