Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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Physics : Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

The document Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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Second Order Linear Equations

 The general form of these equations is

a2(t)y′′ + a1 (t)y′ + a0(t)y = b(t),

where

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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, so y′′ − y = et − e= 0. So it is a solution.

Guess another function: y2(t) = e−t .

Check:  y′ = −e−t , so y′′ = e−t, so y′′ − y = et − et = 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+ c2 e−t ,

then

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Proof: If y1 solves the equation, then

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev                 (I)

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev                     (II)

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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) = ert. Find r.

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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 = ert is a solution.
If there are two real and distinct roots r1 ≠ r2 , then the general solution is y(t) =  Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRevwhere 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:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

(c). We see that y(t) = e2t · (−8 + 7e), 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:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

.General solution is:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

The ICs give

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

(A)+(B) gives

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Plug this in (A):

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

The solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

General solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

The general solutions is

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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


Example 6. Find a 2nd order equation such that c1 e3t + 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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Find the largest interval where solution is valid.

Answer. Rewrite the equation into the proper form:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

so we have

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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.

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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,

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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 = et , g = e−t .

 

Answer.We have

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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, Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev 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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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)

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

From now on, when we say two solutions y1, yof 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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Example 5. If W (f , g) = 3e4t , and f = e2t , find g.

Answer.By definition of the Wronskian, we have

W (f , g) = f g′ − f ′g = e2t g′ − 2e2t g = 3e4t ,

which gives a 1st order equation for g:

g′ − 2g = 3e2t .

Solve it for g, by method of integrating factor :

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

We can choose c = 0, and get g(t) = 3te2t .

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

These two computation must have the same answer, so

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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 y= 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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

The two roots are

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

If b2 − 4ac < 0, the root are complex, i.e., a pair of complex conjugate numbers. We will write r1,2 = λ ± iµ. There are two solutions:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Back to y1, y2 , we have

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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 cy1 + c2 y2 is also a solution for any constants c1 , c2 . In particular, the functions Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev are also solutions.
Write

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev (home work problem).

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev


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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Answer.The characteristic equation is

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Solve these two equations, we get Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev So the solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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, ⇒ r1,2 = −1 ± 10i, ⇒ λ = −1, µ = 10.

So the general solution is

y(t) = e−t(c1 cos 10t + c2 sin 10t),

so

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Fit in the ICs:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

Solution is

y(t) = e−t (cos t + 0.1 sin t).

The graph is given below:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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) = e0.5t (ccos 9t + c2 sin 9t).

A typical graph of the solution looks like:

Linear Ordinary Differential Equations of First and Second Order (Part - 4), UGC - NET Physics Physics Notes | EduRev

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.

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