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**Second Order Linear Equations**

The general form of these equations is

a_{2}(t)y′′ + a_{1} (t)y′ + a_{0}(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: a_{2}, a_{1} , a_{0} are all constants, and g = 0. Let’s write:

We start with an example.

**Example 1**. Solve y′′ − y = 0, (we have here a_{2} = 1, a_{1} = 0, a_{0} = 1).

**Answer**. Let’s guess an answer of the form y_{1}(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: y_{2}(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 = c_{1} y_{1} + c_{2} y_{2} for any arbitrary constants c_{1} , c_{2} (this is called “a linear combination of y_{1}, y_{2} ”) is also a solution.

Check if this claim is true:

y(t) = c_{1}e^{t }+ c_{2} e^{−t} ,

then

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

**Theorem **(The Principle of Superposition) Let y_{1}(t) and y_{2}(t) be solutions of

Then, y = c_{1} y_{1} + c_{2} y_{2} for any constants c_{1} , c_{2} is also a solution.

Proof: If y_{1} solves the equation, then

(I)

(II)

Multiple (I) by c_{1} and (II) by c_{2}, and add them up:

therefore y is also a solution to the equation.

How to ﬁnd the solutions of

a_{2} y′′ + a_{1}y′ + a_{0}y = 0?

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

Since y ≠ 0, we get

a_{2}r_{2} + a_{1}r + a_{0} = 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 r_{1} ≠ r_{2} , then the general solution is y(t) = where c_{1} , c_{2} 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: r _{2} − 5r + 6 =, so (r − 2)(r − 3) = 0, two roots: r_{1} = 2, r_{2} = 3. **

**General solution is:**

Solve these two equations for c_{1}, c_{2} : Plug in c_{2} = −1 − c_{1} into the second equation, we get 2c_{1} + 3(−1 − c_{1} ) = 5, so c_{1} = −8. Then c_{2} = 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 → c_{2} .

**Example 6. Find a 2nd order equation such that c _{1} e^{3t} + c_{2}e^{−t} is its general solution.**From the form of the general solution, we see the two roots are r

Answer.

The characteristic equation could be (r − 3)(r + 1) = 0, or this equation multiplied by any non-zero constant. So r_{2} − 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 t_{0} = 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 y_{1}(t), y_{2} (t) are two solutions of y′′ + p(t)y′ + q(t)y = 0.

Then

(I) We have either W (y_{1}, y_{2} ) ≡ 0 or W (y_{1}, y_{2}) never zero;

II) If W (y_{1}, y_{2}) = 0, the y = c_{1} y_{1} + c_{2}y_{2} is the general solution. They are also cal led to form a fundamental set of solutions. As a consequence, for any ICs y(t_{0}) = y_{0}, there is a unique set of (c_{1} , c_{2} ) that gives a unique solution.

The next Theorem is probably the most important one.

**Theorem **(Abel’s Theorem) Let y_{1}, y_{2} be two (linearly independent) solutions to

y′′ + p(t)y′ + q(t)y = 0

on an open interval I . Then, the Wronskian W (y_{1}, y_{2}) on I is given by

for some constant C depending on y_{1}, y_{2}, but independent on t or on I .

**Example 3. Given**

t^{2} y′′ − t(t + 2)y′ + (t + 2)y = 0.

Find W (y_{1}, y_{2}) 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 t_{0}.

From now on, when we say two solutions y_{1}, y_{2 }of the solution, we mean two linearly independent solutions that can form a fundamental set of solutions.

**Example 4**. If y_{1}, y_{2} are two solutions of

ty′′ + 2y′ + tet y = 0,

and W (y_{1}, y_{2})(1) = 2, ﬁnd W (y_{1}, y_{2} )(5).

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

If W (y_{1}, y_{2})(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 r

By Abel’s Theorem, we have

and we can choose C = 1 and get W (y_{1}, y_{2}) = 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 y_{2}. Solve it:

Choosing c = 0, we get y_{2} = 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 y_{1} = cos t and y_{2 }= sin t both work. By the Wronskian W (y_{1}, y_{2} ) = −2 ≠ 0, we see that these two solutions are linearly independent.

Therefore, the general solution is

y(t) = c_{1} y_{1} + c_{2} y_{2} = c_{1} cos t + c_{2} sin t.

Let’ try to connect this with the characteristics equation:

r_{2} + 1 = 0, r_{2} = −1, r_{1} = i, r_{2} = −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 b^{2} − 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 y_{1}, y_{2} , we have

But these solutions are complex-valued. We want real-valued solutions! To achieve this, we use the Principle of Superposition. If y_{1}, y_{2} are two solutions, then c_{1 }y_{1} + c_{2} y_{2} is also a solution for any constants c_{1} , c_{2} . 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 y_{1}, y_{2} 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) = c_{1} cos 2t + c_{2} sin 2t.

Find c_{1} , c_{2} by initial conditions: since y′ = −2c_{1} sin 2t + 2c_{2} 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

r^{2} + 2r + 101 = 0, ⇒ r_{1,2} = −1 ± 10i, ⇒ λ = −1, µ = 10.

So the general solution is

y(t) = e^{−t}(c_{1} cos 10t + c_{2} 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**. r^{2} − r + 81.25 = 0, ⇒ r = 0.5 ± 9i, ⇒ λ = 0.5, µ = 2.

The general solution is y(t) = e^{0.5t} (c_{1 }cos 9t + c_{2} 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.

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