Second Order Linear Equations
The general form of these equations is
a2(t)y′′ + a1 (t)y′ + a0(t)y = b(t),
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′′ = et , so y′′ − y = et − et = 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) = c1et + c2 e−t ,
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
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) = ert. 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 = ert 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) = e2t · (−8 + 7et ), 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.
.General solution is:
The ICs give
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 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
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:
Example 2. Check if the given pair of functions are linearly dependent or not.
(a). f = et , g = e−t .
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.
(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) = 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 :
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
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 = tet . The general solution is
This is called the method of reduction of order.
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 r1,2 = 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.
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, ⇒ r1,2 = −1 ± 10i, ⇒ λ = −1, µ = 10.
So the general solution is
y(t) = e−t(c1 cos 10t + c2 sin 10t),
Fit in the ICs:
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) = e0.5t (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:
We note that since λ = \(-b \over 2a\) , so the sign of λ follows the sign of −b/a.