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Chapter2. Second-order differential Equations
1. Linear Differential Equations
If we can express a second order differential equation in the form
π¦ β²β² + π π₯ π¦ β² + π(π₯)π¦ = π(π₯)
it is called linear. Otherwise, it is nonlinear. Consider a linear differential equation. If r(x) = 0 it is called
homogeneous, otherwise it is called nonhomogeneous. Some examples are:
Linear Combination: A linear combination of y1; y2 is y = a y1 + b y2.
For a homogeneous linear differential equation any linear combination of solutions is again a solution. The
above result does NOT hold for nonhomogeneous equations.
For example, both π¦ = π ππ π₯ and π¦ = πππ π₯ are solutions π‘π π¦ββ + π¦ = 0, so is π¦ = 2 π ππ π₯ + 5 πππ π₯.
Both π¦ = π ππ π₯ + π₯ and π¦ = πππ π₯ + π₯ are solutions to π¦ββ+ π¦ = π₯, but π¦ = π ππ π₯ + πππ π₯ + 2π₯ is
not.
This is a very important property of linear homogeneous equations, called superposition. It means we can
multiply a solution by any number, or add two solutions, and obtain a new solution.
Linear Independence: Two functions y1; y2 are linearly independent if a y1 + b y2 = 0 a = 0; b = 0.
Otherwise they are linearly dependent. (One is a multiple of the other).
For example, ex and e
2x are linearly independent. e
x and 2e
x are linearly dependent.
General Solution and Basis: Given a second order, linear, homogeneous differential equation, the general
solution is:
π¦ = π π¦1 + π π¦2
where y1; y2 are linearly independent. The set y1; y2 is called a basis, or a fundamental set of the
differential equation.
As an illustration, consider the equation π₯2π¦βββ 5π₯π¦β + 8π¦ = 0. You can easily check that π¦ = π₯2 is a
solution. Therefore 2π₯2; 7π₯2 or βπ₯2 are also solutions. But all these are linearly dependent.
We expect a second, linearly independent solution, and this is y = x4. A combination of solutions is also a
solution, so π¦ = π₯2 + π₯4 or π¦ = 10π₯2 β 5 π₯4 are also solutions.
Therefore the general solution is π¦ = π π₯2 + π π₯4 and the basis of solutions is π₯2 , π₯4 .
2. Reduction of Order
If we know one solution of a second order homogeneous differential equation, we can find the second
solution by the method of reduction of order. Consider the differential equation
π¦ β²β² + π π₯ π¦ β² + π(π₯)π¦ = 0
Suppose one solution y1 is known, then set π¦2 = π’π¦1 and insert in the equation. The result will be
2
π¦1π’β²β² + 2π¦ β²1
+ ππ¦ β²1
π’β² + (π¦ β²β²1
+ ππ¦ β²1
+ ππ¦1) = 0
But y1 is a solution, so the last term is canceled. So we have
π¦1π’β²β² + 2π¦β²
1+ ππ¦β²
1 π’β² = 0
This is still second order, but if we set w = uβ, we will obtain first order equation:
π¦1π€β² + 2π¦β²
1+ ππ¦1 = 0
Solving this, we can find w, then u and then π¦2.
Example: Legendre's equation. The equation
is known as Legendre's equation, after the French mathematician Adrien Marie Legendre
(1752-1833.
Example:
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3. Homogeneous Equations with Constant Coefficients
Up to now we have studied the theoretical aspects of the solution of linear homogeneous differential
equations. Now we will see how to solve the constant coefficient equation π¦ββ+ ππ¦β + ππ¦ = 0 in
practice.
We have the sum of a function and its derivatives equal to zero, so the derivatives must have the same form
as the function. Therefore we expect the function to be ex. If we insert this in the equation, we obtain:
2 + π + π = 0
This is called the characteristic equation of the homogeneous differential equation π¦ββ + ππ¦β + ππ¦ = 0. If we solve the characteristic equation, we will see three different possibilities: Two real roots, double real
root and complex conjugate roots.
Two Real Roots: The general solution is
π¦ = π1 π1π₯ + π2 π2π₯
Example:
Example:
Example:
4
Double Real Root: One solution is ex but we know that a second order equation must have two independent
solutions. Let's use the method of reduction of order to find the second solution.
Let's insert y2 = u e
ax in the equation.
Obviously, uββ = 0 therefore u = c1 + c2 x. The general solution is
Example:
Example:
5
π¦ = π1 ππ₯ + π2 π₯ ππ₯
Example: Solve π¦ββ + 2π¦β + π¦ = 0.
π¦ = ππ₯ . The characteristic equation is 2 + 2+ 1 = 0. Its solution is the double root = +,-1, therefore the
general solution is
π¦ = π1 πβπ₯ + π2 ππ₯
Complex Conjugate Roots: We need the complex exponentials for this case. Euler's formula is
πππ₯ = cos π₯ + π sin(π₯) This can be proved using Taylor series expansions. If the solution of the characteristic equation is
1 = a + i b, and 2 = a - i b
then the general solution of the differential equation will be
π¦ = π1 πππ₯ (cos ππ₯ + π sin ππ₯ ) + π2 πππ₯ (cos ππ₯ β π sin ππ₯ )
By choosing new constants A;B, we can express this as
π¦ = πππ₯ (A cos ππ₯ + π΅ sin ππ₯ )
4. Cauchy-Euler Equation
The equation π₯2π¦ββ + ππ₯π¦β² + ππ¦ = 0 is called the Cauchy-Euler equation. By inspection, we can easily
see that the solution must be a power of x. Let's substitute π¦ = π₯π in the equation and try to determine r.
We will obtain
π π β 1 π₯π + πππ₯π + ππ₯π = 0
π2 + π β 1 π + π = 0
This is called the auxiliary equation. Once again, we have three different cases according to the types of
roots. The general solution is given as follows:
Two real roots: π¦ = π1π₯π1 + π2π₯
π2
Double real root: π¦ = π1π₯π + π2π₯
π πππ₯
Complex conjugate roots where π1, π2 = π Β± π π. π¦ = π₯π(π1 cos π πππ₯ + π2 sin(π πππ₯))
Example:
Example:
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Example:
Example:
Example:
Example:
7
Example:
Example:
8
Example:
5.
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10
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6. Solution of Nonhomogeneous Equation
Thus far, for differential equations of second-order and higher, we have studied only the homogeneous
equation r[y] = 0. In this section we turn to the nonhomogeneous case
r[y] = f(x)
That is, this time we include a nonzero forcing function f(x). Before proceeding with solution techniques,
let us reiterate that the function f(x) (that is, what's left on the right-hand side when the terms involving y
and its derivatives are put on the left-side side) is, essentially, a forcing function, and we will call it that, in
this text, as a reminder of its physical significance.
For instance, we have already met the equation
governing the displacement x(t) of a mechanical oscillator. Here, the forcing function is the applied force
F(t). For the analogous electrical oscillator governed by the equations
on the current i(t), and
on the charge Q(t) on the capacitor, the forcing functions are the time derivative of the applied voltage E(t),
and the applied voltage E(t), respectively.
7. General and Particular Solutions
Consider the nonhomogeneous equation
π¦ β²β² + π π₯ π¦ β² + π(π₯)π¦ = π(π₯)
Let π¦π be a solution of this equation. Now consider the corresponding homogeneous equation
π¦ β²β² + π π₯ π¦ β² + π(π₯)π¦ = 0
Let π¦β be the general solution of this one. If we add π¦β and π¦π , the result will still be a solution for the
nonhomogeneous equation, and it must be the general solution because π¦β contains two arbitrary constants.
This interesting property means that we need the homogeneous equation when we are solving the
nonhomogeneous one. The general solution is of the form
π¦ = π¦β + π¦π
Example 1.
14
Example 2.
Example 3.
Example 4.
15
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8. Method of Undetermined Coefficients
To solve the constant coefficient equation
π2π¦
ππ₯2+ π
ππ¦
ππ₯+ ππ¦ = π(π₯)
Solve the corresponding homogeneous equation, π¦π and π¦β .
Find a candidate for π¦π using the following table:
(You don't have to memorize the table. Just note that the choice consists of r(x) and all its derivatives)
If your choice for π¦π occurs in π¦β , you have to change it. Multiply it by x if the solution corresponds
to a single root, by x2 if it is a double root.
Find the constants in π¦π by inserting it in the equation.
The general solution is π¦ = π¦β + π¦π
Note that this method works only for constant coefficient equations, and only when r(x) is relatively simple.
Example 5.
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9. Method of Variation of Parameters
Consider the linear second order nonhomogeneous differential equation
π¦ β²β² + π π₯ π¦ β² + π(π₯)π¦ = π(π₯)
If a(x); b(x) and c(x) are not constants, or if r(x) is not among the functions given in the table, we can not
use the method of undetermined coefficients. In this case, the variation of parameters can be used if we
know the homogeneous solution.
Let π¦β = π1π¦1 + π2π¦2 be the solution of the associated homogeneous equation
π(π₯)π¦ β²β² + π π₯ π¦ β² + π(π₯)π¦ = 0
Let us express the particular solution as:
π¦π = π£1(π₯)π¦1 + π£2(π₯)π¦2
There are two unknowns, so we may impose an extra condition. Let's choose π£1β² π¦1
β² + π£2β² π¦2
β² = 0 for
simplicity. Inserting π¦π in the equation, we obtain
π£1β² π¦1
β² + π£2β² π¦2
β² =π
π
π£1β² π¦1
β² + π£2β² π¦2
β² = 0
The solution to this linear system is
π£1β² =
βπ¦2π
ππ, π£2
β² =π¦1π
ππ
where W is the Wronskian
Example 6.
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Therefore the particular solution is
References:
1. Greenberg, M.D. Advanced Engineering Mathematics, 2nd edition. Prentice Hall, 1998.
2. O'Neil, P.V. Advanced Engineering Mathematics, 5th edition. Thomson, 2003.
3. Ross, S.L. Introduction to Ordinary Differential Equations, 4th edition. Wiley, 1989.
Example 7.
Example 8.