Unformatted text preview: MEC702 Wk 4 Lect 2 Second Order Linear ODEs Homogeneous Linear ODEs of Second Order
Denition. A second-order ODE is called linear if it can be written
y 00 + p(x)y 0 + q(x)y = r(x) and non-linear if it can not be written in this form.
• The distinctive feature of this equation is that it is linear in y and its derivatives, whereas the
functions p, q and r on the right may be any given functions of x.
• Note that y 00 term in the equation is alone and has no variable or function multiplied to it. Denition. The second- order linear is called homogenous if it can be written
y 00 + p(x)y 0 + q(x)y = 0 If r(x) 6= 0, then the ODE is non-homogenous. Example. Homogeneous Linear ODE:
y 00 + y 0 + y
00 y +y = 0 = 0 3 = 0 y 00 + 25y = e−x cos x y 00 + 3xy 0 + x4 y = ln x 00 2 0 y + x y + 3x y Non-homogeneous Linear ODE: Non-linear ODE:
y 00 + 9y 02 = 0 + 4y − 3xy = 10 y 00 + 3xy 0 + 8y 4 = cos x y 003 0 Denition. A function
y = h(x) is called a solution of a (linear or non-linear) second-order ODE on some interval I if h is dened
and twice dierentiable throughout that interval and the ODE is satised when replacing y by h,
the derivative y 0 by h0 and the second derivative y 00 by h00 .
1 Homogeneous Linear ODEs: Superposition Principle Theorem. (Superposition Principle/Linearity Principle) Given a homogeneous linear ODE and
two separate functions
and y1 = y1 (x) y2 = y2 (x) are both solutions to the ODE. Then their linear combination c1 y1 + c2 y2
is also a solution to the ODE, where c1 , c1 ∈ R are arbitary constants. Don't forget that this highly important theorem holds for homogeneous linear
ODEs only but does not hold for non-homogeneous linear or nonlinear ODEs.
Remark. Initial Value Problems of the Second Order Denition 1. A second-order linear homogeneous IVP is of the form
y 00 + p(x)y 0 + q(x)y = r(x), with two initial conditions. The general y(x0 ) = K0 , y 0 (x0 ) = K1 of the IVP is of the form solution c1 y1 + c2 y2 where y1 and y2 are both solutions to the ODE. The initial conditions are used to nd the values
of c1 and c2 in the general solution to a particular solution to the IVP. Example. Given the IVP
y 00 + y = 0, y 0 (0) = −0.5, y(0) = 3 we see that the two functions
y1 = sin x y2 = cos x are solutions to the ODE. So we have the general solution
y = c1 y1 + c2 y2
= c1 sin x + c2 cos x. Using the initial conditions, we obtain the values
c1 = −0.5, c2 = 3..0 . Hence, we obtain the particular solution for the ODE
y = 3.0 cos x − 0.5 sin x. 2 Basis of Solutions Denition. A basis of solutions of the IVP is a pair of linearly independent solution on an open interval I . Checking for linear independence of two functions y1 and y2 : y2
• If k is a constant, then the two functions are linearly dependent.
• If k is not a constant but another function, then the two functions are linearly independent. Example. The solutions
y1 = sin x y2 = cos x forms the basis of the solutions of the IVP
y 00 + y = 0, y(0) = 3 y 0 (0) = −0.5. Example. Verify that the functions y1 = ex and y2 = e−x are solutions of the ODE
y 00 − y = 0. Then solve the IVP y 00 − y = 0, y(0) = 6 Steps:
• Verify the functions to be solutions.
• Check for linear independence.
• Form a general solution
• Solve for c1 and c2 . 3 y 0 (0) = −2 . Second-Order Homogeneous Linear ODE with Constant Coefcients
We shall now consider second-order homogeneous linear ODEs whose coecients a and b are constant,
y 00 + ay 0 + by = 0 These equations have important applications in mechanical and electrical vibrations, as we shall
see later on.
Recall that for rst order linear ODE with constant coecient k
y 0 + ky = 0 has a solution of the form y = ce−kx . Using this idea, we use the function of the form
y = eλx . Dierentiating, we have
y 0 = λeλx y 00 = λ2 eλx and subsituting into the ODE, we have
0 = y 00 + ay 0 + by
0 = (λ2 eλx ) + a(λeλx ) + b(eλx ) = eλx (λ2 + aλ + b). Since eλx 6= 0 for any value of x, we have to equate other term to 0, hence
0 = λ2 + aλ + b This is known as the characteristic equation.
Remark. If λ is a solution of the characteristic equation, then eλx is a solution to the ODE.
Then from the quadratic equation, we obtain
λ1 = λ2 =
−a + a2 − 4b
−a − a2 − 4b .
2 Hence the solution depends on the discriminant a2 − 4b as such
Two real roots if
a2 − 4b > 0,
(Case II) A real double root if
a2 − 4b = 0,
(Case III) Complex conjugate roots if a2 − 4b < 0. 4 Case I. Two Distinct Real-Roots λ1 and λ2 In this case, a basis of solutions for y + ay + by = 0 on any interval is
00 0 and y1 = eλ1 x y2 = eλ2 x . The corresponding general sollution is
y = c1 eλ1 x + c2 eλ2 x .
Case II. Real Double Root λ = − a2 For this case, since the descriminant is 0, we obtain one real root
λ1 = − ,
2 hence only one solution a y1 = e− 2 x . To nd the second linearly independent solution, we use the method of reduction of order discussed
in Section 2.1 of the book. We obtain the second solution
a y2 = xe− 2 x , hence the general solution is
a y = (c1 + c2 x)e− 2 x .
Case III. Complex Roots − 21 a + iω For this case, the roots are and − 12 a − iω 1
λ = − a ± iω
2 that gives complex solutions of the ODE. However, we obtain a basis of real solutions
y1 = e− ax
2 y2 = e− cos ωx, ax
2 sin ωx where ω > 0 and ω 2 = b − 14 a2 . Hence the general solution for Case III is
y = e− ax
2 (A cos ωx + B sin ωx) where A, B ∈ R. 5...
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