MEC702_Wk_4_Lect_2 - MEC702 Wk 4 Lect 2 Second Order Linear...

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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 =k . y1 • 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 =  p 1 −a + a2 − 4b 2  p 1 −a − a2 − 4b . 2 Hence the solution depends on the discriminant a2 − 4b as such (Case I) 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 a 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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