Higher Order Linear Equations with Constant Coefficients


 Gabriel Williamson
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1 Higher Order Linear Equations with Constant Coefficients The solutions of linear differential equations with constant coefficients of the third order or higher can be found in similar ways as the solutions of second order linear equations. For an nth order homogeneous linear equation with constant coefficients: a n y (n) + a n 1 y (n 1) + + a y + a 1 y + a 0 y = 0, a n 0. It has a general solution of the form y = C 1 y 1 + C y + + C n 1 y n 1 + C n y n where y 1, y,, y n 1, y n are any n linearly independent solutions of the equation. (Thus, they form a set of fundamental solutions of the differential equation.) The linear independence of those solutions can be determined by their Wronskian, i.e., W(y 1, y,, y n 1, y n )(t) 0. Note 1: In order to determine the n unknown coefficients C i, each nth order equation requires a set of n initial conditions in an initial value problem: y(t 0 ) = y 0, y (t 0 ) = y 0, y (t 0 ) = y 0, and y (n 1) (t 0 ) = y (n 1) 0. Note : The Wronskian W(y 1, y,, y n 1, y n )(t) is defined to be the determinant of the following n n matrix y1 y' 1 y" 1 : ( n y1 y y : y' y" y' y" 1) ( n 1) ( n 1) n y y : n n n. 008, 014 Zachary S Tseng B41
2 Such a set of linearly independent solutions, and therefore, a general solution of the equation, can be found by first solving the differential equation s characteristic equation: a n r n + a n 1 r n a r + a 1 r + a 0 = 0. This is a polynomial equation of degree n, therefore, it has n real and/or complex roots (not necessarily distinct). Those necessary n linearly independent solutions can then be found using the four rules below. (i). If r is a distinct real root, then y = e r t is a solution. (ii). If r = λ ± µi are distinct complex conjugate roots, then y = e λ t cos µt and y = e λ t sin µt are solutions. (iii). If r is a real root appearing k times, then y = e r t, y = te r t, y = t e r t,, and y = t k 1 e r t are all solutions. (iv). If r = λ ± µi are complex conjugate roots each appears k times, then y = e λ t cos µt, y = e λ t sin µt, y = t e λ t cos µt, y = t e λ t sin µt, y = t e λ t cos µt, y = t e λ t sin µt, : : : : y = t k 1 e λ t cos µt, and y = t k 1 e λ t sin µt, are all solutions. 008, 014 Zachary S Tseng B4 
3 Example: y (4) y = 0 The characteristic equation is r 4 1 = (r + 1)(r + 1)(r 1) = 0, which has roots r = 1, 1, i, i. Hence, the general solution is y = C 1 e t + C e t + C 3 cos t + C 4 sin t. Example: y (5) 3 y (4) + 3 y (3) y = 0 The characteristic equation is r 5 3r 4 + 3r 3 r = r (r 1) 3 = 0, which has roots r = 0 (a double root), and 1 (a triple root). Hence, the general solution is y = C 1 e 0 t + C t e 0 t + C 3 e t + C 4 t e t + C 5 t e t = C 1 + C t + C 3 e t + C 4 t e t + C 5 t e t. Example: y (4) + 4 y (3) + 8 y + 8 y + 4 y = 0 The characteristic equation is r 4 + 4r 3 + 8r + 8r + 4 = (r + r + ) = 0, which has roots r = 1 ± i (repeated). Hence, the general solution is y = C 1 e t cos t + C e t sin t + C 3 t e t cos t + C 4 t e t sin t. 008, 014 Zachary S Tseng B43
4 Example: What is a 4th order homogeneous linear equation whose general solution is y = C 1 e t + C e t + C 3 e 3t + C 4 e 4t? The solution implies that r = 1,, 3, and 4 are the four roots of the characteristic equation. Therefore, r 1, r, r 3, and r 4 are its factors. Consequently, the characteristic equation is Hence, an equation is (r 1)(r )(r 3)(r 4) = 0 r 4 10r r 50r + 4 = 0 y (4) 10 y (3) + 35 y 50 y + 4 y = 0. Note: The above answer is not unique. Every nonzero constant multiple of the above equation also has the same general solution. However, the indicated equation is the only equation in the standard form that has the given general solution. 008, 014 Zachary S Tseng B44
5 Exercises B4.1: 1 8 Find the general solution of each equation. 1. y (3) + 5y = 0. y (3) + 7y = 0 3. y (4) 18 y + 81y = 0 4. y (4) 3 y 4 y = 0 5. y (4) + 3 y + 56 y = 0 6. y (5) + 5y (4) + 10y (3) + 10y + 5y + y = 0 7. y (5) + y (4) + 5y (3) = 0 8. y (6) y = Solve each initial value problem. 9. y (3) + 4 y 5 y = 0, y(0) = 4, y (0) = 7, y (0) = y (3) + 3 y + 3 y + y = 0, y(0) = 7, y (0) = 7, y (0) = y (4) 10y + 9y = 0, y(0) = 5, y (0) = 1, y (0) = 1, y (3) (0) = y (4) + 13y + 36y = 0, y(0) = 0, y (0) = 3, y (0) = 5, y (3) (0) = Same as #10, except with initial conditions y(6561) = 7, y (6561) = 7, y (6561) = Same as #1, except with initial conditions y(47) = 0, y (47) = 3, y (47) = 5, y (3) (47) = Find a 3rd order homogeneous linear equation which has a particular solution y = e t e t + 5t e t. 16. Find a 5th order homogeneous linear equation whose general solution is y = C 1 e t + C t e t + C 3 t e t + C 4 cos t + C 5 sin t. 008, 014 Zachary S Tseng B45
6 17. Find a 6th order homogeneous linear equation whose general solution is y = C 1 + C t + C 3 e t cos t + C 4 e t sin t + C 5 t e t cos t + C 6 t e t sin t. [Hint: the polynomial, with leading coefficient 1, that has complex conjugate roots λ ± µi has the form r λr + (λ + µ ).] 18. Find a 6th order homogeneous linear equation whose general solution is y = C 1 cos t + C sin t + C 3 t cos t + C 4 t sin t + C 5 t cos t + C 6 t sin t. 008, 014 Zachary S Tseng B46
7 Answers B4.1: 1. y = C 1 + C cos 5t + C 3 sin 5t 3 t t 3t y= C1 e + Ce cos t+ C3e sin t 3. y = C 1 e 3t + C e 3t + C 3 t e 3t + C 4 t e 3t 4. y = C 1 e t + C e t + C 3 cos t + C 4 sin t 5. y = C 1 cos 4t + C sin 4t + C 3 t cos 4t + C 4 t sin 4t 6. y = C 1 e t + C t e t + C 3 t e t + C 4 t 3 e t + C 5 t 4 e t 7. y = C 1 + C t + C 3 t + C 4 e t cos t + C 5 e t sin t t t t t t t y= C1e + Ce + C3e cos t+ C4e sin t+ C5e cos + C6e sin t 9. y = 5 e t + e 5t 10. y = 7e t + t e t 11. y = 4e t e t + e 3t 1. y = cos t 3sin t cos 3t + sin 3t 13. y = 7e t (t 6561) e t y = cos (t 47) 3sin (t 47) cos 3(t 47) + sin 3(t 47) 15. y (3) + 3y 4y = y (5) + 3y (4) + 4y (3) + 4y + 3y + y = y (6) + 8y (5) + 6y (4) + 40y (3) + 5y = y (6) + 1y (4) + 48y + 64y = , 014 Zachary S Tseng B47
8 Lastly, here is an example of an application of a very simple 4th order nonhomgeneous linear equation that might be familiar to many engineering students. The static deflection of a uniform beam Consider a horizontal beam of length L, of uniform cross section and made of homogeneous material, acted upon by a vertical force. The beam is positioned along the xaxis, with its left end at the origin. The deflection of the beam (its vertical displacement relative to the horizontal axis) at any point x is given by, according to the Euler Bernoulli beam equation *, EI u (4) = W(x), 0 < x < L. The positive constants E and I are, respectively, the Young s modulus of elasticity of the material of the beam, and the beam s crosssectional moment of inertia about the horizontal axis (i.e., the second moment of the crosssectional area). The value of the product EI is a measurement of the beam s stiffness. The forcing function W describes the load/force that the beam bears per unit length. If the beam is not bearing external load, then W(x) = w, which is the weightdensity of the beam itself (acting downwards, which is the positive direction per our usual convention). The equation then becomes EI u (4) = w. It is a (very) simple 4th order nonhomgeneous linear equation. It could be solved simply by integrating both sides four times with respect to x. However it is certainly more illustrative for our purpose to solve it using the general procedure that we have learned, namely by characteristic equation and undetermined coefficients methods to obtain both parts of the solution of this nonhomogeneous linear equation, in the form of u = u c + U. * 4 The equation is originally d d u ( EI ) = W ( x). When EI is constant, it simplifies to d u EI = W ( x). 4 dx dx dx 008, 014 Zachary S Tseng B48
9 Its characteristic equation is EI r 4 = 0, which has zero as a (quadruple) root. The complementary solution is, therefore, u c = C 1 + C x + C 3 x + C 4 x 3. The particular solution can be found using the method of Undetermined Coefficients that we have already learned. What form would the particular solution of the equation take? The general solution of this deflection equation is w u ( x) + x 4EI 3 4 = C1+ C x+ C3 x + C4 x. The graph of the deflection function is called the deflection curve or elastic curve of the beam. Another interesting aspect of this problem is that this equation does not come with initial conditions. Instead, it comes paired with four boundary conditions describing the physical conditions at the two ends, at x = 0 and x = L, of the beam. For example, if the beam is securely embedded into walls on both ends, the deflection function above must satisfy the boundary conditions: u(0) = u (0) = u(l) = u (L) = 0. In this case the four conditions tell us that the displacement, u(x), is zero at both ends, and that the slope of the beam, u (x), is also zero at the two ends hence the beam is securely fixed at both ends (by embedding into the walls). Notice that there are four conditions, which are necessary to solve the four unknown coefficients present in the general solution of this fourth order equation. Being a fourth order equation, the boundary conditions in a beam problem frequently involve u and uʺʹ. Physically, u represents the bending moment and uʺʹ represents the shear force experienced by the beam at a given point. 008, 014 Zachary S Tseng B49
10 For a simply supported beam, as another example, where a beam is either pinned or hinged at both ends (having no displacement nor resistance to rotation at the two ends), the required boundary conditions are, therefore, u(0) = u (0) = u(l) = u (L) = 0. We will study simple boundary value problems, in a quite different context, later in the course. 008, 014 Zachary S Tseng B410
11 Exercises B4.: 1. Deflection curves of uniformly loaded beams Based on the earlier calculation, a beam bearing a uniformly distributed load of density w has a deflection curve in the form 3 w 4 u ( x) = C1 + C x+ C3 x + C4 x + x. 4EI Find the deflection curve of each set of boundary conditions below. (i) Beam with both ends embedded: u(0) = u (0) = u(l) = u (L) = 0. (ii) Simply supported beam: u(0) = u (0) = u(l) = u (L) = 0. (iii) Cantilever beam (left end embedded, right end free): u(0) = u (0) = u (L) = u (L) = 0.. Consider a beam of length L bearing a load that is proportional to the distance from the left endpoint, i.e. W(x) = wx, for some positive constant w. (i) Find the general solution of its deflection curve. (ii) Find the deflection curve given u(0) = u (0) = u(l) = u (L) = 0. (iii) Find the deflection curve given u(0) = u (0) = u(l) = u (L) = 0. (iv) Find the deflection curve given u(0) = u (0) = u (L) = u (L) = Consider a cantilever beam of length L bearing a load that is proportional to the distance from the right endpoint, i.e. W(x) = w(l x), for some positive constant w. Find its deflection curve given that u(0) = u (0) = u (L) = u (L) = , 014 Zachary S Tseng B411
12 Answers B4.: w (i) u ( x) = ( L x Lx + x ) 4EI w u ( x) = L x Lx + x 4EI w 3 u ( x) = 6L x 4Lx + x 4EI (ii) ( ) 4 (iii) ( ) 3 w 5. (i) u ( x) = C1 + C x+ C3 x + C4 x + x 10EI w (ii) u ( x) = ( L x 3L x + x ) 10EI w (iii) u ( x) = ( 7L x 10L x + 3x ) 360EI w (iv) u ( x) = ( 0L x 10L x + x ) 10EI 3. The general solution is 3 wl 4 w 5 u( x) = C1 + C x+ C3 x + C4 x + x x. 4EI 10EI The cantilever beam s deflection curve is w u( x) = ( 10L x 10L x + 5Lx x ). 10EI 008, 014 Zachary S Tseng B41
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