Auxiliary Equation with Complex Roots; Hyperbolic Functions MATH 374 Lecture 17 Auxiliary Equation with Complex Roots; Hyperbolic Functions

5.5: The Auxiliary Equation: Complex Roots Given the nth order linear homogeneous differential equation with constant coefficients, f(D)y = 0 (1) suppose the auxiliary equation f(m) = 0 has real coefficients and m1 = a + i b (a, b real with b  0) is a root. 2

Auxiliary Equation with Complex Roots f(D)y = 0 (1) Auxiliary Equation with Complex Roots Then, by Remark 4 in our Section 5.3 & 5.4 notes, m2 = a - i b is also a root of f(m) = 0. It follows that f(D) = g(D)(D – (a+ib))(D – (a-ib)), so from Example 2 in our Section 5.3 & 5.4 notes, we see that y = c1 e(a+ib)x + c2 e(a-ib)x (2) is a solution of (1). 3

Auxiliary Equation with Complex Roots f(D)y = 0 (1) Auxiliary Equation with Complex Roots y = c1 e(a+ib)x + c2 e(a-ib)x (2) Using (7) in our Section 5.3 & 5.4 notes, we can rewrite (2): y = c1eax(cos bx + i sin bx) + c2eax(cos bx – i sin bx) = (c1+c2)eax cos bx + i(c1 – c2)eax sin bx. Letting c3 = c1 + c2 and c4 = i(c1 – c2), we see that y = c3 eax cos bx + c4eax sin bx (3) is a solution of (1) corresponding to the roots a § i b of f(m) = 0. Note: eax cos bx and eax sin bx are linearly independent! 4

Auxiliary Equation with Complex Roots f(D)y = 0 (1) Auxiliary Equation with Complex Roots What we’ve just shown is: Theorem 5.4: If for the linear homogeneous equation (1), the auxiliary equation f(m) = 0 has real coefficients and m1 = a+ib (a, b real and b  0) is a root, then so is m2 = a-ib and y = c1 eax cos bx + c2eax sin bx where c1 and c2 are constants is a solution to (1). Remark: For repeated complex or real roots, a theorem similar to Theorem 5.2 holds!

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4 -4

Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0. Solution: The auxiliary equation is m3 + m2 + 4m + 4 = 0. -1 1 4 -4 ) (m+1)(m2 + 4) =0 ) m = -1, § 2i ) y = c1 e-x + c2 e0·x cos 2x + c3 e0·x sin 2x = c1 e-x + c2 cos 2x + c3 sin 2x

Example 2: Solve [(D-5)2 + 9]2 y = 0 Solution: Note that for the auxiliary equation [(m-5)2 + 9]2 = 0, the roots are m = 5 § 3i, 5 § 3i. Hence, y = c1 e5x cos 3x + c2 x e5x cos 3x + c3 e5x sin 3x + c4 x e5x sin 3x. Note: In general, for a,b real, a § bi are roots of (m-a)2 + b2 = 0.

5.6: Hyperbolic Functions Definition: The hyperbolic sine and hyperbolic cosine are the functions y = sinh x y = cosh x

Properties of sinh x and cosh x cosh2 x – sinh2 x = 1 d/dx[cosh x] = sinh x and d/dx[sinh x] = cosh x. cosh x is even. sinh x is odd. y1 = cosh(ax) and y2 = sinh(ax) are linearly independent solutions of (D2 – a2)y = 0. (a real, a  0) (1)

Properties of sinh x and cosh x y1 = cosh(ax) and y2 = sinh(ax) are linearly independent solutions of (D2 – a2)y = 0. (a real, a  0) (1) Properties of sinh x and cosh x Note: From property 4, it follows that solutions of (1) can be written in the form y = c1 cosh(ax) + c2 sinh(ax) (2) instead of y = c1 eax + c2 e-ax (3) as before. Solution form (2) is useful when solving IVP with initial data taken at x = 0. From the definition of cosh x and sinh x, it is clear that (2) and (3) are equivalent! (check)