Joshua G. answered • 06/19/24
Biostatistics PhD Who Earned an A in Differential Equations Course
To solve d4y/dx4 - 2d3y/dx3- 15d2y/dx2+36dy/dx = 3 + 36x, we must first solve the associated homogeneous linear equation with constant coefficients d4y/dx4 - 2d3y/dx3- 15d2y/dx2+36dy/dx = 0. This associated homogeneous linear equation with constant coefficients further implies the auxiliary equation m4 - 2m3 - 15m2 + 36m = 0. By inspection, we can factor out m from the left side of the preceding equation to get m(m3 - 2m2 - 15m + 36) = 0. Factorization of f(m) = m3 - 2m2 - 15m + 36 requires substantially more work. Since f(m) has two sign changes and f(-m) = (-m)3 - 2(-m)2 - 15(-m) + 36 = -m3 - 2m2 + 15m + 36 has one sign change, we know by Descartes's Rule of Signs that f(m) has either two real zeros or no real zeros and exactly one negative real zero. The Rational Zero Theorem further tells us that ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, and ±36 are the possible rational zeros of f(m). Plugging in 1,2, and 3 for m into f(m) reveals that 1 and 2 are not zeros of f(m) but 3 is a zero of f(m). Synthetic division then enables us to deduce that f(m) = (m - 3)(m2 + m - 12). Since m2 + m - 12 = (m+a)(m+b) = m2 + (a+b)m + ab if and only if a+b = 1 and ab = -12, we can further deduce that a and b must have opposite signs and the positive number must have the larger absolute value. Since the sum of two odd numbers is even and the sum of two even numbers is even, we can also deduce that one of the numbers is odd while the other number is even. Trial and error finally reveals that 4 + -3 = 1 and 4(-3) = -12. Thus, m2 + m - 12 = (m + 4)(m - 3), f(m) = (m - 3)2(m + 4), and the auxiliary equation can be written as m(m - 3)2(m + 4) = 0. This finally tells us that the general solution for d4y/dx4 - 2d3y/dx3- 15d2y/dx2+36dy/dx = 0 is yc = c1e0x + c2xe3x + c3e3x + c4e-4x = c1 + c2xe3x + c3e3x + c4e-4x.
Since 3 + 36x is a linear function of x, m = 0 is a solution of the auxiliary equation, and the derivative of any constant is 0, we'll suppose that yp = Ax + Bx2 is a particular solution of d4y/dx4 - 2d3y/dx3- 15d2y/dx2+36dy/dx = 3 + 36x. Since dyp/dx = A + 2Bx, dyp2/dx2 = 2B, dyp3/dx3 = 0, and dyp4/dx4 = 0, we must have that 0 - 2(0) - 15(2B) + 36(A + 2Bx) = -30B + 36A + 72Bx = 3 + 36x. This further implies that -30B + 36A = 3 and 72B = 36. Since 72B = 36 implies B = 36/72 = 1/2, we can further find that -30(1/2) + 36A = 3, -15 + 36A = 3, 36A = 3 + 15 = 18, and A = 18/36 = 1/2. Thus, yp = (1/2)x + (1/2)x2 and the general solution for d4y/dx4 - 2d3y/dx3- 15d2y/dx2+36dy/dx = 3 + 36x is yc + yp = c1 + c2xe3x + c3e3x + c4e-4x + (1/2)x + (1/2)x2.
