5.2: 18, 22 5.2-18 ====== y''' - 3y'' + 4y' -2y = 0 is not actually used in this problem. You are not supposed to solve for y, because given are three atoms which generate the general solution y. y=linear combination of three atoms y=linear combination of exp(x), exp(x) cos x, exp(x) sin x y = c1 exp(x) + c2 exp(x) cos x + c3 exp(x) sin x Given initial conditions y(0)=1, y'(0)=0, y''(0)=0, then c1, c2, c3 are uniquely determined. Your problem is to find c1, c2, c3, using linear algebra methods. First, write three equations, one for each of y, y', y'', which involve the symbols c1, c2, c3, x. Substitute x=0 in each equation, then collect on c1, c2, c3 to obtain a system of 3 equations in 3 unknowns. Oberve that substitution x=0 eliminates symbol x from the equations, leaving only three symbols c1, c2, c3. c1 exp(0) + c2 exp(0) cos 0 + c3 exp(0) sin 0 = y(0) = 1, c1 exp(0) + c2 exp(0) (cos 0 - sin 0) + c3 exp(0) (sin 0 + cos 0) = 0, c1 exp(0) + remaining terms similar to above = 0. The system is then of the form Au=b where A:=Matrix([[1,1,0],[1,1,1],[1,*,*]]); b:=<1,0,0>; u:=; Solve it for c1, c2, c3 (the answer is unique). 5.2-22 ====== The equation y''-4y=12 is never used, because both a particular solution y_p = -3 and the general solution y_h of the homogeneous problem y'' -4y=0 has been supplied. Superpositon says that y=y_h+y_p, or y = c1 exp(2x) + c2 exp(-2x) + (-3) The initial conditions y(0)=0, y'(0)=10 apply to the equation immediately above, not to the equation for y_h, which is the common error. As in 5.2-18, differentiate the general solution y, to get two equations, one for y and one for y', each in terms of c1, c2, x. Substitute x=0 on both to eliminate the symbol x, leaving only symbols c1, c2, to obtain c1 exp(0) + c2 exp(0) - 3 = 0, 2 c1 exp(0) - 2 c2 exp(0) + 0 = 10. This is a system Au=b where A:=Matrix([[1,1],[2,-2]]); b:=<3,10>; u:=; Solve it for c1, c2. the answer is unique. === end ===