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1. (D + 2)(D + 1)y = e 2x + 3e 3x + x 2 e x yp = Ae 2x + Be 3x + (C + Dx + Ex 2 )e x 2. (D + 2)(D + 1)y = sin 2x + sin 3x yp = A cos 2x + B sin 2x + C cos 3x + D sin 3x 3. D 2 (D + 2)(D + 1)y = 3x 2 + cos(3x) yp = x 2 (A + Bx + Cx 2 ) + D cos 3x + E sin 3x 4. (D 2 + 4)y = x sin 2x + e x yp = x[(A + Bx) cos 2x + (C + Dx) sin 2x] + Ee x 5. (D + 2)(D + 1)y = e 2x cos(2x) yp = Ae 2x cos 2x + Be 2x sin 2x (Hint: In this problem, we suppose that y(x) is the unknown function and x is the independent variable. D is an operator denoting the opertion of differentiation with respect to x, so that Dy = dy dx We know it is easy to get the characteristic equation for differential equations written by using the operator D, e.g. the characteristic equation of is = y′ (D − 2) 2 (D + 3)y = 0 (r − 2) 2 (r + 3) = 0). 2
3(15 points): Solve the following initial value problem (x is the independent variable): Solution: The characteristic equation is y ′′ + 4y = 8x 2 ; y(0) = 1, y ′ (0) = 0. r 2 + 4 = 0 and its roots are ±2i. So the complementary function is yc = c1 cos 2x + c2 sin 2x The first guess of a particular solution is of the form yp = A + Bx + Cx 2 according to the right hand of the nonhomogeneous equation. Compare yc and yp and we find there is no duplication. So the appropriate form of a particular solution is yp = A + Bx + Cx 2 Substitution yp and its derivatives into the nonhomogeneous equation yields 2C + 4(A + Bx + Cx 2 ) = 8x 2 Collecting coefficients and equating the coefficients of like terms yield 2C + 4A = 0; 4B = 0; 4C = 8 So A = −1, B = 0, C = 2. So the general solution of the nonhomogeneous equation is y = c1 cos 2x + c2 sin 2x + (−1 + 2x 2 ) The initial conditions y(0) = 1 and y ′ (0) = 0 yield 1 = c1 cos(2 · 0) + c2 sin(2 · 0) + (−1 + 2 · 0 2 ) 0 = −2c1 sin(2 · 0) + 2c2 cos(2 · 0) + 4 · 0 So c1 = 2, c2 = 0. Plugging c1 and c2 into the general solution yields the solution of the initial value problem y = −1 + 2x 2 + 2 cos 2x 3
Page 1: MATH 2171 Test 2 Summer II, 2005 Na
Page 5 and 6: 5.(10 points): Transform the given
Page 7: 7.(5 points) Use the Wronskian to p

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