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MTHE 225 HOMEWORK SET 11 1. Solve the differential equation: 2y 00 + 3y 0 − 2y = 14x2 − 4x − 11. 2. Solve the differential equation: y 00 − y 0 − 2y = x + e2x . 3. Solve the differential equation: y 00 + y = 3 sin x. 4. Solve the differential equation: 4y 00 + 4y 0 + y = 3xex . 1 SOLUTION 1. (Non-homogeneous Linear DE with Constant Coefficients) • For yc : Auxiliary equation: 2m2 + 3m − 2 = 0 ⇒ (2m − 1)(m + 2) = 0 ⇒ m = 1 ⇒ yc = C1 e 2 x + C2 e−2x . 1 , −2 2 • For yp : Let yp = A2 x2 + A1 x + A0 . ⇒ yp0 = 2A2 x + A1 ⇒ yp00 = 2A2 . Now plug these in the given DE. 2(2A2 ) + 3(2A2 x + A1 ) − 2(A2 x2 + A1 x + A0 ) = 14x2 − 4x − 11 ⇒ (−2A2 )x2 + (6A2 − 2A1 )x + (4A2 + 3A1 − 2A0 ) = 14x2 − 4x − 11 Now compare the coefficients. −2A2 = 14 ⇒ A2 = −7. 6A2 − 2A1 = −4 ⇒ −42 − 2A1 = −4 ⇒ A1 = −19. 4A2 + 3A1 − 2A0 = −11 ⇒ −28 − 57 − 2A0 = −11 ⇒ A0 = −37. Therefore, yp = −7x2 − 19x − 37. 1 Hence the general solution of the DE is y(x) = yc + yp = C1 e 2 x + C2 e−2x − 7x2 − 19x − 37. 2. (Non-homogeneous Linear DE with Constant Coefficients) • For yc : Auxiliary equation: m2 − m − 2 = 0 ⇒ (m − 2)(m + 1) = 0 ⇒ m = −1, 2 ⇒ yc = C1 e−x + C2 e2x . • For yp : not Be2x z }| { Let yp = (A1 x + A0 ) + Bxe2x ⇒ yp0 = A1 + B(e2x + 2xe2x ) = A1 + B(1 + 2x)e2x ⇒ yp00 = B(2e2x + (1 + 2x)2e2x ) = B(4 + 4x)e2x . Now plug these in the given DE. (B(4 + 4x)e2x ) − (A1 + B(1 + 2x)e2x ) − 2((A1 x + A0 ) + Bxe2x ) = x + e2x ⇒ (−2A1 )x + (−A1 − 2A0 ) + B(4 + 4x − 1 − 2x − 2x)e2x = x + e2x 2 ⇒ (−2A1 )x + (−A1 − 2A0 ) + B(3)e2x = x + e2x Now compare the coefficients. 1 −2A1 = 1 ⇒ A1 = − . 2 1 1 −A1 − 2A0 = 0 ⇒ − 2A0 = 0 ⇒ A0 = . 2 4 1 3B = 1 ⇒ B = . 3 1 1 1 Therefore, yp = − x + + xe2x . 2 4 3 1 1 1 Hence the general solution of the DE is y(x) = yc + yp = C1 e−x + C2 e2x − x + + xe2x . 2 4 3 3. (Non-homogeneous Linear DE with Constant Coefficients) • For yc : √ Auxiliary equation: m2 +1 = 0 ⇒ m2 −( −1)2 = 0 ⇒ m2 −i2 = 0 ⇒ (m−i)(m+i) = 0 ⇒ m = ±i. (Recall any complex/imaginary root(s) of a polynomial comes in a pair of form α ± βi) ⇒ yc = C1 eαx cos (βx) + C2 eαx sin (βx) = C1 e0x cos (1x) + C2 e0x sin (1x) = C1 cos x + C2 sin x. • For yp : not (A cos x+B sin x) z }| { Let yp = x(A cos x + B sin x) ⇒ yp0 = (A cos x + B sin x) + x(−A sin x + B cos x) ⇒ yp00 = (−A sin x + B cos x) + (−A sin x + B cos x) + x(−A cos x − B sin x) ⇒ yp00 = 2(−A sin x + B cos x) + x(−A cos x − B sin x) Now plug these in the given DE. 2(−A sin x + B cos x) + x(−A cos x − B sin x) + x(A cos x + B sin x) = 3 sin x ⇒ −2A sin x − 2B cos x = 3 sin x Now compare the coefficients. 3 −2A = 3 ⇒ A = − 2 −2B = 0 ⇒ B = 0 3 Therefore, yp = − x cos x 2 3 Hence the general solution of the DE is y(x) = yc + yp = C1 cos x + C2 sin x − x cos x. 2 4. (Non-homogeneous Linear DE with Constant Coefficients) 3 • For yc : 1 1 Auxiliary equation: 4m2 + 4m + 1 = 0 ⇒ (2m + 1)2 = 0 ⇒ m = − , − . 2 2 1 1 ⇒ yc = C1 e− 2 x + C2 xe− 2 x . • For yp : Let yp = (Ax + B)ex ⇒ yp0 = Aex + (Ax + B)ex = (Ax + A + B)ex ⇒ yp00 = Aex + (Ax + A + B)ex = (Ax + 2A + B)ex Now plug these in the given DE. 4(Ax + 2A + B)ex + 4(Ax + A + B)ex + (Ax + B)ex = 3xex ⇒ (9Ax + 12A + 9B)ex = 3xex ⇒ 9Ax + 12A + 9B = 3x Now compare the coefficients. 1 9A = 3 ⇒ A = 3 1 4 12A + 9B = 0 ⇒ 12 · + 9B = 0 ⇒ B = − 3 9 1 4 x Therefore, yp = x− e 3 9 Hence the general solution of the DE is y(x) = yc +yp = C1 e 4 − 21 x + C2 xe − 12 x + 1 4 x− 3 9 ex .