Please send the answers within Half hour. Thanks. Follow the Steps:
Posted: Wed May 04, 2022 10:47 am
Please send the answers within Half hour. Thanks.
Follow the Steps:
Q4 (7 points) Given that o(m) = m² (m + 1)² (m² + 2m + 5)² = 0 is the auxiliary equation associated with the linear differential equation (D) y = 4 + 6e ² sin 2x + cos(2x), state the form for the particular solution y(x) predicted by the method of undetermined coefficients. DO NOT EVALUATE THE COEFFICIENTS IN y₂(2).
7. The differential equation dy 2x2d²y dx² - 5x- - 4y = 2x³ dx is linear, but it does not have constant coefficients. (a) Show that a general solution of the associated homogeneous equation can be obtained by setting y = rm and finding values for m. (b) Find a particular solution of the nonhomogeneous equation by setting yp(x) = Ar³, and finding A. (c) What is a general solution of the nonhomogeneous equation? (d) For what values of x is your solution valid? (a) When we substitute y = xm into the homogeneous equation, 0 = 2x²m(m1)x™-2 - 5xmxm-1 - 4x = xm (2m² - 7m - 4) = xm (2m + 1)(m – 4). Thus, m = -1/2, 4 and r² and 1/√ are solutions of the homogeneous equation. A general solution is yh(x) = C₁x + C₂/√T. (b) When we substitute yp(x) = Ar³ into the nonhomogeneous equation 2x³ = 2x² (6 Ar) - 5x(3 Ax²) - 4(Ax³) = x³ (12A - 15A - 4A)x³ = -7Ax³ A A particular solution is yp(x) = -2r³/7. (c) A general solution of the nonhomogeneous equation is 2x3 y(x) = C₁x¹ + C₂/√x − 27²³ (d) The solution is valid for x > 0. || 1 VIN
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7. The differential equation dy 2x2d²y dx² - 5x- - 4y = 2x³ dx is linear, but it does not have constant coefficients. (a) Show that a general solution of the associated homogeneous equation can be obtained by setting y = rm and finding values for m. (b) Find a particular solution of the nonhomogeneous equation by setting yp(x) = Ar³, and finding A. (c) What is a general solution of the nonhomogeneous equation? (d) For what values of x is your solution valid? (a) When we substitute y = xm into the homogeneous equation, 0 = 2x²m(m1)x™-2 - 5xmxm-1 - 4x = xm (2m² - 7m - 4) = xm (2m + 1)(m – 4). Thus, m = -1/2, 4 and r² and 1/√ are solutions of the homogeneous equation. A general solution is yh(x) = C₁x + C₂/√T. (b) When we substitute yp(x) = Ar³ into the nonhomogeneous equation 2x³ = 2x² (6 Ar) - 5x(3 Ax²) - 4(Ax³) = x³ (12A - 15A - 4A)x³ = -7Ax³ A A particular solution is yp(x) = -2r³/7. (c) A general solution of the nonhomogeneous equation is 2x3 y(x) = C₁x¹ + C₂/√x − 27²³ (d) The solution is valid for x > 0. || 1 VIN