We consider the non-homogeneous initial value problem zy" – 4ry' + 6y = -12z-1, y(1) = -5, y'(1) = -9 By looking for solutions in the form y = z' in the homogeneous Euler-Cauchy problem Az?y" + Bry' + Cy = 0, we obtain a auxiliary equation Ar? + (B - A)r +C = 0 which is the analog of the auxiliary equation in the const: coefficient case. (1) For this problem find the auxiliary equation: (2) Find the roots of the auxiliary equation: (enter your results as a comma separated list ) (3) Find a fundamental set of solutions y1, 2 (enter your results as a comma separated list). Then the complementary solution is y. = C1yı + c2y2- (4) For the non-homogeneous problem we need to apply the variation of parameters formula to find a particular solution yp. Carrying this out you obtain Yp = Recall that the general solution is y = Ye + Yp. Find the unique solution satisfying y(1) = -5, y'(1) --9
We consider the non-homogeneous initial value problem zy" – 4ry' + 6y = -12z-1, y(1) = -5, y'(1) = -9 By looking for solutions in the form y = z' in the homogeneous Euler-Cauchy problem Az?y" + Bry' + Cy = 0, we obtain a auxiliary equation Ar? + (B - A)r +C = 0 which is the analog of the auxiliary equation in the const: coefficient case. (1) For this problem find the auxiliary equation: (2) Find the roots of the auxiliary equation: (enter your results as a comma separated list ) (3) Find a fundamental set of solutions y1, 2 (enter your results as a comma separated list). Then the complementary solution is y. = C1yı + c2y2- (4) For the non-homogeneous problem we need to apply the variation of parameters formula to find a particular solution yp. Carrying this out you obtain Yp = Recall that the general solution is y = Ye + Yp. Find the unique solution satisfying y(1) = -5, y'(1) --9
Calculus: Early Transcendentals
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![We consider the non-homogeneous initial value problem z'y" – 4xy' + 6y = -12z1, y(1) = -5, y(1) = -9
By looking for solutions in the form y = z' in the homogeneous Euler-Cauchy problem Az?y" + Bry' + Cy = 0, we obtain a auxiliary equation Ar? + (B – A)r +C=0 which is the analog of the auxiliary equation in the constant
coefficient case.
(1) For this problem find the auxiliary equation:
(2) Find the roots of the auxiliary equation:
(enter your results as a comma separated list)
(3) Find a fundamental set of solutions y1, y2:
(enter your results as a comma separated list ). Then the complementary solution is yc = C1Y1 + c2y2-
(4) For the non-homogeneous problem we need to apply the variation of parameters formula to find a particular solution yp. Carrying this out you obtain
Yp =
Recall that the general solution is y = Ye + Yp. Find the unique solution satisfying y(1) = -5, y'(1) = -9
y =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8edd889e-338a-4275-872c-5b2b400b4e22%2F107714d7-741c-48f5-814e-c7725a4e8921%2Fdz81wo8_processed.png&w=3840&q=75)
Transcribed Image Text:We consider the non-homogeneous initial value problem z'y" – 4xy' + 6y = -12z1, y(1) = -5, y(1) = -9
By looking for solutions in the form y = z' in the homogeneous Euler-Cauchy problem Az?y" + Bry' + Cy = 0, we obtain a auxiliary equation Ar? + (B – A)r +C=0 which is the analog of the auxiliary equation in the constant
coefficient case.
(1) For this problem find the auxiliary equation:
(2) Find the roots of the auxiliary equation:
(enter your results as a comma separated list)
(3) Find a fundamental set of solutions y1, y2:
(enter your results as a comma separated list ). Then the complementary solution is yc = C1Y1 + c2y2-
(4) For the non-homogeneous problem we need to apply the variation of parameters formula to find a particular solution yp. Carrying this out you obtain
Yp =
Recall that the general solution is y = Ye + Yp. Find the unique solution satisfying y(1) = -5, y'(1) = -9
y =
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