Nonhomogeneous linear differential equation

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The discussion centers on solving the nonhomogeneous linear differential equation y'' - 6y' + 11y - 6 = e^(4x). The Wronskian determinant is calculated as W = 10e^(6x), and solutions for u', v', and w' are derived as u' = (1/4)e^(8x), v' = (-1/9)e^(9x), and w' = (-1/7)e^(7x). Integration of these results yields y = (1/10)e^(2x) - (1/30)e^(3x) - (1/10)e^(x). The original poster notes a discrepancy in the number of solutions, indicating a potential error in their Wronskian calculation, suggesting W = 2e^(6x).
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Homework Statement


solve for y(x).

y"'-6y"+11y'-6=e^{4x}



Homework Equations



Wronskian determinant. Method of variations.

The Attempt at a Solution

Supposing that [u', v', w'] are the solutions, wronskian det=W is 10e^{6x}
By use of x_k=\frac{det(M_{k})}{det(x)}, I got u'=\frac{1}{4}e^{8x},v'=\frac{-1}{9}e^{9x}, w'=\frac{-1}{7}e^{7x}. Integration gives y=\frac{1}{10}e^{2x}-\frac{1}{30}e^{3x}-\frac{e^{x}}{10}.
 
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The answer has four whereas i came up with only 3 since the roots are 1,2, and 3 respectively. I think W=2e^{6x}
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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