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Solve the given third-order differential equation by variation of parameters.

y''' + y' = cot(x)

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Answer: To solve the third-order differential equation y''' + y' = cot(x) by variation of parameters, we first need to find the solution to the associated homogeneous equation, which is:

y''' + y' = 0

The characteristic equation is r^3 + r = 0, which can be factored as r(r^2 + 1) = 0. This gives us the roots r = 0, r = i, and r = -i. Therefore, the general solution to the homogeneous equation is:

y_h = c1 + c2 cos(x) + c3 sin(x)

To find a particular solution to the non-homogeneous equation using variation of parameters, we assume that the solution has the form:

y_p = u1(x) + u2(x) cos(x) + u3(x) sin(x)

where u1, u2, and u3 are functions to be determined.

We can find the derivatives of y_p:

y'_p = u1'(x) + u2'(x) cos(x) - u2(x) sin(x) + u3'(x) sin(x) + u3(x) cos(x)

y''_p = u1''(x) + u2''(x) cos(x) - 2u2'(x) sin(x) - u2(x) cos(x) + u3''(x) sin(x) + 2u3'(x) cos(x) - u3(x) sin(x)

y'''_p = u1'''(x) + u2'''(x) cos(x) - 3u2''(x) sin(x) - 3u2'(x) cos(x) - u2(x) sin(x) + u3'''(x) sin(x) + 3u3''(x) cos(x) - 3u3'(x) sin(x)

Substituting these derivatives into the non-homogeneous equation, we get:

u1'''(x) + u2'''(x) cos(x) - 3u2''(x) sin(x) - 3u2'(x) cos(x) - u2(x) sin(x) + u3'''(x) sin(x) + 3u3''(x) cos(x) - 3u3'(x) sin(x) + u1'(x) + u2'(x) cos(x) - u2(x) sin(x) + u3'(x) sin(x) + u3(x) cos(x) = cot(x)

Grouping the terms with the same functions together, we get:

u1'''(x) + u1'(x) = 0

u2'''(x) cos(x) - 3u2''(x) sin(x) - u2(x) sin(x) + u2'(x) cos(x) + u2'(x) cos(x) = cot(x) cos(x)

u3'''(x) sin(x) + 3u3''(x) cos(x) - 3u3'(x) sin(x) + u3'(x) sin(x) + u3(x) cos(x) = cot(x) sin(x)

The first equation is a first-order differential equation, which can be solved by integrating both sides:

u1'(x) + u1(x) = c1

where c1 is a constant of integration. The solution to this equation is:

u1(x) = c1 + c2 e^(-x)

where c2 is another constant of integration.

Explanation:

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