# -sin(t)y''+cos(t)y'+(sin(t)-2csc(t))y=t^2sin^2t?

show that y1t=sint is a solution to complementary equation.

use the method of reduction of order to construct a second solution y2(t) to the complementary equation and show y1t) and y2(t) form a fundamental set of solutions to the complementary eqation.

Relevance

Note: (-sin t) y'' + 2(cos t) y' + (sin t - 2 csc t) y = t^2 sin t

should (more than likely) be the DE; otherwise y = sin t is not a homogeneous solution.

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Letting y = sin t:

(-sin t) y'' + 2(cos t) y' + (sin t - 2 csc t) y

= (-sin t) (-sin t) + 2(cos t) (cos t) + (sin t - 2 csc t) (sin t)

= 2( sin^2(t) + cos^2(t)) - 2

= 2 - 2

= 0.

So, y = sin t is a solution to the complementary (homogeneous) equation.

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To use Reduction of Order, assume that y = (sin t) z.

Differentiating yields

y' = (cos t)z + (sin t)z'

y'' = (-sin t)z + 2(cos t)z' + (sin t)z''.

Substituting this into (-sin t) y'' + 2(cos t) y' + (sin t - 2 csc t) y = t^2 sin^2(t) yields

(-sin t) [(-sin t)z + 2(cos t)z' + (sin t)z''] + 2(cos t) [(cos t)z + (sin t)z'] + (sin t - 2 csc t) ((sin t) z) = t^2 sin^2(t).

Simplifying:

-sin^2(t) * z'' = t^2 sin^2(t).

==> z'' = -t^2.

Integrate both sides twice:

z' = (-1/3)t^3 + A

z = (-1/12)t^4 + At + B.

So, a general solution is

y = [(-1/12)t^4 + At + B] sin t

...= (At sin t + B sin t) - (1/12)t^4 sin t.

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Take y2 = t sin t

(since y = (-1/12)t^4 sin t is a particular solution, having no arbitrary constants).

We verify that y1 and y2 form a fundamental set of solutions to the complementary

equation by checking that their Wronskian is nonzero (as a function in t):

|..sin t......t sin t..|

|(sin t)'...(t sin t)'| =

|sin t.........t sin t.......|

|cos t...sin t + t cos t| = sin^2(t), which is a nonzero function (in t).

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I hope this helps!

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• Anonymous
7 years ago

The answer to said question is:

0.00003046483

We hope this answer has helped you emotionaly and physically.

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