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.
- kbLv 77 years agoFavorite Answer
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.
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
So, y = sin t is a solution to the complementary (homogeneous) equation.
To use Reduction of Order, assume that y = (sin t) z.
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).
-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.
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).
I hope this helps!
- Anonymous7 years ago
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