uberathlete
New member
- Joined
- Jan 16, 2006
- Messages
- 48
Hi everyone. I've done a differential equation problem but am unsure if I'm right. I would appreciate it if someone could look over what I did ...
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Find x(t) for:
. . .[x(t)]^2 - 2[x'(t)]^2 = c
...where c is some constant.
Differentiate both sides with respect to t:
. . .2x'(t) - 4x"(t) = 0
. . .1/2 x'(t) = x"(t)
. . .x"(t) - 1/2 x'(t) = 0
Let u'(t) = x"(t) and u(t) = x'(t). Then:
. . .u'(t) - 1/2 u(t) = 0
. . .d [ e^[-t/2) u(t) ] / dt = 0
. . .e^[(-t/2)]t u(t) = c
...where c is a constant
. . .u(t) = x'(t) = c e^[(t/2)]
Integrate both sides to get:
. . .x(t) = 2c e^[(t/2)] + k
...where k is a constant of integration
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There. Hope it's right. It looks right though.
-------------------------------------------------------------------------------
Find x(t) for:
. . .[x(t)]^2 - 2[x'(t)]^2 = c
...where c is some constant.
Differentiate both sides with respect to t:
. . .2x'(t) - 4x"(t) = 0
. . .1/2 x'(t) = x"(t)
. . .x"(t) - 1/2 x'(t) = 0
Let u'(t) = x"(t) and u(t) = x'(t). Then:
. . .u'(t) - 1/2 u(t) = 0
. . .d [ e^[-t/2) u(t) ] / dt = 0
. . .e^[(-t/2)]t u(t) = c
...where c is a constant
. . .u(t) = x'(t) = c e^[(t/2)]
Integrate both sides to get:
. . .x(t) = 2c e^[(t/2)] + k
...where k is a constant of integration
-------------------------------------------------------------------------------
There. Hope it's right. It looks right though.