Answer:
The second linearly independent solution is
g(t) = = -(4/9)(3t + 1)
Step-by-step explanation:
Given the differential equation
tx'' - (3t + 1)x' + 3x = 0 ...................(1)
and a solution
f(t) = 4e^(3t)
We want to find a second linearly independent solution g(t), using the method of reduction of order.
Let this second solution be
x = uf(t)
x = u. 4e^(3t) ...................................(2)
x' = u'. 4e^(3t) + u. 12e^(3t) ...........(3)
x'' = u''. 4e^(3t) + u'. 12e^(3t) + u'. 12e^(3t) + u. 36e^(3t)
= 4u''e^(3t) + 24u'e^(3t) + 36ue^(3t) .............(4)
Using the values of x, x', and x'' in (2), (3), and (4) in (1), we have
t[4u''e^(3t) + 24u'e^(3t) + 36ue^(3t)] - (3x + 1)[u'. 4e^(3t) + u. 12e^(3t)] + 3u. 4e^(3t) = 0
4tu''e^(3t) + (12t - 4)u'e^(3t) = 0......(5)
Let w = u'
then w' = u''
(5) now becomes
4tw'e^(3t) + (12t - 4)we^(3t) = 0
4tw'e^(3t) = (4 - 12t)we^(3t)
w'/w = (4 - 12t)/4t = (1/t) - 3
Integrating this, assuming all constants of integration are 0, we have
lnw = lnt - 3t
w = e^(lnt - 3t) = e^(lnt)e^(-3t)
w = te^(-3t)
But remember w = u'
=> u' = te^(-3t)
Integrating this, by part, taking constant of integration as 0, we have
u = -(1/9)(3t + 1)e^(-3t)
Using this in (2)
x = 4 [-(1/9)(3t + 1)e^(-3t)]e^(3t)
= -(4/9)(3t + 1)
And this is what we are looking for.
<em>this is the formula we will use because the signs match the one in the question</em>.
, now we need to solve for
, but
.
and
are on both sides, so dive them and they will be canceled out.
and you will have your value for