Resolved: "A differential equation and a nontrivial solution f..."

banganX

Answered question

2021-02-05

A differential equation and a nontrivial solution f are given below. Find a second linearly independent solution using reduction of order. Assume that all constants of integration are zero.
$t x - (2 t + 1) x + 2 x = 0, t > 0, f (t) = 3 \cdot e^{2 t}$

Answer & Explanation

lamusesamuset

Skilled2021-02-06Added 93 answers

Consider the second linearly independent solution as $g (t) = v (t), f (t) = 3 v e^{2 t}$
$g^{'} (t) = 6 v e^{2 t} + 3 e^{2 t} v^{'}$
$g^{″} (t) = 12 v e^{2 t} + 6 e^{2 t} v^{'} + 6 e^{2 t} v^{'} + 3 e^{2 t} v^{″}$ $= 12 v e^{2 t} + 12 e^{2 t} v^{'} + 3 e^{2 t} v^{″}$
$t (12 v e^{2 t} + 12 e^{2 t} v^{'} + 3 e^{2 t} v^{″}) - (2 t + 1) (6 v e^{2 t} + 3 e^{2 t} v^{'}) + 2 (3 v e^{2 t}) = 0$
$12 t v e^{2 t} + 12 t e^{2 t} v^{'} + 3 t e^{2 t} v^{″} - 12 t v e^{2 t} - 6 t e^{2 t} v^{'} - 6 v e^{2 t} - 3 e^{2 t} v^{'} + 6 v e^{2 t} = 0$
$3 t e^{2 t} v^{″} + (6 t - 3) e^{2 t} v^{'} = 0$
$3 t v^{″} + (6 t - 3) v^{'} = 0$
Let $w = v^{'}$ , then
$3 t w^{'} + (6 t - 3) w = 0$
$3 t w^{'} = - (6 t - 3) w$
$\frac{d w}{w} = \frac{3 - 6 t}{3 t} d t$
$\frac{d w}{w} = (\frac{1}{t} - 2) d t$
$\int \frac{d w}{w} = \int (\frac{1}{2} - 2) d t$
$\ln w = \ln t - 2 t + C$
$w = e^{\ln t - 2 t + C} = A t e^{- 2 t}$
$w = v^{'}$ , $v^{'} = A t e^{- 2 t}$
$v (t) = - \frac{A}{2} t e^{- 2 t} + \int \frac{A}{2} e^{- 2 t} d t$
$v (t) = - \frac{A}{2} t e^{- 2 t} - \frac{A}{4} e^{- 2 t} + C$
Choose $A = - 2$ and $C = 0$ , thus $v (t) = t e^{- 2 t} + \frac{1}{2} e^{- 2 t}$
$g (t) = 3 (t e^{- 2 t} + \frac{1}{2} e^{- 2 t}) e^{2 t}$ $= 3 t + \frac{3}{2}$
The second linearly independent solution is $g (t) = 3 t + \frac{3}{2}$

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