Answered: "Use the method of reduction of order to..."

midtlinjeg

Answered question

2021-09-04

Use the method of reduction of order to find the general of the differential equation
$t (t + 3) y^{″} - 3 (t + 2) y^{'} + 3 y = 0$

Answer & Explanation

Sally Cresswell

Skilled2021-09-05Added 91 answers

Solve $t \frac{d^{2} y (t)}{{d t}^{2}} (t + 3) - 3 \frac{d y (t)}{d t} (t + 2) + 3 y (t) = 0 :$
Let $y (t) = (t + 2) v (t), w h i c h g i v e s \frac{d y (t)}{d t} = \frac{d v (t)}{d t} (t + 2) + v (t) and \frac{d^{2} y (t)}{{d t}^{2}} = \frac{d^{2} v (t)}{{d t}^{2}} (t + 2) + 2 \frac{d v (t)}{d t} : - 3 (t + 2) (\frac{d v (t)}{d t} (t + 2) + v (t)) + t (t + 3) (\frac{d^{2} v (t)}{{d t}^{2}} (t + 2) + 2 \frac{d v (t)}{d t}) + 3 (t + 2) v (t) = 0$
Simplify:
$t \frac{d^{2} v (t)}{{d t}^{2}} (t^{2} + 5 t + 6) + \frac{d v (t)}{d t} (- t^{2} - 6 t - 12) = 0$
Let $\frac{d v (t)}{d t} = u (t)$ , which gives $\frac{d^{2} v (t)}{{d t}^{2}} = \frac{d u (t)}{d t} :$
Solve for $\frac{d u (t)}{d t} :$
$\frac{d u (t)}{d t} = \frac{(t^{2} + 6 t + 12) u (t)}{t (t^{2} + 5 t + 6)}$

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