find tbe general solution of the D.E y^(′′) −5y^′ +4y=e^t cos(t)


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Question Number 79626 by M±th+et£s last updated on 26/Jan/20

$f i n d t b e g e n e r a l s o l u t i o n o f t h e D . E$ $y^{^{″}} - 5 y^{^{'}} + 4 y = e^{t} c o s (t)$
	Answered by mind is power last updated on 27/Jan/20

	$y^{″} - 5 y^{'} + 4 y = 0$ $\Rightarrow y = {a e}^{t} + {b e}^{4 t}$ $f (t) = e^{t} (a c o s (t) + b s i n (t))$ $\Rightarrow f^{'} (t) = e^{t} (- a s i n (t) + b c o s (t) + a c o s (t) + b s i n (t))$ $f^{″} (t) = e^{t} (- 2 a s i n (t) + 2 b c o s (t))$ $e^{t} (3 a s i n (t) - 3 b c o s (t) - a c o s (t) - b s i n (t)) = e^{t} c o s (t)$ $3 a - b = 0$ $- (3 b + a) = 1 \Rightarrow 3 b + a = - 1$ $\Rightarrow a = - \frac{1}{10}, b = - \frac{3}{10}$ $f = - \frac{e^{t} (c o s (t) + 3 s i n (t))}{10}$ $g e n e r a l s o l u t i o n {a e}^{t} + {b e}^{4 t} - \frac{e^{t} (c o s (t) + 3 s i n (t))}{10}$
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