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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}