Question-131488 – Tinku Tara

Question Number 131488 by Ahmed1hamouda last updated on 05/Feb/21

Answered by Ñï= last updated on 18/Feb/21

y^{″} + 2 y^{'} + 5 y = 6 e^{2 x} + {x s i n}^{2} x + e^{x} c o s 2 x

y_{p} = \frac{1}{D^{2} + 2 D + 5} (6 e^{2 x} + {x s i n}^{2} x + e^{x} c o s 2 x)

= \frac{6}{2^{2} + 2 \times 2 + 5} e^{2 x} + e^{x} (D^{2} - 2 D + 5) \frac{1}{{(D^{2} + 5)}^{2} - 4 D^{2}} c o s 2 x + \frac{1}{D^{2} + 2 D + 5} {x s i n}^{2} x

= \frac{6}{13} e^{2 x} + e^{x} (1 - 2 D) c o s 2 x + \frac{1}{D^{2} + 2 D + 5} I m (e^{2 i x} x)

= \frac{6}{13} e^{2 x} + e^{x} (c o s 2 x + 4 s i n 2 x) + I m (\frac{1}{D^{2} + 2 D + 5} e^{2 i x} x)

= \frac{6}{13} e^{2 x} + e^{x} (c o s 2 x + 4 s i n 2 x) + I m (e^{2 i x} \frac{1}{{(D + 2 i)}^{2} + 4 i + 5} x)

= \frac{6}{13} e^{2 x} + e^{x} (c o s 2 x + 4 s i n 2 x) + I m (e^{2 i x} \frac{1}{4 i + 1} \cdot \frac{1}{\frac{D^{2}}{4 i + 1} + \frac{4 i D}{4 i + 1} + 1} x)

= \frac{6}{13} e^{2 x} + e^{x} (c o s 2 x + 4 s i n 2 x) + I m [e^{2 i x} \frac{1}{4 i + 1} \cdot (x - \frac{4 i}{4 i + 1})]

= \frac{6}{13} e^{2 x} + e^{x} (c o s 2 x + 4 s i n 2 x) - \frac{4 x}{15} c o s 2 x - \frac{x}{15} s i n 2 x - \frac{16}{15} s i n 2 x - \frac{4}{15} c o s 2 x

λ^{2} + 2 λ + 5 = 0 \Rightarrow λ_{1} = - 1 + 2 i, λ = - 1 - 2 i

y = C_{1} e^{- x} s i n 2 x + C_{2} e^{- x} c o s 2 x + \frac{6}{13} e^{2 x} + e^{x} (c o s 2 x + 4 s i n 2 x) - \frac{4 x}{15} c o s 2 x - \frac{x}{15} s i n 2 x - \frac{16}{15} s i n 2 x - \frac{4}{15} c o s 2 x

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