"Applications of Differential and Difference Equations Solve (D2−2D+1)y=xexsin⁡x" solved and explained

Name: Applications of Differential and Difference Equations Solve (D2−2D+1)y=xexsin⁡x
Uploaded: 2022-02-23T17:22:57+00:00
Description: Applications of Differential and Difference Equations Solve (D2−2D+1)y=xexsin⁡x

maduregimc

Answered question

2022-01-18

Applications of Differential and Difference Equations
Solve

(D^{2} - 2 D + 1) y = x e^{x} \sin x

Answer & Explanation

otoplilp1

Beginner2022-01-19Added 41 answers

(D^{2} - 2 D + 1) y = x e^{x} \sin x

The auxiliary equation is

m^{2} - 2 + 1 = 0

m = 1, 1

C . F = (A x + B) e^{x}

P . I = \frac{1}{D^{2} - 2 D + 1} \times e^{x} \sin x

= e^{x} \frac{1}{{(D + 1)}^{2} - 2 (D + 1) + 1} \times \sin x

= e^{x} [\frac{1}{D^{2}}] \times \sin x

= e^{x} [x \frac{1}{D^{2}} \sin x] - e^{x} \frac{2 D}{{(D^{2})}^{2}} \sin x

= - x e^{x} \sin x - 2 e^{x} \cos x

= - e^{x} (x \sin x + 2 \cos x)

y = c f + π

y = (A x + b) e^{x} - e^{x} (x \sin x + 2 \cos x)

Navreaiw

Beginner2022-01-20Added 34 answers

The auxiliary equation is

m^{2} - 2 + 1 = 0

m = 1, 1

C . F = (A x + B) e^{x}

= \frac{1}{D^{2} - 2 D + 1} x e^{x} \sin x

= e^{x} \frac{1}{{(D + 1)}^{2} - 2 (D + 1) + 1} x \sin x

= e^{x} [\frac{1}{D^{2}}] x \sin x

= e^{x} [x \frac{1}{D^{2}} \sin x] - e^{x} \frac{2 D}{{(D^{2})}^{2}} \sin x

= - x e^{x} \sin x - 2 e^{x} \cos x

= - e^{x} (x \sin x + 2 \cos x)

∴ y = C . F + P . I .

y = (A x + B) e^{x} - e^{x} (x \sin x + 2 \cos x)

alenahelenash

Expert2022-01-24Added 556 answers

y_{p} = \frac{1}{D^{2} - D + 1} x e^{x} \sin x

= e^{x} \frac{1}{(D + 1)^{2} - (2 D + D) + 1} x \sin x

= e^{x} \frac{1}{D^{2} + 2 D + 1 - 2 D 2 + 1} x \sin x

= e^{x} \frac{1}{D^{2}} x \sin x

In that case x and sin x,

\sin x = v

Note:

y_{p} = \frac{1}{F (D)} [x \times V] = x [\frac{1 \times [v]}{F (D)} - \frac{F^{'} (D) x [v]}{[F (D)]^{2}}]

= e^{x} {(x \frac{\sin x}{D^{2}}) - (\frac{2 D \times \sin x}{(D^{2})^{2}})}

= e^{x} {(\frac{x \sin x}{(- 1)}) - (\frac{2 D \times \sin x}{D^{4}})}

= e^{x} {(- x \sin x) - (\frac{2}{D^{3}} \sin x)}

= e^{x} {(- x \sin x) - \frac{2 \sin x}{D \times D^{2}}}

= e^{x} (- x \sin x - \frac{2 \sin x}{D \times (- 1)}) - (Put D^{2} = - (1)^{2})

= e^{x} (- x \sin x - 2 \cos x)

y_{p} = - e^{x} (x \sin x + 2 \cos x)

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