y-3y-x-3-3x-

Question Number 91770 by john santu last updated on 03/May/20

y^{″} + 3 y = x^{3} + 3 x

Commented by Joel578 last updated on 03/May/20

Let the particular solution is in form

y_{p} = {A x}^{3} + {B x}^{2} + C x + D

\Rightarrow y_{p}^{^{'}} = 3 {A x}^{2} + 2 B x + C

\Rightarrow y_{p}^{^{″}} = 6 A x + 2 B

\Rightarrow y_{p}^{^{″}} + 3 y_{p} = [6 A x + 2 B] + 3 [{A x}^{3} + {B x}^{2} + C x + D]

= 3 {A x}^{3} + 3 {B x}^{2} + (6 A + 3 C) x + (2 B + 3 D) = x^{3} + 3 x

Comparing coeff . on both sides, we get

A = \frac{1}{3}, B = 0, C = \frac{1}{3}, D = 0

Complete solution

y = c_{1} \cos (\sqrt{3} x) + c_{2} \sin (\sqrt{3} x) + \frac{1}{3} x^{3} + \frac{1}{3} x

Commented by john santu last updated on 03/May/20

thank you

Answered by niroj last updated on 03/May/20

y^{^{″}} + 3 y = x^{3} + 3 x

\frac{d^{2} y}{{dx}^{2}} + 3 y = x^{3} + 3 x

(D^{2} + 3) y = x^{3} + 3 x

A . E ., m^{2} + 3

m = 0 \bar{+} \sqrt{3} i

CF = A \cos \sqrt{3} x + Bsin \sqrt{3} x

PI = \frac{x^{3} + 3 x}{D^{2} + 3} = \frac{1}{3} (\frac{1}{1 + \frac{D^{2}}{3}}) (x^{3} + 3 x)

= \frac{1}{3} {(1 + \frac{D^{2}}{3})}^{- 1} (x^{3} + 3 x)

= \frac{1}{3} [1 - \frac{D^{2}}{3} + \frac{D^{3}}{3} - . .] (x^{3} + 3 x)

= \frac{1}{3} (1 - \frac{D^{2}}{3} + \frac{D^{3}}{3}) (x^{3} + 3 x)

= \frac{1}{3} [(x^{3} + 3 x) - \frac{1}{3} . 18 x + \frac{1}{3} 18]

= \frac{1}{3} [x^{3} + 3 x - 6 x + 6]

= \frac{1}{3} (x^{3} - 3 x + 6)

y = Acos \sqrt{3} x + Bsin \sqrt{2} x + \frac{1}{3} (x^{3} - 3 x + 6) / / .

Commented by john santu last updated on 03/May/20

f a n t a s t i c

Commented by niroj last updated on 03/May/20

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Commented by abdomathmax last updated on 03/May/20

y = \frac{1}{3} (x^{3} - 3 x + 6) \Rightarrow y^{^{'}} = x^{2} - 1 a n d y^{^{″}} = 2 x

y^{^{″}} + 3 y = 2 x + x^{3} - 3 x + 6 = x^{3} - x + 6 \Rightarrow y_{p} i s n o t

s o l u t i o n f o r t h i s e q u a t i o n i t h i n k t h e e r r o r ? c o m e s

f r o m t h i s \frac{1}{1 + \frac{d^{2}}{3}} = 1 - \frac{d^{2}}{3} + \frac{d^{4}}{9} + \dots .

Commented by niroj last updated on 03/May/20

{(1 + D)}^{- 1} = 1 - D + D^{2} - \dots .

{(1 + D)}^{- 2} = 1 - 2 D + {3 D}^{2} - . .

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