(D^2 +1)^2 y = x^2 cos x


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Question Number 91147 by john santu last updated on 28/Apr/20

${(D^{2} + 1)}^{2} y = x^{2} \cos x$
Commented by MWSuSon last updated on 28/Apr/20
I've corrected the mistake.
	Answered by MWSuSon last updated on 28/Apr/20
	$Auxillary equation (m^2 +1)(m^2 +1)=0 m_1 =±i m_2 =±i y_c =(C_1 +xC_2 )cos (x)+(C_3 +xC_4 )sin (x) y_p =(1/((D^2 +1)^2 ))x^2 cosx let cosx=Re{e^(ix) }=Re{cosx+isinx} y_p =e^(ix) (1/([(D+i)^2 +1]^2 ))x^2 =e^(ix) (1/([(D^2 +i2D−1)+1]^2 ))x^2 =e^(ix) (1/((D^2 +i2D)^2 ))x^2 =e^(ix) (1/(D^4 +i4D^3 −4D^2 ))x^2 =e^(ix) (1/(D^2 ))((1/(D^2 +i4D−4))x^2 ) =e^(ix) (1/D^2 )(D^2 +i4D−4)^(−1) x^2 =e^(ix) (1/D^2 )[−(1/4)(1+(−((D^2 +i4D)/4)))^(−1) x^2 ] skipping the expansion part =e^(ix) (1/D^2 )[−(x^2 /4)−(1/8)−((ix)/2)+(1/2)] =e^(ix) [(x^2 /4)−(x^4 /(48))−((ix^3 )/(12))−(x^2 /(16))] =e^(ix) [((3x^2 )/(16))−(x^4 /(48))−((ix^3 )/(12))] =(cosx+isinx)[((3x^2 )/(16))−(x^4 /(48))−((ix^3 )/(12))] taking only the real part of the expansion y_p =((3x^2 cosx)/(16))−((x^4 cosx)/(48))+((x^3 sinx)/(12)) y=(C_1 +xC_2 )cos (x)+(C_3 +xC_4 )sin (x)+((3x^2 cosx)/(16))−((x^4 cosx)/(48))+((x^3 sinx)/(12)) it got really messy doing this i might have made mistakes with the signs.$
	$A u x i l l a r y e q u a t i o n$ $(m^{2} + 1) (m^{2} + 1) = 0$ $m_{1} = \pm i$ $m_{2} = \pm i$ $y_{c} = (C_{1} + {x C}_{2}) \cos (x) + (C_{3} + {x C}_{4}) \sin (x)$ $y_{p} = \frac{1}{{(D^{2} + 1)}^{2}} x^{2} c o s x$ $l e t c o s x = R e {e^{i x}} = R e {c o s x + i s i n x}$ $y_{p} = e^{i x} \frac{1}{{[{(D + i)}^{2} + 1]}^{2}} x^{2}$ $= e^{i x} \frac{1}{{[(D^{2} + i 2 D - 1) + 1]}^{2}} x^{2}$ $= e^{i x} \frac{1}{{(D^{2} + i 2 D)}^{2}} x^{2}$ $= e^{i x} \frac{1}{D^{4} + i 4 D^{3} - 4 D^{2}} x^{2}$ $= e^{i x} \frac{1}{D^{2}} (\frac{1}{D^{2} + i 4 D - 4} x^{2})$ $= e^{i x} \frac{1}{D^{2}} {(D^{2} + i 4 D - 4)}^{- 1} x^{2}$ $= e^{i x} \frac{1}{D^{2}} [- \frac{1}{4} {(1 + (- \frac{D^{2} + i 4 D}{4}))}^{- 1} x^{2}]$ $s k i p p i n g t h e e x p a n s i o n p a r t$ $= e^{i x} \frac{1}{D^{2}} [- \frac{x^{2}}{4} - \frac{1}{8} - \frac{i x}{2} + \frac{1}{2}]$ $= e^{i x} [\frac{x^{2}}{4} - \frac{x^{4}}{48} - \frac{{i x}^{3}}{12} - \frac{x^{2}}{16}]$ $= e^{i x} [\frac{3 x^{2}}{16} - \frac{x^{4}}{48} - \frac{{i x}^{3}}{12}]$ $= (c o s x + i s i n x) [\frac{3 x^{2}}{16} - \frac{x^{4}}{48} - \frac{{i x}^{3}}{12}]$ $t a k i n g o n l y t h e r e a l p a r t o f t h e e x p a n s i o n$ $y_{p} = \frac{3 x^{2} c o s x}{16} - \frac{x^{4} c o s x}{48} + \frac{x^{3} s i n x}{12}$ $y = (C_{1} + {x C}_{2}) \cos (x) + (C_{3} + {x C}_{4}) \sin (x) + \frac{3 x^{2} c o s x}{16} - \frac{x^{4} c o s x}{48} + \frac{x^{3} s i n x}{12}$ $i t g o t r e a l l y m e s s y d o i n g t h i s i m i g h t h a v e$ $m a d e m i s t a k e s w i t h t h e s i g n s .$
	Commented by john santu last updated on 28/Apr/20

	$w h a t t y p e t h e d i f f e q s i r ?$
	Commented by MWSuSon last updated on 28/Apr/20
	fourth order constant coefficient
	Answered by MWSuSon last updated on 28/Apr/20

	$i n t r e s t i n g c o m p l i m e n t a r y f u n c t i o n$
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