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Question Number 91147 by john santu last updated on 28/Apr/20
$$\left({D}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {y}\:=\:{x}^{\mathrm{2}} \mathrm{cos}\:{x}\: \\ $$
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.](Q91191.png)
$${Auxillary}\:{equation} \\ $$$$\left({m}^{\mathrm{2}} +\mathrm{1}\right)\left({m}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${m}_{\mathrm{1}} =\pm{i} \\ $$$${m}_{\mathrm{2}} =\pm{i} \\ $$$${y}_{{c}} =\left({C}_{\mathrm{1}} +{xC}_{\mathrm{2}} \right)\mathrm{cos}\:\left({x}\right)+\left({C}_{\mathrm{3}} +{xC}_{\mathrm{4}} \right)\mathrm{sin}\:\left({x}\right) \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{\left({D}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{x}^{\mathrm{2}} {cosx} \\ $$$${let}\:{cosx}={Re}\left\{{e}^{{ix}} \right\}={Re}\left\{{cosx}+{isinx}\right\} \\ $$$${y}_{{p}} ={e}^{{ix}} \frac{\mathrm{1}}{\left[\left({D}+{i}\right)^{\mathrm{2}} +\mathrm{1}\right]^{\mathrm{2}} }{x}^{\mathrm{2}} \\ $$$$={e}^{{ix}} \frac{\mathrm{1}}{\left[\left({D}^{\mathrm{2}} +{i}\mathrm{2}{D}−\mathrm{1}\right)+\mathrm{1}\right]^{\mathrm{2}} }{x}^{\mathrm{2}} \\ $$$$={e}^{{ix}} \frac{\mathrm{1}}{\left({D}^{\mathrm{2}} +{i}\mathrm{2}{D}\right)^{\mathrm{2}} }{x}^{\mathrm{2}} \\ $$$$={e}^{{ix}} \frac{\mathrm{1}}{{D}^{\mathrm{4}} +{i}\mathrm{4}{D}^{\mathrm{3}} −\mathrm{4}{D}^{\mathrm{2}} }{x}^{\mathrm{2}} \\ $$$$={e}^{{ix}} \frac{\mathrm{1}}{{D}^{\mathrm{2}} \:}\left(\frac{\mathrm{1}}{{D}^{\mathrm{2}} +{i}\mathrm{4}{D}−\mathrm{4}}{x}^{\mathrm{2}} \right) \\ $$$$={e}^{{ix}} \frac{\mathrm{1}}{{D}^{\mathrm{2}} }\left({D}^{\mathrm{2}} +{i}\mathrm{4}{D}−\mathrm{4}\right)^{−\mathrm{1}} {x}^{\mathrm{2}} \\ $$$$={e}^{{ix}} \frac{\mathrm{1}}{{D}^{\mathrm{2}} }\left[−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\left(−\frac{{D}^{\mathrm{2}} +{i}\mathrm{4}{D}}{\mathrm{4}}\right)\right)^{−\mathrm{1}} {x}^{\mathrm{2}} \right] \\ $$$${skipping}\:{the}\:{expansion}\:{part} \\ $$$$={e}^{{ix}} \frac{\mathrm{1}}{{D}^{\mathrm{2}} }\left[−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}−\frac{{ix}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$={e}^{{ix}} \left[\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}^{\mathrm{4}} }{\mathrm{48}}−\frac{{ix}^{\mathrm{3}} }{\mathrm{12}}−\frac{{x}^{\mathrm{2}} }{\mathrm{16}}\right] \\ $$$$={e}^{{ix}} \left[\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{16}}−\frac{{x}^{\mathrm{4}} }{\mathrm{48}}−\frac{{ix}^{\mathrm{3}} }{\mathrm{12}}\right] \\ $$$$=\left({cosx}+{isinx}\right)\left[\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{16}}−\frac{{x}^{\mathrm{4}} }{\mathrm{48}}−\frac{{ix}^{\mathrm{3}} }{\mathrm{12}}\right] \\ $$$${taking}\:{only}\:{the}\:{real}\:{part}\:{of}\:{the}\:{expansion} \\ $$$${y}_{{p}} =\frac{\mathrm{3}{x}^{\mathrm{2}} {cosx}}{\mathrm{16}}−\frac{{x}^{\mathrm{4}} {cosx}}{\mathrm{48}}+\frac{{x}^{\mathrm{3}} {sinx}}{\mathrm{12}} \\ $$$${y}=\left({C}_{\mathrm{1}} +{xC}_{\mathrm{2}} \right)\mathrm{cos}\:\left({x}\right)+\left({C}_{\mathrm{3}} +{xC}_{\mathrm{4}} \right)\mathrm{sin}\:\left({x}\right)+\frac{\mathrm{3}{x}^{\mathrm{2}} {cosx}}{\mathrm{16}}−\frac{{x}^{\mathrm{4}} {cosx}}{\mathrm{48}}+\frac{{x}^{\mathrm{3}} {sinx}}{\mathrm{12}} \\ $$$${it}\:{got}\:{really}\:{messy}\:{doing}\:{this}\:{i}\:{might}\:{have} \\ $$$${made}\:{mistakes}\:{with}\:{the}\:{signs}. \\ $$
Commented by john santu last updated on 28/Apr/20
$${what}\:{type}\:{the}\:{diff}\:{eq}\:{sir}? \\ $$
Commented by MWSuSon last updated on 28/Apr/20
fourth order constant coefficient
Answered by MWSuSon last updated on 28/Apr/20
$${intresting}\:{complimentary}\:{function} \\ $$