Question Number 61180 by mathsolverby Abdo last updated on 30/May/19
$${solve}\:{y}^{''} \:+\mathrm{3}{y}^{'} −{y}\:={sin}\left(\mathrm{2}{x}\right) \\ $$
Answered by tanmay last updated on 30/May/19
![(d^2 y/dx^2 )+3(dy/dx)−y=sin2x C.F determination (d^2 y/dx^2 )+3(dy/dx)−y=0 y=Ae^(mx) Am^2 e^(mx) +3Ame^(mx) −Ae^(mx) =0 (m^2 +3m−1)=0 [Ae^(mx) ≠0] m=((−3±(√(9+4)))/2)=((−3±(√(13)))/2) C.F=C_1 e^((((−3+(√(13)))/2))x) +C_2 e^((((−3−(√(13)))/2))x) P.I (D^2 +3D−1)y=sin2x y=((D^2 −1−3D)/((D^2 −1)^2 −9D^2 ))×sin2x [Dsin2x=2cos2x and D^2 sin2x=−4sin2x] y=((−4sin2x−sin2x−6cos2x)/((−2^2 −1)^2 −9(−2^2 ))) complete solution y=C_1 e^((((−3+(√(13)))/2))x) +C_2 e^((((−3−(√(13)))/2))x) +((−5sin2x−6cos2x)/(25−36)) y=C_1 e^((((−3+(√(13)))/2))x) +C_2 e^((((−3−(√(13)))/2))+((5sin2x+6cos2x)/(11)))](https://www.tinkutara.com/question/Q61239.png)
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{3}\frac{{dy}}{{dx}}−{y}={sin}\mathrm{2}{x} \\ $$$${C}.{F}\:{determination} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{3}\frac{{dy}}{{dx}}−{y}=\mathrm{0} \\ $$$${y}={Ae}^{{mx}} \: \\ $$$${Am}^{\mathrm{2}} {e}^{{mx}} +\mathrm{3}{Ame}^{{mx}} −{Ae}^{{mx}} =\mathrm{0} \\ $$$$\left({m}^{\mathrm{2}} +\mathrm{3}{m}−\mathrm{1}\right)=\mathrm{0}\:\:\:\:\:\:\:\left[{Ae}^{{mx}} \neq\mathrm{0}\right] \\ $$$${m}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}+\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${C}.{F}={C}_{\mathrm{1}} {e}^{\left(\frac{−\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right){x}} +{C}_{\mathrm{2}} {e}^{\left(\frac{−\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}\right){x}} \\ $$$${P}.{I} \\ $$$$\left({D}^{\mathrm{2}} +\mathrm{3}{D}−\mathrm{1}\right){y}={sin}\mathrm{2}{x} \\ $$$${y}=\frac{{D}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{D}}{\left({D}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9}{D}^{\mathrm{2}} }×{sin}\mathrm{2}{x}\:\:\:\:\left[{Dsin}\mathrm{2}{x}=\mathrm{2}{cos}\mathrm{2}{x}\:{and}\:{D}^{\mathrm{2}} {sin}\mathrm{2}{x}=−\mathrm{4}{sin}\mathrm{2}{x}\right] \\ $$$$ \\ $$$${y}=\frac{−\mathrm{4}{sin}\mathrm{2}{x}−{sin}\mathrm{2}{x}−\mathrm{6}{cos}\mathrm{2}{x}}{\left(−\mathrm{2}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9}\left(−\mathrm{2}^{\mathrm{2}} \right)} \\ $$$${complete}\:{solution} \\ $$$${y}={C}_{\mathrm{1}} {e}^{\left(\frac{−\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right){x}} +{C}_{\mathrm{2}} {e}^{\left(\frac{−\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}\right){x}} +\frac{−\mathrm{5}{sin}\mathrm{2}{x}−\mathrm{6}{cos}\mathrm{2}{x}}{\mathrm{25}−\mathrm{36}} \\ $$$${y}={C}_{\mathrm{1}} {e}^{\left(\frac{−\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right){x}} +{C}_{\mathrm{2}} {e}^{\left(\frac{−\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}\right)+\frac{\mathrm{5}{sin}\mathrm{2}{x}+\mathrm{6}{cos}\mathrm{2}{x}}{\mathrm{11}}} \\ $$