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Question Number 86242 by niroj last updated on 27/Mar/20

Find y=CF+PI in following differential equation: (d^2 y/dx^2 )+3(dy/dx)+2y= e^(2x) sinx .

$$\:\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{C}}\mathrm{F}+\boldsymbol{\mathrm{P}}\mathrm{I}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }+\mathrm{3}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}+\mathrm{2}\boldsymbol{\mathrm{y}}=\:\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \:\boldsymbol{\mathrm{sinx}}\:. \\ $$$$\: \\ $$$$ \\ $$

Answered by TANMAY PANACEA. last updated on 27/Mar/20

y=e^(mx) m^2 e^(mx) +3me^(mx) +2e^(mx) =0 e^(mx) (m+1)(m+2)=0 e^(mx) ≠0 so m=−1,−2 C.F Ae^(−x) +Be^(−2x) P.I=((e^(2x) sinx)/(D^2 +3D+2)) =e^(2x) .((sinx)/((D+2)^2 +3(D+2)+2)) =e^(2x) .((sinx)/(D^2 +7D+12)) =e^(2x) .((D^2 +12−7D)/((D^2 +12)^2 −49D^2 )).sinx =e^(2x) .((−sinx+12sinx−7cosx)/((−1^2 +12)^2 −49(−1^2 ))) =e^(2x) .((11sinx−7cosx)/(121+49))=e^(2x) .((11sinx−7cosx)/(170)) y=Ae^(−x) +Be^(−2x) +e^(2x) .((11sinx−7cosx)/(170))

$${y}={e}^{{mx}} \: \\ $$$${m}^{\mathrm{2}} {e}^{{mx}} +\mathrm{3}{me}^{{mx}} +\mathrm{2}{e}^{{mx}} =\mathrm{0} \\ $$$${e}^{{mx}} \left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)=\mathrm{0} \\ $$$${e}^{{mx}} \neq\mathrm{0}\:\:{so}\:{m}=−\mathrm{1},−\mathrm{2} \\ $$$${C}.{F} \\ $$$${Ae}^{−{x}} +{Be}^{−\mathrm{2}{x}} \\ $$$${P}.{I}=\frac{{e}^{\mathrm{2}{x}} {sinx}}{{D}^{\mathrm{2}} +\mathrm{3}{D}+\mathrm{2}} \\ $$$$={e}^{\mathrm{2}{x}} .\frac{{sinx}}{\left({D}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}\left({D}+\mathrm{2}\right)+\mathrm{2}} \\ $$$$={e}^{\mathrm{2}{x}} .\frac{{sinx}}{{D}^{\mathrm{2}} +\mathrm{7}{D}+\mathrm{12}} \\ $$$$={e}^{\mathrm{2}{x}} .\frac{{D}^{\mathrm{2}} +\mathrm{12}−\mathrm{7}{D}}{\left({D}^{\mathrm{2}} +\mathrm{12}\right)^{\mathrm{2}} −\mathrm{49}{D}^{\mathrm{2}} }.{sinx} \\ $$$$={e}^{\mathrm{2}{x}} .\frac{−{sinx}+\mathrm{12}{sinx}−\mathrm{7}{cosx}}{\left(−\mathrm{1}^{\mathrm{2}} +\mathrm{12}\right)^{\mathrm{2}} −\mathrm{49}\left(−\mathrm{1}^{\mathrm{2}} \right)} \\ $$$$={e}^{\mathrm{2}{x}} .\frac{\mathrm{11}{sinx}−\mathrm{7}{cosx}}{\mathrm{121}+\mathrm{49}}={e}^{\mathrm{2}{x}} .\frac{\mathrm{11}{sinx}−\mathrm{7}{cosx}}{\mathrm{170}} \\ $$$${y}={Ae}^{−{x}} +{Be}^{−\mathrm{2}{x}} +{e}^{\mathrm{2}{x}} .\frac{\mathrm{11}{sinx}−\mathrm{7}{cosx}}{\mathrm{170}} \\ $$

Commented by niroj last updated on 27/Mar/20

great job sir.

$$\mathrm{great}\:\mathrm{job}\:\mathrm{sir}. \\ $$

Commented by TANMAY PANACEA. last updated on 27/Mar/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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