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Find-y-CF-PI-in-following-differential-equation-d-2-y-dx-2-3-dy-dx-2y-e-2x-sinx-




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 .
Findy=CF+PIinfollowingdifferentialequation:d2ydx2+3dydx+2y=e2xsinx.
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=emxm2emx+3memx+2emx=0emx(m+1)(m+2)=0emx0som=1,2C.FAex+Be2xP.I=e2xsinxD2+3D+2=e2x.sinx(D+2)2+3(D+2)+2=e2x.sinxD2+7D+12=e2x.D2+127D(D2+12)249D2.sinx=e2x.sinx+12sinx7cosx(12+12)249(12)=e2x.11sinx7cosx121+49=e2x.11sinx7cosx170y=Aex+Be2x+e2x.11sinx7cosx170
Commented by niroj last updated on 27/Mar/20
great job sir.
greatjobsir.
Commented by TANMAY PANACEA. last updated on 27/Mar/20
thank you sir
thankyousir

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