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Question Number 82390 by jagoll last updated on 21/Feb/20

$${(D^2-1)}^{2}y=t^3$$$$findsolution$$

Answered by TANMAY PANACEA last updated on 21/Feb/20

$$y=e^{mt}so\frac{dy}{dt}={me}^{mt}and\frac{d^{2}y}{{dt}^2}=m^2{e}^{mt}$$$${(m^2-1)}^2=0$$$$m=1,1,-1,-1$$$$complementaryfunction$$$$y={Ae}^t+{Bte}^t+{Ce}^{-t}+{Fte}^{-t}$$$$$$$$particularintregal$$$$nowformula$$$${(1+x)}^{-2}=1-2x+3x^2-4x^3+...$$$$y=\frac{t^3}{{(D^2-1)}^2}=\frac{t^3}{{(1-D^2)}^2}$$$$y={(1-D^2)}^{-2}{t}^3$$$$=(1+2D^2+3D^4+otherstermsignored)t^3$$$$=(t^3+2\times 3\times 2t)[\frac{{dt}^3}{dt}=3t^2\to \frac{d\left(3t^2\right)}{dt}=6t]$$$$=t^3+12t$$$$completesolution$$$$y={Ae}^t+{Bte}^t+{Ce}^{-t}+{Fte}^{-t}+t^3+12t$$

Commented by jagoll last updated on 21/Feb/20

$$thankyousir$$

Commented by TANMAY PANACEA last updated on 21/Feb/20

$$mostwelcome$$

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