Answer to Question #225228 in Differential Equations for Asif

Question #225228

Find the particular solution of the differential equation 8d2x/dt2 + 16dx/dt -5x =0, when x(0) = 16 and x'(0) = -8.


1
Expert's answer
2021-08-12T18:15:50-0400

Given IVP:


8x(t)+16x(t)5x(t)=08x''(t)+16x'(t)-5x(t)=0

x(0)=16,x(0)=8x(0)=16,\quad x'(0)=-8

The characteristic equation is as follows

8k2+16k5=08k^2+16k-5=0

Roots:

k1=126/4,k1=1+26/4k_1=-1-\sqrt{26}/4,\quad k_1=-1+\sqrt{26}/4

General solution of IVP:

x(t)=Aek1t+Bek2t=et(Ae26/4t+Be26/4t)x(t)=Ae^{k_1t}+Be^{k_2t}=e^{-t}(Ae^{-\sqrt{26}/4t}+Be^{\sqrt{26}/4t})

The initial conditions give

x(0)=A+B=16x(0)=(A+B)+26/4(A+B)=8x(0)=A+B=16\\ x'(0)=-(A+B)+\sqrt{26}/4(-A+B)=-8A=8/13(13+26),B=8/13(13+26)A=-8/13(-13+\sqrt{26}), \quad B=8/13(13+\sqrt{26})

Finally,

x(t)=et(8/13(13+26)e26/4t+8/13(13+26)e26/4t)x(t)=e^{-t}\Big(-8/13(-13+\sqrt{26})e^{-\sqrt{26}/4t}\\+8/13(13+\sqrt{26})e^{\sqrt{26}/4t}\Big)


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