Answer to Question #225228 in Differential Equations for Asif

Question

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

Answer 1

Given IVP:

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

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

The characteristic equation is as follows

$$8k^2+16k-5=0$$

Roots:

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

General solution of IVP:

$$x(t)=Ae^{k_1t}+Be^{k_2t}=e^{-t}(Ae^{-\sqrt{26}/4t}+Be^{\sqrt{26}/4t})$$

The initial conditions give

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

Finally,

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