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Question Number 757 by 123456 last updated on 07/Mar/15

k(d^2 i/dt^2 )+l(di/dt)+ri=v  i(0)=0  i′(0)=0  k,l,r,v are constants

$${k}\frac{{d}^{\mathrm{2}} {i}}{{dt}^{\mathrm{2}} }+{l}\frac{{di}}{{dt}}+{ri}={v} \\ $$$${i}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${i}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${k},{l},{r},{v}\:{are}\:{constants} \\ $$

Commented by prakash jain last updated on 08/Mar/15

Homogeneous Solution  kx^2 +lx+r=0  x=((−l±(√(l^2 −4kr)))/(2k))  i_h =c_1 e^(x_1 t) +c_2 e^(x_2 t)  if (x_1 ≠x_2 )  i_h =c_1 e^(x_1 t) +c_2 te^(x_1 t)  if (x_1 =x_2 )  Particular Solution  i_p =(v/r)  i(t)=i_h +i_p

$$\mathrm{Homogeneous}\:\mathrm{Solution} \\ $$$${kx}^{\mathrm{2}} +{lx}+{r}=\mathrm{0} \\ $$$${x}=\frac{−{l}\pm\sqrt{{l}^{\mathrm{2}} −\mathrm{4}{kr}}}{\mathrm{2}{k}} \\ $$$${i}_{{h}} ={c}_{\mathrm{1}} {e}^{{x}_{\mathrm{1}} {t}} +{c}_{\mathrm{2}} {e}^{{x}_{\mathrm{2}} {t}} \:\mathrm{if}\:\left({x}_{\mathrm{1}} \neq{x}_{\mathrm{2}} \right) \\ $$$${i}_{{h}} ={c}_{\mathrm{1}} {e}^{{x}_{\mathrm{1}} {t}} +{c}_{\mathrm{2}} {te}^{{x}_{\mathrm{1}} {t}} \:\mathrm{if}\:\left({x}_{\mathrm{1}} ={x}_{\mathrm{2}} \right) \\ $$$$\mathrm{Particular}\:\mathrm{Solution} \\ $$$${i}_{{p}} =\frac{{v}}{{r}} \\ $$$${i}\left({t}\right)={i}_{{h}} +{i}_{{p}} \\ $$

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