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Question

Model each application with a differential equation. All rates are assumed to be with respect to time. The voltage drop across an inductor is proportional to the rate at which the current is changing with respect to time.

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Let V(t)V(t) be the voltage,and I(t)I(t) the current on the inductor.,We know that the voltage can be expressed as

V(t)=LdI(t)dt\pmb{V(t)}=L\, \frac{dI(t)}{dt}

where LL is the inductivity of the inductor.,The voltage change can thus be modeled as

dV(t)dt=Ld2I(t)dt2\frac{dV(t)}{dt}=L\, \frac{d^2I(t)}{dt^2}

Since we have a voltage drop we will have a minus sign in front of dVdt\frac{dV}{dt}.,From the condition that the voltage drop across the inductor is proportional to the rate of change of the current with respect to time,we get the equation

dV(t)dt=kdI(t)dtLd2I(t)dt2=kdI(t)dt-\frac{dV(t)}{dt}=k\, \frac{dI(t)}{dt}\Leftrightarrow -L\frac{d^2I(t)}{dt^2}=k \, \frac{dI(t)}{dt}

which can also be written as:

Ld2I(t)dt2+kdI(t)dt=0\boxed{L\, \frac{d^2I(t)}{dt^2}+k\, \frac{dI(t)}{dt}=0}

where kk is the proportionality constant.

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