Discussion Forum

 The solution  of the differential equation  $\frac{d\theta}{dt}=-k(\theta-\theta_{0})$ where k is constant , is ...................

$\theta=\theta_{0}+ae^{kt}$

Answer:

Explanation:

 We have differential equation 

$\frac{d\theta}{dt}=-k(\theta-\theta_{0})$ , where k is constant 

$\Rightarrow$    $\frac{d\theta}{dt}+k\theta=k\theta_{0}$

 which is linear  differential equation in the form of 

$\Rightarrow$    $\frac{dy}{dx}+Py=Q$

$\therefore$    $IF= e^{\int kdt  }=e^{kt}$

 therefore required solution,

    $(\theta)(e^{kt})=\int(e^{kt}\times k \theta_{0})dt$

 $\Rightarrow$   $\theta e^{kt}=e^{kt}\theta_{0}+a$

$\Rightarrow$     $\theta =\theta_{0}+ae^{-kt}$