Answer: At $t=0^-$, the inductor acts as short circuit and hence the current at $t=0^-$is given as ,

$i(0^-)=25 \times \dfrac{140}{140+60}$

$\therefore i(0^-)=17.5A$

Since the current through the inductor cannot change instantaneously,

$\therefore i(0^+)=17.5A$

For (t>0), the circuit is simplified by source transformation, 

By writing the KVL equation for (t>0),

$-15i-0.3\dfrac{di}{dt}=0$

$\therefore \dfrac{di}{dt}+50i=0$

Comparing with the differential equation $\dfrac{di}{dt}+Pi=0$, we get 

$P=50$

The solution of the differential equation is given by ,

$i(t)=ke^{-Pt}=ke^{-50t}$

At $t=0 , \space i(0)=17.5A$

$\therefore k=17.5$

Hence , $i(t)=17.5e^{-50t}$    for  (t>0)