So i solved this by first finding differential equations and then by taking laplace transform of them here are my working-
for left loop $$5i_{1}\left( t\right) +4.9\dfrac{d}{dt}i_{1}\left( t\right) -1.4\dfrac{d}{dt}i_{2}\left( t\right)=20\qquad ....(i)$$ for right loop $$5i_{2}\left( t\right) +0.4\dfrac{d}{dt}i_{2}\left( t\right) -1.4\dfrac{d}{dt}i_{1}\left( t\right) =0\qquad ....(ii)$$ Laplace Transform of these :- $$5I_{1}\left( s\right) +4.9\left[ s\cdot I_{1}\left( s\right) -i_{1}\left( 0\right) \right] -1.4\left[ sI_{2}\left( s\right) -i_{2}\left( 0\right) \right] =\dfrac{20}{s}\qquad ....(iii)\\ 5I_{2}\left( s\right) +0.4\left[ sI_{2}\left( s\right) -i_{2}\left( 0\right) \right] -1.4\left[ sI_{1}\left( s\right) -i_{1}\left( 0\right) \right] =0\qquad ....(iv)$$ Now pluging in initial current value zero and solving for I_1 and I_2 and i get... $$I_{1}\left( s\right) =\dfrac{8\left( 2s+25\right) }{s\left( 53s+50\right) };\qquad I_{2}\left( s\right) =\dfrac{56}{53s+50}\qquad ....(v)$$ and corresponding Inverse Laplace is- $$i_{1}\left( t\right) =\left[ 4-\dfrac{196}{53}e^{-\left( \dfrac{50}{53}\right) t}\right] u\left( t\right) ;\qquad i_{2}\left( t\right) =\dfrac{56}{53}e^{-\left( \dfrac{50}{53}\right) t}u\left( t\right) \qquad ....(vi)$$
Question:- So i think i made some kind of mistake in my calculation because these values are not consistent with initial current i.e. i(0) = 0 , when in m putting t=0 in both of these equation i m not getting i_1=0 or i_2 =0, definitely i made some mistake in calculation can someone please give me some hints.
