Hello,
I am attempting this problem but have got stuck on finding the natural response to the circuit at t>0.
When the independent current source is turned of we get a short circuit from the top to bottom. I am struggling to come up with the appropriate KCL and KVL equations.

 

When the independent current source is turned of we get a short circuit from the top to bottom.

No we don't. Presumably you mean turned on....?

Through which elements does the current flow at t=0+ when the switch first closes? The source will adopt whatever voltage is required to match its terminal load conditions.

Through which elements does the current flow at t=∞?

 

At t=0+ i(0+)=i(0-)=0
At t=\(\infty\) i=2

 

And - at t=0+ all of the 2A flows into the left hand branch.

 

For the LHS I used v(t)=v(\(\infty\))+(v(0)-v(\(\infty\)))e\(^{-t/RC}\)
And for the RHS i(t)=i(\(\infty\))+(i(0)-i(\(\infty\)))e\(^{-tR/L}\)

The result is the same as the answer given however in both cases I get the exponent = -2t rather than -5t.

 

The solution is somewhat convoluted ...

Denote the current in the left branch as i1(t) and the voltage at the top node as v1(t)

\(v_1(t)=10i_1+20\int i_1 dt=4i+2\frac{di}{dt}\)

also

\(i=2-i_1\) & \(\frac{di}{dt}=-\frac{di_1}{dt}\)

\(v_1(t)=10i_1+20\int i_1 dt=4(2-i_1)-2\frac{di_1}{dt}\)

leading to

\(10i_1+20\int i_1 dt=4(2-i_1)-2\frac{di_1}{dt}\)

and

\(\frac{di_1}{dt}+7i_1+10\int i_1 dt=4\)

\(\frac{d^2i_1}{dt^2}+7\frac{di_1}{dt}+10 i_1=0\)

Initial conditions: i1(0)=2, i(0)=0, v1(0)=20

The real roots of the CE are -5 and -2

So

\(i_1(t)=K_1e^{-2t}+K_2e^{-5t}\)

Using i1(0)=2

\(K_1+K_2=2\)

Using v1(0)=20 and i1 as defined above, along with the relationship

\(v_1(t)=8-4i_1-2\frac{di_1}{dt}\)

The values of K1 & K2 can then be found and so on ...

As it turns out the value of K1=0 which accounts for the existence of the exponential solution being in -5t only.

This problem is far more readily solved by the use of Laplace transforms.

 

Your right - it is convoluted. Haven't got to Laplace Transforms yet. That's towards the end of the book. I notice Khan Academy has included them in the differentials section. This is a good free learning resource for anybody interested.