Laplace transforms for transient analysis
The voltage, \$\small V(s)\$ at the top node, for each branch, is given by:

$$\small V(s)= \frac{19}{s}\small -I_1(s)$$

$$\small V(s)=I(s)\left(\frac{s}{100}\small+10\right)-0.017$$

$$\small V(s)=\left((I_1(s)-I(s)\right)\frac{500}{s}$$

where \$\small I_1(s)\$ is the source current.

Solving simutaneously: \$\small I(s)=\large\frac{1.727}{s}+\frac{7.72}{s+861.8}-\frac{7.72}{s+638.2}\$ 

Giving the time function: \$\small I(t)=1.727+7.72e^{-861.8t}-7.72e^{-638.2t}\$.

By inspection of the diagram, \$\small I(t) =1.727 A\$ is the correct current for \$\small t=0 \$ and \$\small t=\infty\$.