Solving an RL Circuit with an Inductor in Parallel to Resistor-Inductor in Series
I had mentioned that I'd probably take a different approach than user600016's approach, using KCL instead of KVL. Both work for me, but I'm usually more comfortable with KCL as I tend to make fewer errors as I develop the equations. That's a personal detail and may not apply to anyone else. But since it is a different approach than the one taken by user600016, it maybe worth the time to add it here as it provides another way of looking at things. And more ways are better than fewer.
So, let's draw out your schematic as it should have been drawn out in your question, so that we get labeled part numbers. (You should get into the practice of this.)



simulate this circuit &ndash; Schematic created using CircuitLab
I agree, of course, with your initial conditions regarding the currents at \$t=0^-\:\text{s}\$ when \$V_0=0\:\text{V}\$. These currents will be the same at \$t=0^+\:\text{s}\$ when \$V_0=20\:\text{V}\$. We can also say that at \$t=0^+\:\text{s}\$, \$V_1=20\:\text{V}\$ and \$V_2=20\:\text{V}\$, since there are no currents in either resistor and therefore no voltage drops across them.
KCL develops the following two equations:
$$\begin{align*}
\frac{V_1}{R_1}+\frac{1}{L_1}\int V_1\:\text{d}t+\frac{V_1}{R_2}&amp;=\frac{V_0}{R_1}+\frac{V_2}{R_2}\\\\
\frac{V_2}{R_2}+\frac{1}{L_2}\int V_2\:\text{d}t&amp;=\frac{V_1}{R_2}
\end{align*}$$
(I've placed the out-flowing currents on the left side of the equation and the in-flowing currents on the right side.)
Taking the time derivatives (to remove those integrals):
$$\begin{align*}
\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\frac{\text{d}\,V_1}{\text{d}t}+\frac{V_1}{L_1}&amp;=\frac{1}{R_2}\frac{\text{d}\,V_2}{\text{d}t}&amp;&amp;\rightarrow &amp;\frac{\text{d}\,V_1}{\text{d}t}+\frac{R_1\mid \mid R_2}{L_1}V_1&amp;=\frac{R_1}{R_1+R_2}\frac{\text{d}\,V_2}{\text{d}t}\\\\
\frac{1}{R_2}\frac{\text{d}\,V_2}{\text{d}t}+\frac{V_2}{L_2}&amp;=\frac{1}{R_2}\frac{\text{d}\,V_1}{\text{d}t}&amp;&amp;\rightarrow &amp;\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2&amp;=\frac{\text{d}\,V_1}{\text{d}t}
\end{align*}$$
Substitution of the second equation into the first (right sides) and then taking the time derivative (again):
$$\begin{align*}
\left(\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2\right)+\frac{R_1\mid \mid R_2}{L_1}\int\left(\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2\right)\:\text{d}t&amp;=\frac{R_1}{R_1+R_2}\frac{\text{d}\,V_2}{\text{d}t}\\\\
\left(\frac{\text{d}^2\,V_2}{\text{d}t^2}+\frac{R_2}{L_2}\frac{\text{d}\,V_2}{\text{d}t}\right)+\frac{R_1\mid \mid R_2}{L_1}\left(\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_2}{L_2}V_2\right)&amp;=\frac{R_1}{R_1+R_2}\frac{\text{d}^2\,V_2}{\text{d}t^2}\\\\
\frac{\text{d}^2\,V_2}{\text{d}t^2}+\left(\frac{R_1}{L_1}+\frac{R_1+R_2}{L_2}\right)\frac{\text{d}\,V_2}{\text{d}t}+\frac{R_1\, R_2}{L_1\, L_2}V_2&amp;=0\:\frac{\text{V}}{\text{s}^2}\\\\\text{Similarly, you'll wind up with the same expression for }V_1,\\\\\frac{\text{d}^2\,V_1}{\text{d}t^2}+\left(\frac{R_1}{L_1}+\frac{R_1+R_2}{L_2}\right)\frac{\text{d}\,V_1}{\text{d}t}+\frac{R_1\, R_2}{L_1\, L_2}V_1&amp;=0\:\frac{\text{V}}{\text{s}^2}
\end{align*}$$
As mentioned before, the initial conditions at \$t=0^+\:\text{s}\$, \$V_1=20\:\text{V}\$ and \$V_2=20\:\text{V}\$. But we also need the initial conditions for \$\frac{\text{d}\,V_1}{\text{d}t}\$ and \$\frac{\text{d}\,V_2}{\text{d}t}\$, which will make the difference for us. Let's assume that \$I_1\$ is the current in \$L_1\$ and that \$I_2\$ is the current in \$L_2\$:
$$\begin{align*}
V_1&amp;=V_0-R_1\left(I_1+I_2\right)&amp;&amp;&amp;V_2&amp;=V_1-R_2\,I_2\\\\&amp;&amp;&amp;\therefore\\\\
\frac{\text{d}\,V_1}{\text{d}t}&amp;=\frac{\text{d}\,V_0}{\text{d}t}-R_1\left(\frac{\text{d}\,I_1}{\text{d}t}+\frac{\text{d}\,I_2}{\text{d}t}\right)&amp;&amp;&amp;\frac{\text{d}\,V_2}{\text{d}t}&amp;=\frac{\text{d}\,V_1}{\text{d}t}-R_2\frac{\text{d}\,I_2}{\text{d}t}\\\\&amp;=-R_1\left(\frac{\text{d}\,I_1}{\text{d}t}+\frac{\text{d}\,I_2}{\text{d}t}\right)&amp;&amp;&amp;&amp;=\frac{\text{d}\,V_1}{\text{d}t}-R_2\frac{V_2}{L_2}\\\\&amp;=-R_1\left(\frac{V_1}{L_1}+\frac{V_2}{L_2}\right)
\end{align*}$$
We can resolve this at \$t=0^+\:\text{s}\$. Clearly, \$\frac{\text{d}\,V_1}{\text{d}t}=-4\:\Omega\left(\frac{20\:\text{V}}{2\:\text{H}}+\frac{20\:\text{V}}{2\:\text{H}}\right)=-80\:\frac{\text{V}}{\text{s}}\$ and from that we can now solve for \$\frac{\text{d}\,V_2}{\text{d}t}=-80\:\frac{\text{V}}{\text{s}}-6\:\Omega\,\frac{20\:\text{V}}{2\:\text{H}}=-140\:\frac{\text{V}}{\text{s}}\$.
Now you have all four conditions at \$t=0^+\:\text{s}\$: \$V_1\$, \$\frac{\text{d}\,V_1}{\text{d}t}\$, \$V_2\$, and \$\frac{\text{d}\,V_2}{\text{d}t}\$. So you can now easily solve both 2nd order linear diff-eqs for each voltage node and you therefore have \$V_{1\left(t\right)}\$ and \$V_{2\left(t\right)}\$ for \$t\ge 0\:\text{s}\$.
From there, it's a cake walk to get the currents:
$$\begin{align*}I_1&amp;=\frac{V_0-V_1}{R_1}-\frac{V_1-V_2}{R_2}&amp;&amp;&amp;I_2&amp;=\frac{V_1-V_2}{R_2}\end{align*}$$
Post your results and I'll share mine. (Just as an way to check your results, you should find that the peak current in \$L_2\$ will be \$\approx 1.165\:\text{A}\$ and will occur at \$t\approx 358.35\:\text{ms}\$. Those were computed from the formulas using sympy. Below, you can then see the LTspice output, as well. It confirms the equations.)


Appendix
Enough time has passed without a response. So, for others, I'll go ahead and complete the above.
First, let's set up \$\tau_{_1}=\frac{L_1}{R_1}\$, \$\tau_{_2}=\frac{L_2}{R_2}\$, and \$\tau_{_3}=\frac{L_2}{R_1}\$. (I think you can see why these may be the interesting \$\tau\$ values.) Let's also set \$a=\tau_{_1}\,\tau_{_2}\,\tau_{_3}\$ and \$b=\tau_{_1}\,\tau_{_2}+\tau_{_1}\,\tau_{_3}+\tau_{_2}\,\tau_{_3}\$, so that putting these in parallel, as if they were resistances, provides a new \$\tau_{_4}=\frac{\tau_{_1}\,\tau_{_2}\,\tau_{_3}}{\tau_{_1}\,\tau_{_2}+\tau_{_1}\,\tau_{_3}+\tau_{_2}\,\tau_{_3}}=\frac{a}{b}\$. Let's set \$c=\tau_{_3}\$ and set \$\tau_{_5}^2=\sqrt{b^2-4\,a\,c}\$ so that \$\omega_{_0}=\frac{\sqrt{b^2-4\,a\,c}}{2\,a}=\frac{\tau_{_5}^2}{2\,a}\$ (and that \$\tau_{_0}=\frac{1}{\omega_{_0}}\$.)
With the above in hand, the generalized (and specific) answers are (for \$t\ge 0\:\text{s}\$):
$$\begin{align*}
V_{1\mathrel{\mkern+3mu\left(t\right)}}&amp;=V_\text{0}\cdot e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\cosh\left(\frac{t}{\tau_{_0}}\right)+\frac{2\,\tau_{_1}\,\tau_{_3}-b}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\\\\&amp;=20\:\text{V}\cdot e^{^{\frac{-7}{2}t}}\left[\cosh\left(\frac{5}{2}t\right)-\frac15\cdot\sinh\left(\frac{5}{2}t\right)\right]\\\\
V_{2\mathrel{\mkern+3mu\left(t\right)}}&amp;=V_\text{0}\cdot e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\cosh\left(\frac{t}{\tau_{_0}}\right)+\frac{-b}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\\\\&amp;=20\:\text{V}\cdot e^{^{\frac{-7}{2}t}}\left[\cosh\left(\frac{5}{2}t\right)-\frac75\cdot\sinh\left(\frac{5}{2}t\right)\right]\\\\
I_{R1\mathrel{\mkern+3mu\left(t\right)}}&amp;=\frac{V_0-V_{1\mathrel{\mkern+3mu\left(t\right)}}}{R_1}=\frac{V_\text{0}}{R_1}\cdot\left\{ 1-e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\cosh\left(\frac{t}{\tau_{_0}}\right)+\frac{2\,\tau_{_1}\,\tau_{_3}-b}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\right\}\\\\&amp;=5\:\text{A}\cdot\left\{ 1- e^{^{\frac{-7}{2}t}}\left[\cosh\left(\frac{5}{2}t\right)-\frac15\cdot\sinh\left(\frac{5}{2}t\right)\right]\right\}\\\\
I_{2\mathrel{\mkern+3mu\left(t\right)}}&amp;=\frac{V_{1\mathrel{\mkern+3mu\left(t\right)}}-V_{2\mathrel{\mkern+3mu\left(t\right)}}}{R_2}=\frac{V_\text{0}}{R_2}\cdot e^{^{\frac{-t}{2\,\tau_{_4}}}}\left[\frac{2\,\tau_{_1}\,\tau_{_3}}{\tau_{_5}^2}\cdot\sinh\left(\frac{t}{\tau_{_0}}\right)\right]\\\\&amp;=4\:\text{A}\cdot e^{^{\frac{-7}{2}t}}\cdot\sinh\left(\frac{5}{2}t\right)\\\\
I_{1\mathrel{\mkern+3mu\left(t\right)}}&amp;=I_{R1\mathrel{\mkern+3mu\left(t\right)}}-I_{2\mathrel{\mkern+3mu\left(t\right)}}
\end{align*}$$