Electronic – Finding the voltage across a inductor in a charged RLC-circuit

Well, I've the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

In the circuit I want to find the voltage \$\text{V}_{\text{Y}_1}\left(t\right)\$ as a function of time, after I open the switch (this switch was closed 'forever' theoretically seen). And the question is: is my derivation for the voltage across the inductor correctly?

My work:

When the switch opens, the direction of the current in the inductor does not change, so the current will go in the direction of the red arrow in the circuit I've drawn down here:

^{simulate this circuit}

When the switch is open we can find the current in the circuit using Faraday's law:

$$\text{V}_{\text{R}_1}\left(t\right)+\text{V}_{\text{R}_2}\left(t\right)+\text{V}_\text{C}\left(t\right)=-\text{V}_\text{L}\left(t\right)\tag1$$

Using the voltage and current relations in a resistor, inductor and capacitor we can write equation \$(1)\$ in terms of the current, \$\text{I}\left(t\right)\$, that will run trough the circuit when the switch is open (this current is the same trough all the components because it is a series circuit):

$$\text{I}'\left(t\right)\cdot\text{R}_1+\text{I}'\left(t\right)\cdot\text{R}_2+\text{I}\left(t\right)\cdot\frac{1}{\text{C}}=-\text{I}''\left(t\right)\cdot\text{L}\tag2$$

In order to solve this DE, we need two initial conditions:

1. The first one is pretty simple, when the switch is closed the maximum current that will run trough the coil is equal to \$\text{V}/\text{R}_1\$. So when I open the switch at \$t=0\$ that current will be running trough the inductor:
   $$\text{I}\left(0\right)=\frac{\text{V}}{\text{R}_1}\tag3$$
2. At the point \$\text{I}'\left(0\right)\$ we get that:
   $$\text{I}'\left(0\right)=0\tag4$$

So, now we are able to solve that DE that is written in equation \$(2)\$. I solve it using Mathematica.

Now, in order to find the voltage across the coil we can use the voltage current relation in an inductor:

$$\text{V}_\text{L}\left(t\right)=\text{L}\cdot\text{I}_\text{L}'\left(t\right)\tag5$$

So, in the end we get that the voltage at the node \$\text{Y}_1\$ is given by:

$$\text{V}_{\text{Y}_1}\left(t\right)=\text{L}\cdot\text{I}'\left(t\right)\tag6$$

Where \$\text{I}'\left(t\right)\$ is the derivative of the solution to our DE out of equation \$(2)\$.

If I plot my solution, using the values that are given in the circuit I get the following graph:

QUESTION: is my method of derivation right or did I make a mistake?