Electronic – State variable equations for a RLC circuit

Writing nodal equation at ground node, $$-7 =\frac{V_1}{10} + \frac{V_2}{5} + \frac{V_2}{2}$$ $$or,\ \mathbf{0.1V_1 + 0.7V_2 = -7}$$ Writing nodal equation at \$V_1\$ correctly will also produce the same equation. You actually missed the term \$0.7V_2\$ in it.

Now, $$V_2 = V_1+6I_x$$ but, $$I_x = -0.1V_1$$ then, $$V_2 = 0.4V_1$$ $$\mathbf{0.4V_1 - V_2 = 0}$$

Writing these equations in matrix format, $$ \left[ \begin{array}{cc} 0.4 & -1 \\ 0.1 & 0.7\end{array} \right]\left[ \begin{array}{c} V_1 \\ V_2\end{array} \right]=\left[ \begin{array}{c} 0 \\ -7\end{array} \right] $$

I am getting an extra -1. Correct me if I am wrong.

You are on the right track but your solution to the differential equation is not quite right. I solved it using Wolfram Alpha and this is what I get:

$$v_c(t)=\frac{R_2V}{R_1+R_2}-\frac{R_2V}{R_1+R_2}e^{-\frac{R_1+R_2}{CR_1R_2}t} $$

Or $$v_c(t)=\frac{R_2V}{R_1+R_2}\bigg(1-e^{-\dfrac{t}{\tau}}\bigg) $$

*Where \$V\$ is the voltage from the source (94V).

Which makes sense because when \$t\rightarrow\infty\$ (capacitor is charged) all you have is a voltage divider formed by the two resistors.

Since when the step voltage is applied, the capacitor is a short, \$R_2\$ gets shorted out and all the the current has to flow through \$R_1\$ at that instant. That's why you are given \$i_{R_1}(0^+)\$. Then

$$ R_1 =\frac{V}{i_{R_1}(0^+)}$$

Where \$V\$ is the voltage from the source (94V). (@Transistor answered this in his answer)

You are also given the power on \$R_2\$ when steady state. You got this one right.

Now, in order to find \$C\$, it would be helpful to use the time constant \$\tau\$, which is the time it takes the capacitor to reach 63.2% of its final value.

$$ \tau=\frac{CR_1R_2}{R_1+R_2}$$

And

$$ C=\bigg(\frac{R_1+R_2}{R_1R_2}\bigg)\tau$$

You already have the values for \$R_1\$ and \$R_2\$, and the only unknown is \$\tau\$. From the original question I can't see anything that would force \$C\$ to be a specific value, that is, there is no given constraint on the time constant, other than \$\tau>0\$ so that the problem is nontrivial. No constraint for \$\tau\$ means no constraint for \$C\$. So any value of \$C\$ does the job.

Edit:

Answering the OP question in from the comments.

If you let \$t=\tau\$ then you should get a value for \$v_c(t)\$ equal to 63.2% of it final value. So at \$t=\tau\$:

$$ v_c(\tau)=\frac{R_2V}{R_1+R_2}(1-e^{-1})$$

Or

$$ v_c(\tau)=\frac{R_2V}{R_1+R_2}(0.632)$$

Whatever that gives you won't help you in finding a value for \$C\$. As you can see when you let \$t=\tau\$, the \$C\$ term isn't there anymore (it's embedded in \$\tau\$) so can't solve for \$C\$ from there.

The only place to solve for \$C\$ is in the \$\tau\$ equation.