RL Energy Exchange

$\textcolor{#20A900}{Fun Problem}$
As @Karan Chatrath sir has provided a very nice solution through Laplace transformation ,I have solved it analytically.

The basic equation is, $q$ is the charge through circuit $V_1-\dot{q}-\ddot{q}-V_2=0$
On substituting values $sin(5t) -\dot{q}-\ddot{q}-sin(3t) =0$
On solving the above differential equation $\large q(t)=c_1e^{-t} +c_2+0.1sin(3t) -\frac{1}{26}sin(5t) +\frac{1}{30}cos(3t) -\frac{1}{130}cos(5t)$
Where $c_1, c_2$ are arbitrary constants.
Using the initial conditions $\textcolor{#333333}{q(0) =0}$ $\textcolor{#333333}{\dot{q}(0)=0}$ we can easily calculate $c_1$ and $c_2$
I get $c_1=+0.107692$ $c_2=-0.13333$
Use this value to find the correct differential equation of $q(t)$
Differentiate that expression of $q(t)$ to get $I(t)$
Now to calculate heat dissipated in resistor, use this $\large E_{resistor}=\int_{0}^{t}I^{2}(t) Rdt$
Now, the energy stored in inductor $\large E_{inductor}= \frac{1}{2}LI^{2}(t)$
After this I have plotted the graph, as shown below
The $1^{st}$ graph is of inductor. $($ $y$ axis denotes to $E$ (energy) and $x$ axis denotes to $t$ (time) $)$

The largest value of $t$ at which $\large E_R(t) =E_L(t)$

$\large\boxed{\textcolor{#3D99F6}{t=1.2534}}$

I will also elaborate my solution, if anyone ask.

The circuit equation reads:

$L\dot{I} + RI = V_1-V_2$ $\implies \dot{I} + I = \sin\left(5t\right)-\sin\left(3t\right)$

$I(0)=0$

Laplace transform on both sides gives:

$I(s) = \left(\frac{1}{s+1}\right)\left(\frac{5}{s^2+25} - \frac{3}{s^2+9}\right)$

Inverse Laplace Transform yields:

$I(t)= \frac{3\,\cos\left(3\,t\right)}{10}-\frac{5\,\cos\left(5\,t\right)}{26}-\frac{7\,{\mathrm{e}}^{-t}}{65}-\frac{\sin\left(3\,t\right)}{10}+\frac{\sin\left(5\,t\right)}{26}$

Having found $I$ , we are asked to solve for the largest value of $t$ for which the following holds:

$E_L=E_R$ $\implies \frac{I(t)^2}{2} = \int_{0}^{t} I(z)^2R \ dz$

Another equally cumbersome route is:

$L\dot{I} + RI = V_1-V_2$ $LI\dot{I} + I^2R = V_1I-V_2I$ $LI \ dI + I^2R \ dt = V_1I \ dt-V_2I \ dt$ $\frac{LI^2}{2} + E_R = \int_{0}^{t}V_1(z)I(z) \ dz-\int_{0}^{t}V_2(z)I(z) \ dz$ $E_L + E_R = \int_{0}^{t}\left(V_1(z)-V_2(z)\right)I(z) \ dz$ $E_R = \int_{0}^{t}\left(V_1(z)-V_2(z)\right)I(z) \ dz - \frac{I(t)^2}{2}$ $\implies\boxed{E_L + E_R = E_{V_1} - E_{V_2}}$

The above boxed equation holds true for all instants of time. Plotting $E_L$ and $E_R$ gives:

The energy dissipated by the resistor constantly increases as the circuit is supplied with energy from the sources all the time. What makes this problem interesting is doing an energy balance. We know from above that:

$E_L + E_R = E_{V_1} - E_{V_2}$

One can see that the two curves perfectly overlap, thus demonstrating the above equation.