I have the following system of differential equations that I'm really struggling to solve
$L_1(di_2/dt)+(R_1+R_2)i_2+R_1i_3=E(t)$
$L_2(di_3/dt)+R_1i_2+R_1i_3 = E(t)$
$R_1=6\Omega$, $R_2=5\Omega$, $L_1=1h$, $L_2=1h$, $E(t)=48sin(t)V$, $i_2(0)=0$, $i_3(0)=0$
I'm really struggling with this system and I was wondering if anyone has any ideas in how to solve it.
Thanks.
Hint If the two equations are ${\scr E_1}$ and ${\scr E_2}$, find a number $\theta$ such that ${\scr E_1} + \theta {\scr E_2}$ has the form
$$\frac{d}{d t}(L_1 i_2+\theta L_2 i_3) + \lambda (L_1 i_2 + \theta L_2 i_3) = (1 + \theta)E(t)$$
This equation has the simple form $\frac{d f}{ d t} + \lambda f = g$ and it is easy to solve.
The necessary condition on theta is that the determinant $$\left|\begin{array}&L_1&\theta L_2\\(R_1+R_2)+ \theta R_1&R_1+\theta R_1\end{array}\right| = 0$$ This gives two different solutions for $\theta$. The number $\lambda$ is given by $$\lambda = \frac{1}{L_1}(R_1 + R_2 + \theta R_1)$$ After finding $f = L_1 i_2 + \theta L_2 i_3$ for each value of $\theta$, one can compute $i_2$ and $i_3$.
Edit: with your choice of $E(t) = 48 \sin(t)$, each function $f$ will have the form $f(t) = (A \cos(t) + B\sin(t) + C) e^{-\lambda t}$
Edit: Using the given numerical values, the values of $\theta$ are $\theta_1=\frac{2}{3}$ and $\theta_2 = - \frac{3}{2}$, which give the values $\lambda_1=15$ and $\lambda_2=2$. Hence the equations $$\frac{d}{d t}(i_2 + \frac{2}{3}i_3) + 15(i_2 + \frac{2}{3}i_3) = \frac{5}{3} E(t)$$ $$\frac{d}{d t}(i_2 - \frac{3}{2}i_3) + 2(i_2 - \frac{3}{2}i_3) = -\frac{1}{2} E(t)$$
