I have read this:Systems of Differential Equations and higher order Differential Equations
and I am trying to apply it to the following set of equations but with not much success
$$L_{11} \frac{dI_1}{dt}+L_{12}\frac{dI_2}{dt}+I_1R_1=\varepsilon(t)$$
$$L_{22}\frac{dI_2}{dt}+L_{12}\frac{dI_1}{dt}+I_2R_2=0$$
I isolated $I_2$ from the second equation and took the derivative with respect to $t$ and I got:
$$\frac{dI_2}{dt}=-\frac{1}{R_2}\left(L_{22}\frac{d^2I_2}{dt^2}+L_{12}\frac{d^2I_1}{dt^2}\right)$$
but when I substitute $\frac{dI_2}{dt}$ in the first equation I get:
$$L_{11} \frac{dI_1}{dt}-\frac{L_{12} L_{22}}{R_2}\frac{d^2I_2}{dt^2}-\frac{L_{12}^2}{R_2}\frac{d^2I_1}{dt^2}+I_1R_1=\varepsilon(t)$$
so i take the derivative of the expression for $\frac{dI_2}{dt}$ above but i keep getting higher and higher orders differential equations
Best Answer
We have
$$\tag 1 L_{11} I_1' + L_{12} I_2' + R_1 I_1 = \varepsilon(t) \\ L_{22} I_2' + L_{12} I_1' + R_2 I_2 = 0$$
Solve the second equation for $I_1'$ and take the derivative
$$\tag 2 I_1' = -\dfrac{1}{L_{12}} \left(L_{22}I_2' + R_2 I_2\right) \\ I_1'' = -\dfrac{1}{L_{12}}\left( L_{22} I_2'' + R_2 I_2'\right)$$
Take the derivative of the first equation and isolate $I_2''$
$$\tag 3 I_2'' = \dfrac{1}{L_{12}} \left(\varepsilon'(t) - L_{11} I_1'' - R_1 I_1'\right) $$
Substitute the two relations in $(2)$ into $(3)$ and you end up with a higher order equation for $I_2$ to solve then use it to find $I_1$.