91Ó°ÊÓ

In mathematics, solving a differential equation often starts with finding the homogeneous solution. This involves setting the right-hand side of the differential equation to zero, which simplifies our task. Here, we deal with the equation:

\[ L \frac{d^2I}{dt^2} + R \frac{dI}{dt} + \frac{1}{C} I = 0 \]

For our specific problem, after substituting the values for L, R, and C, we get:

\[ \frac{d^2I}{dt^2} + 6 \frac{dI}{dt} + 9 I = 0 \]

The homogeneous solution is found by solving this simplified equation. Assume the solution of the form \[ I_h = e^{rt} \] and substitute into the equation to find the characteristic equation

\[ r^2 + 6r + 9 = 0 \].

Factoring gives \[ (r+3)^2 = 0 \], so the roots are \[ r = -3 \]. Hence, the homogeneous solution becomes:

\[ I(h) = (A + Bt)e^{-3t} \].

To solve the differential equation completely, we need the particular solution, which addresses the non-zero right-hand side of our equation:

\[ \frac{d^2I}{dt^2} + 6 \frac{dI}{dt} + 9 I = 100 \sin t \].

We guess a solution of the form:

\[ I_p = A \sin t + B \cos t \].

Substituting this guess into the differential equation and comparing coefficients helps us find the values of A and B. After working through the algebra:

\[ (-A \sin t - B \cos t) + 6 (A \cos t - B \sin t) + 9(A \sin t + B \cos t) = 100 \sin t \],

we simplify and solve the system of equations to get:

\[ 15A + 6B = 100 \]
\[ 6A + 15B = 0 \].

Solving these gives:

\[ A = \frac{500} {63}, \ B = -\frac{200}{63} \].

The particular solution then is:

\[ I_p = \frac{500}{63} \sin t - \frac{200}{63} \cos t \].

Initial conditions are vital to determine the constants in our general solution. Here, it is specified that \[ I(0) = 0 \].

Applying this condition to our general solution which combines both the homogeneous and particular solutions:

\[ I(t) = (A + Bt)e^{-3t} + \frac{500}{63} \sin t - \frac{200}{63} \cos t \],

we substitute t = 0:

\[ 0 = (A + B \cdot 0)e^{-3 \cdot 0} + \frac{500}{63} \sin 0 - \frac{200}{63} \cos 0 \],

which simplifies to:

\[ 0 = A - \frac{200}{63} \],

therefore, \[ A = \frac{200}{63} \].

This final value of A is used to write our specific solution.

A non-homogeneous differential equation is one that includes a term independent of the function we aim to solve. Here our equation is:

\[ \frac{d^2I}{dt^2} + 6 \frac{dI}{dt} + 9 I = 100 \sin t \],

where the term \[ 100 \sin t \] makes it non-homogeneous.

Solving such equations typically involves two steps:

• Finding the homogeneous solution, \[ I_h \], by solving the equation where this independent term is set to zero.
• Finding the particular solution, \[ I_p \], that directly responds to the non-homogeneous part.

Combining these two solutions gives us the general solution. For our equation, this results in:

\[ I(t) = (\frac{200}{63} + Bt)e^{-3t} + \frac{500}{63} \sin t - \frac{200}{63} \cos t \].