91Ó°ÊÓ

When solving differential equations, initial conditions are crucial to finding a unique solution. They specify the value of the solution and/or its derivatives at a certain point.
For a second-order differential equation like this one, you'll typically be given two initial conditions. Here, they're given as:

• \(y\left(\frac{\pi}{3}\right) = 0\)
• \(y'\left(\frac{\pi}{3}\right) = 2\)

These conditions allow us to determine the constants in the general solution.
Once the general solution is established, plug in these values to set up a system of equations. Solving this system lets you pinpoint the specific solution for your problem.

To find the structure of the solution for a linear homogeneous differential equation with constant coefficients, derive the characteristic equation.
For the given differential equation, \(y'' + y = 0\), we assume a solution in the form \(y = e^{rt}\).
Substitute this assumed solution into your differential equation to get:

• \(r^2 + 1 = 0\)

This is referred to as the characteristic equation because its solutions, the roots \(r\), determine the nature of the general solution. The method involves finding these roots, which can be real or complex, and using them to form the solution's structure.

The general solution to a linear homogeneous differential equation depends on the nature of the roots of its characteristic equation.
In our case, we resolved the characteristic equation \(r^2 + 1 = 0\). The solution leads to complex roots, \(r = i\) and \(r = -i\).
With complex roots of the form \(a \pm bi\), the general solution takes the form:

• \(y(t) = C_1 \cos(bt) + C_2 \sin(bt)\)

Here, since our roots are purely imaginary (0 \(\pm\) i), it simplifies to:

• \(y(t) = C_1 \cos(t) + C_2 \sin(t)\)

The constants \(C_1\) and \(C_2\) are then adjusted to meet the initial conditions given in the problem.

Complex roots occur when the discriminant (\(b^2 - 4ac\)) of a quadratic equation is negative, leading to imaginary numbers.
For characteristic equations like \(r^2 + 1 = 0\), the presence of \(-1\) inside the square root indicates complex roots.
If the roots are of the form \(a \pm bi\), the solution to the differential equation is expressed using trigonometric functions as:

• \(y(t) = e^{at}(C_1 \cos(bt) + C_2 \sin(bt))\)

However, as our equation yield \(i\) and \(-i\), which are purely imaginary, the exponential part \(e^{at}\) simply equals 1, focusing the solution entirely on the trigonometric part:

• \(y(t) = C_1 \cos(t) + C_2 \sin(t)\)

This transformation helps solve the differential equation using sine and cosine, relating the imaginary components to these cyclic functions effectively.