91Ó°ÊÓ

In mathematics, a homogeneous differential equation is one in which every term is a function of the dependent variable (typically \(y\)) and its derivatives. There's no separate term added. For our example, the homogeneous version is \( \frac{d^2 y}{dx^2} - \frac{d y}{dx} - 2 y = 0 \).

The goal here is to find a general solution, \(y_h\). These equations are crucial because they help us understand the behavior of the system without external forces acting on it.

To solve this, we convert the differential equation into a characteristic equation by assuming solutions of the form \(y = e^{rx}\). This gives us the characteristic equation: \(r^2 - r - 2 = 0\).

Solving this characteristic equation will provide the roots, which will then help us form the general solution to the homogeneous part.

A non-homogeneous differential equation includes an additional function that is not dependent on \(y\) or its derivatives. For our example, this extra function is \(5e^{-x}\).

The non-homogeneous equation takes the form: \(\frac{d^2 y}{dx^2} - \frac{d y}{dx} - 2 y = 5 e^{-x}\)

These equations can be tricky because the additional term (in this case, \(5 e^{-x}\)) changes how the system behaves compared to the homogeneous part alone.

To find the general solution, we need to:

• First, solve the homogeneous equation
• Then, guess a particular solution that fits the non-homogeneous term

Combining these will give us a complete solution to the original problem.

The characteristic equation arises from trying to solve a linear homogeneous differential equation by assuming solutions of the form \ y = e^{rx} \.

For our homogeneous equation: \( \frac{d^2 y}{dx^2} - \frac{d y}{dx} - 2 y = 0 \, we assume such a solution.

This gives us the characteristic equation: \( r^2 - r - 2 = 0 \)

The characteristic equation is a quadratic equation, and solving it will give roots \ r \ which are crucial for forming the solution to the differential equation.

For example, solving \ r^2 - r - 2 = 0 \ using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), we get roots \( r = 2 \) and
\ r = -1 \ (our \ a=-1 \, \ b=-1 \, \ c=-2 \).

These roots help form the general solution to the homogeneous equation: \ y_h = C_1 e^{2x} + C_2 e^{-x} \.

To find a solution that fits the non-homogeneous equation, we guess a particular solution, \(y_p\).

For the term \(5 e^{-x}\), we assume \(y_p = A e^{-x} \).

Substituting \(y_p \) into the original differential equation involves taking the first and second derivatives of \ y_p \.

Compute: \[ \frac{d y_p}{dx} = -A e^{-x} \]

\[ \frac{d^2 y_p}{dx^2} = A e^{-x} \]

Substitute these into the original differential equation:
\[ A e^{-x} + A e^{-x} - 2 A e^{-x} = 5 e^{-x} \]

This simplifies down to:
\[ 0 = 5 e^{-x}, \] giving \ A = -5 \.

Thus, \(y_p = -5 e^{-x} \).

The full general solution to the original non-homogeneous differential equation is then: \ y = y_h + y_p = C_1 e^{2x} + (C_2 - 5)e^{-x} \.