Ejemplo de ecuaciones lineales homogéneas

How to identify a second order homogeneous differential equation?

When dealing with differential equations, it is crucial to identify the type of equation we are working with. A second order differential equation is recognized by the presence of the second derivative of the unknown function, in this case symbolized as ( y'' ). In a second order homogeneous differential equation with constant coefficients, these coefficients do not depend on the variable ( x ), and the equation equals zero. The main focus of this class is to learn how to solve these equations to advance the study of more complex differential equations.

What is the characteristic equation and how is it derived?

Solving a homogeneous second order differential equation involves finding its general solution and for this, the characteristic equation is used. This is derived by assuming a solution of the form ( y = e^{rx} ), where ( r ) is the value we are looking for.

To derive the characteristic equation:

1. We find the first and second derivatives of our assumed solution:

   • The first derivative: ( y' = re^{rx} )
   • Second derivative: ( y'' = r^2 e^{rx} )

2. We substitute these derivatives into the original differential equation:

   • ( r^2 e^{rx} + 2r e^{rx} - e^{rx} = 0 )

3. We factor the common term ( e^{rx} ):

   • ( e^{rx} (r^2 + 2r - 1) = 0 )

4. Since ( e^{rx} ) is never zero, the characteristic equation is:

   • ( r^2 + 2r - 1 = 0 )

How to solve the quadratic equation to find ( r )?

The derived characteristic equation is a standard quadratic equation that we can solve using the quadratic formula:

[ r = \frac{-b \sqrt{b^2 - 4ac}}{2a} ]

In our case, with:

• ( a = 1 ),
• ( b = 2 ),
• ( c = -1 ),

The application of the formula leads us to:

[ r = \frac{-2 \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}}{2 \cdot 1} ]

Simplifies to:

[ r = \frac{-2 \pm \sqrt{8}}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} ]

Thus, the solutions for ( r ) are ( r_1 = -1 + \sqrt{2} ) and ( r_2 = -1 - \sqrt{2} ).

What are the linearly independent solutions and the general solution?

For a second order differential equation, such as this one, we obtain two linearly independent solutions. With our values of ( r ):

• Solution 1: ( y_1 = e^{(-1 + \sqrt{2})x} )
• Solution 2: ( y_2 = e^{(-1 - \sqrt{2})x} )

The general solution is a linear combination of these two solutions:

[ y = c_1 e^{(-1 + \sqrt{2})x} + c_2 e^{(-1 - \sqrt{2})x} ]

Where ( c_1 ) and ( c_2 ) are arbitrary constants, to be determined with specific initial conditions for particular problems. This general solution applies to all homogeneous second order differential equations with constant coefficients and provides the basis for more complex methods in the study of differential equations.