Note on coefficients \( p(x) \), \( q(x) \), and \( f(x) \):
cannot contain \( y \) (else nonlinear)
First, let's investigate homogeneous equation
right side = 0
\( f(x) = 0 \)
To understand the structure of the solution, let's look at a very simple one:
This shows that the solution is a linear combo of the functions \( y_1 = x \) , \( y_2 = 1 \)
solution is sum of scalar multiples of each
This means the two functions (which are linearly independent) span the solution space
In fact, the two functions form a basis for the solution space
solution to \( y'' + p(x)y' + q(x)y = 0 \)
is \( y = c_1y_1 + c_2y_2 \) , \( y_1, y_2 \) span and form a basis for solution space
If \( y_1 \) and \( y_2 \) are linearly independent on an interval of \( x \), then on that interval \( c_1 y_1 + c_2 y_2 = 0 \) is possible if and only if \( c_1 = c_2 = 0 \).
If \( y_1, y_2 \) are linearly independent on an interval, then
and
unique solution of \( c_1 = c_2 = 0 \) if and only if
Wronskian of \( y_1, y_2 \)
means the function \( y \) that satisfies the equation.
(for 2nd order, there are two)
\( y \) is a function that's equal to its 2nd derivative
Both \( y_1 = e^x \) and \( y_2 = e^{-x} \) are solutions to \( y'' - y = 0 \) and so are all linear combos:
\[ y = c_1 e^x + c_2 e^{-x} \]
general solution
\( y_1 = e^x \) and \( y_2 = e^{-x} \) are linearly indp because their Wronskian is never zero
We guessed and checked to solve \( y'' - y = 0 \).
A more systematic way to solve \( y'' + py' + qy = 0 \) is to use the characteristic polynomial or characteristic equation.
For constant coefficient equation, the solutions are ALWAYS in the form \( y = e^{rx} \) because \( e^{rx} \) has derivatives that are constant multiples of itself.
\( y'' + ay' + by = 0 \) is ALWAYS solved by \( y = e^{rx} \) (where \( r \) is const).
Sub in:
\[ r^2 e^{rx} + are^{rx} + be^{rx} = 0 \]
\[ (r^2 + ar + b)e^{rx} = 0 \]
\( e^{rx} \neq 0 \) for any \( r \)
So, \[ r^2 + ar + b = 0 \]
Characteristic Equation
Solve for \( r \): two solution values \( r_1, r_2 \)
Solutions of diff. eq: \( y_1 = e^{r_1 x} \), \( y_2 = e^{r_2 x} \)
general solution: \( y = c_1 e^{r_1 x} + c_2 e^{r_2 x} \)
Constant coefficient
Assume a solution of the form:
Substitute into the equation above:
Notice coefficients match the original differential equation.
The two linearly independent solutions:
General solution is a linear combination of them:
To find these constants, we need two initial conditions.
Typically: \( y(x_0) = y_0 \) and \( y'(x_0) = y'_0 \)
For example, suppose \( y(0) = 1 \) and \( y'(0) = 0 \).
General solution: \( y = c_1e^{-2x} + c_2e^{-x} \)
Substitute in \( y(0) = 1 \):
\[ c_1e^0 + c_2e^0 = 1 \]\[ c_1 + c_2 = 1 \quad \text{--- (1)} \]To use \( y'(0) = 0 \), we need \( y' \).
sub in \( y'(0) = 0 \)
solve this system (Gaussian elimination)
characteristic eq has same coefficients
if roots are repeated
here,