Wronskian Method
Wronskian Method for Second-Order Differential Equations
The Wronskian method is a technique used to find a particular solution to a non-homogeneous second-order differential equation of the form: $$ y'' + a(x)y' + b(x)y = f(x) $$
Given such an equation, if the corresponding homogeneous equation has a general solution of the form:
$$ y_o = c_1 v_1(x) + c_2 v_2(x) $$
then a particular solution to the non-homogeneous equation can be written as:
$$ y_p = c_1(x) v_1(x) + c_2(x) v_2(x) $$
where \( c_1(x) \) and \( c_2(x) \) are functions defined by the integrals:
$$ c_1(x) = - \int v_2(x) \frac{f(x)}{W(x)} \, dx $$
$$ c_2(x) = \int v_1(x) \frac{f(x)}{W(x)} \, dx $$
The function \( W(x) \) in the denominator is called the Wronskian and is computed as:
$$ W(x) = v_1(x) v_2'(x) - v_2(x) v_1'(x) $$
Note. The Wronskian arises from the system: $$ \begin{cases} c_1'(x) v_1 + c_2'(x) v_2 = 0 \\ \\ c_1'(x) v_1' + c_2'(x) v_2' = f(x) \end{cases} $$ The corresponding coefficient matrix is: $$ \begin{pmatrix} v_1 & v_2 \\ v_1' & v_2' \end{pmatrix} $$ and the Wronskian is its determinant: $$ W(x) = v_1(x) v_2'(x) - v_2(x) v_1'(x) $$
A Worked Example
Consider the non-homogeneous second-order differential equation:
$$ y'' - 2y' - 3y = \frac{4x - 1}{x^2} e^{3x} $$
The general solution to the associated homogeneous equation is:
$$ y_o = c_1 e^{-x} + c_2 e^{3x} $$
Note. The corresponding homogeneous equation is: $$ y'' - 2y' - 3y = 0 $$ Solving the characteristic equation: $$ z^2 - 2z - 3 = 0 $$ yields two distinct roots: $$ z = \frac{2 \pm \sqrt{(-2)^2 + 4 \cdot 3}}{2} = \begin{cases} z_1 = -1 \\ \\ z_2 = 3 \end{cases} $$ Hence, the general solution to the homogeneous equation is: $$ y_o = c_1 e^{-x} + c_2 e^{3x} $$
We now look for a particular solution of the form:
$$ y_p = c_1(x) v_1(x) + c_2(x) v_2(x) $$
where \( v_1 = e^{-x} \) and \( v_2 = e^{3x} \), as derived from the homogeneous solution:
$$ y_p = c_1(x) e^{-x} + c_2(x) e^{3x} $$
The functions \( c_1(x) \) and \( c_2(x) \) are computed using:
$$ \begin{cases} c_1(x) = - \int e^{3x} \cdot \frac{f(x)}{W(x)} \, dx \\ \\ c_2(x) = \int e^{-x} \cdot \frac{f(x)}{W(x)} \, dx \end{cases} $$
To evaluate these integrals, we first compute the Wronskian:
$$ W(x) = e^{-x} \cdot \frac{d}{dx} e^{3x} - e^{3x} \cdot \frac{d}{dx} e^{-x} $$
$$ W(x) = e^{-x} \cdot 3e^{3x} + e^{3x} \cdot e^{-x} = 3e^{2x} + e^{2x} = 4e^{2x} $$
Substituting this expression into the formulas for \( c_1(x) \) and \( c_2(x) \):
$$ \begin{cases} c_1(x) = - \int \frac{4x - 1}{4x^2} e^{4x} \, dx \\ \\ c_2(x) = - \int \frac{4x - 1}{4x^2} \, dx \end{cases} $$
Carrying out the integrations yields:
$$ \begin{cases} c_1(x) = -\frac{e^{4x}}{4x} \\ \\ c_2(x) = -\left( \frac{1}{4x} + \log|x| \right) \end{cases} $$
Substituting these into the expression for the particular solution gives:
$$ y_p = c_1(x) e^{-x} + c_2(x) e^{3x} $$
$$ y_p = -\frac{e^{3x}}{4x} + \left( \frac{1}{4x} + \log|x| \right) e^{3x} $$
$$ y_p = e^{3x} \log|x| $$
Thus, the general solution to the original non-homogeneous equation is:
$$ y = y_o + y_p = c_1 e^{-x} + c_2 e^{3x} + e^{3x} \log|x| $$
where \( c_1 \) and \( c_2 \) are arbitrary real constants.
And so on.
