Try \(y (x) = e^{Mx}⇒M\) satisfies \(P_2 M^2 + P_1 M + P_0 = 0\) (if \(M\) repeated, \(y = xe^{Mx}\))

Try \(y (x) = x^m⇒m\) satisfies \(αm (m - 1) + \beta m + γ= 0\) (if \(m\) is repeated, \(y (x) = x^m \ln x\))

If we can find one solution \(y_1 (x)\) we can find another by writing it as \(y_2 (x) = y_1 (x) v (x)\)

Substituting into equation \(ℒy_2 = 0\) and simplify using the fact that \(y_1\) is a solution,

\[P_2 y_1 v^{\prime \prime} + (2 P_2 y_1' + P_1 y_1) v' = 0\]which is a separable first-order ODE for \(v'\). One further integration then gives \(v\) and thus the general solution \(y (x) = v (x) y_1 (x)\).

To solve \(ℒy = f\) for a general function \(f\), suppose \(y (x) = c_1 y_1 (x) + c_2 y_2 (x)\) with linearly independent \(y_1, y_2\) is the general solution of the homogeneous problem.

We seek a solution as \(y (x) = c_1 (x) y_1 (x) + c_2 (x) y_2 (x)\) and suppose that

\(\displaystyle c_1' y_1 + c_2' y_2 = 0\) | (1) |

Then \(y' = \cancel{c_1' y_1} + c_1 y_1' + \cancel{c_2' y_2} + c_2 y_2'\)

\(y'' = c_1' y_1' + c_1 y_1'' + c_2' y_2' + c_2 y_2''\)

So

\[ℒy = P_2 (c_1' y_1' + c_2' y_2') + c_1 (\cancel{P_2 y_1'' + P_1 y_1' + P_0 y_1}) + c_2 (\cancel{P_2 y_2'' + P_1 y_2' + P_0 y_2}) = f\]So

\(\displaystyle c_1' y_1' + c_2' y_2' = \frac{f}{P_2}\) | (2) |

(1)&(2) together are

\[\left( \begin{array}{ll} y_1 & y_2\\ y_1' & y_2' \end{array} \right) \left( \begin{array}{c} c_1'\\ c_2' \end{array} \right) = \left( \begin{array}{c} 0\\ f / P_2 \end{array} \right)\]Invert to find

\[\left( \begin{array}{c} c_1'\\ c_2' \end{array} \right) = \frac{1}{W} \left( \begin{array}{cc} y_2' & - y_2\\ - y_1' & y_1 \end{array} \right) \left( \begin{array}{c} 0\\ f / P_2 \end{array} \right) = \frac{f}{P_2 W} \left( \begin{array}{c} - y_2\\ y_1 \end{array} \right)\]Integrating

\(\displaystyle c_1 (x) = - \int^x \frac{f (ξ) y_2 (ξ)}{P_2 (ξ) W (ξ)} \text{d} ξ, \quad c_2 (x) = \int^x \frac{f (ξ) y_1 (ξ)}{P_2 (ξ) W (ξ)} \text{d} ξ\) | (3) |

Then \(y (x) = c_1 (x) y_1 (x) + c_2 (x) y_2 (x)\) is a particular solution to (N).

**Example. **\(y'' + y = \tan x\)

First note that two homogeneous solutions for \(y'' + y = 0\) are \(y_1 (x) = \cos x, y_2 (x) = \sin (x)\)

Wronskian \(W (x) = y_1 y_2' - y_2 y_1' = \cos^2 x + \sin^2 x = 1 \neq 0\)

From (3)\begin{align*} c_1 (x) & = - \int \tan x \sin x \mathrm{d} x = \sin (x) - \log (\sec x + \tan x)\\ c_2 (x) & = \int \tan x \cos x \mathrm{d} x = - \cos x \end{align*}Thus a particular integral

\[y (x) = c_1 (x) y_1 (x) + c_2 (x) y_2 (x) = - \cos (x) \log (\sec x + \tan x)\]We can try to incorporate homogeneous boundary conditions directly into this method, by choosing \(y_1 (x)\) and \(y_2 (x)\) to satisfy one each of the boundary conditions.

\(ℒy = f (x)\) on \(a<x<b, \quad y (a) = 0, y (b) = 0\)

Let \(y_1 (x)\) satisfy \(ℒy_1 = 0\) with \(y_1 (a) = 0\) and \(y_2 (x)\) satisfy \(ℒy_2 = 0\) with \(y_2 (b) = 0\)

Then for \(y (x) = c_1 (x) y_1 (x) + c_2 (x) y_2 (x)\) to satisfy the boundary conditions, we need \(c_2 (a) = 0\) and \(c_1 (b) = 0\), this can be achieved by setting the limits of integration in (3),

\[c_1 (x) = \int_x^b \frac{f (ξ) y_2 (ξ)}{P_2 (ξ) W (ξ)} \text{d} ξ, \quad c_2 (x) = \int^x_a \frac{f (ξ) y_1 (ξ)}{P_2 (ξ) W (ξ)} \text{d} ξ\]The solution is then

\[y (x) = \int_x^b \frac{f (ξ) y_2 (ξ) y_1 (x)}{P_2 (ξ) W (ξ)} \text{d} ξ + \int^x_a \frac{f (ξ) y_1 (ξ) y_2 (x)}{P_2 (ξ) W (ξ)} \text{d} ξ\]Write as a single integral

\[y (x) = \int_a^b g (x, ξ) f (ξ) \mathrm{d} ξ\]where $g (x, \xi) = \begin{cases}
\frac{y_1 (\xi) y_2 (x)}{P_2 (\xi) W (\xi)} & a < \xi < x < b\\
\frac{y_1 (x) y_2 (\xi)}{P_2 (\xi) W (\xi)} & a < x < \xi < b
\end{cases}$ is called *Green's function*.

**Example. **Find a formula for the solution to

As linearly independent solutions of \(y'' + y = 0\), take

\[y_1 (x) = \sin x, \quad y_2 (x) = \cos x\]to satisfy the boundary conditions.

The formula is \(y (x) = \int_0^{\fracπ{2}} g (x, ξ) f (ξ) \mathrm{d} ξ\) [\(W = - \sin^2 x - \cos^2 x = - 1\)]

where \(g (x, ξ) = \left\{ \begin{array}{ll} - \sin ξ \cos x & ξ < x\\ - \sin x \cos ξ & x < ξ \end{array} \right.\)

It is not always possible to find \(y_1, y_2\) to satisfy each of the boundary conditions:

**Example.****(Nonexistence/nonuniqueness of solution)**If we try the same example but replace \(y \left( \fracπ2\right) = 0\) with \(y' \left( \fracπ2\right) = 0\), we run into a problem.

\(y_1 (x) = \sin x\) automatically satisfies both boundary conditions.

The method just described fails.

Variation of parameters still gives a particular integral \(y_{\text{P}} (x) = c_1 (x) \sin x + c_2 (x) \cos x\) from (3)

\[c_1 (x) = - \int_x^{\fracπ{2}} \cos ξ f (ξ) \mathrm{d} ξ \quad c_2 (x) = \int_0^x \sin ξ f (ξ) \mathrm{d} ξ\](Choosing to make \(c_2 (0) = c_1 \left( \fracπ2\right) = 0\))

The general solution is \(y (x) = c_1 (x) \sin x + c_2 (x) \cos x + A \sin x + B \cos x\)

Boundary conditions \(\begin{array}{l} y (0) = 0⇒0 = B\\ y' \left( \fracπ2\right) = 0⇒0 = - c_2 \left( \fracπ2\right) - B \end{array} ⇒\)need \(c_2 \left( \fracπ2\right) = 0 ⇒\)need

\[\int_0^{\fracπ{2}} \sin ξ f (ξ) \mathrm{d} ξ = 0\]This is a *solvability condition.*