Second Order ODE/y'' - f(x) y' + (f(x) - 1) y = 0

Theorem

The second order ODE:

$(1): \quad y - \map f x y' + \paren {\map f x - 1} y = 0$

has the general solution:

$\ds y = C_1 e^x + C_2 e^x \int e^{-2 x + \int \map f x \rd x} \rd x$

Proof

Note that:

$1 - \map f x + \paren {\map f x - 1} = 0$

so if $y = y' = y$ we find that $(1)$ is satisfied.

So:

\(\ds y_1\)

\(=\)

\(\ds e^x\)

\(\ds \leadsto \ \ \)

\(\ds {y_1}'\)

\(=\)

\(\ds e^x\)

Derivative of Exponential Function

\(\ds \leadsto \ \ \)

\(\ds {y_1}\)

\(=\)

\(\ds e^x\)

Derivative of Exponential Function

and so:

$y_1 = e^x$

is a particular solution of $(1)$.

$(1)$ is in the form:

$y + \map P x y' + \map Q x y = 0$

where:

$\map P x = -\map f x$

From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:

$\map {y_2} x = \map v x \, \map {y_1} x$

where:

$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.

We have that:

\(\ds \int P \rd x\)

\(=\)

\(\ds \int \paren {-\map f x} \rd x\)

\(\ds \leadsto \ \ \)

\(\ds e^{-\int P \rd x}\)

\(=\)

\(\ds e^{-\int \paren {-\map f x} \rd x}\)

\(\ds \)

\(=\)

\(\ds e^{\int \map f x \rd x}\)

Hence:

\(\ds v\)

\(=\)

\(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\)

Definition of $v$

\(\ds \)

\(=\)

\(\ds \int \dfrac 1 {e^{2 x} } e^{\int \map f x \rd x} \rd x\)

as $y_1 = e^x$

\(\ds \)

\(=\)

\(\ds \int e^{-2 x + \int \map f x \rd x} \rd x\)

as $y_1 = e^x$

and so:

\(\ds y_2\)

\(=\)

\(\ds v y_1\)

Definition of $y_2$

\(\ds \)

\(=\)

\(\ds e^x \int e^{-2 x + \int \map f x \rd x} \rd x\)

From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$\ds y = C_1 e^x + C_2 e^x \int e^{-2 x + \int \map f x \rd x} \rd x$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $10$