Linear Second Order ODE/y'' + y = cosecant x

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Theorem

The second order ODE:

$(1): \quad y + y = \csc x$

has the general solution:

$y = C_1 \sin x + C_2 \cos x - x \cos x + \sin x \map \ln {\sin x}$


Proof

It can be seen that $\paren 1$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:

$y + p y' + q y = \map R x$

where:

$p = 0$
$q = 1$
$\map R x = \csc x$


First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$\paren 2: \quad y + y = 0$

From Linear Second Order ODE: $y + y = 0$, this has the general solution:

$y_g = C_1 \sin x + C_2 \cos x$


It remains to find a particular solution $y_p$ to $\paren 1$.


Expressing $y_g$ in the form:

$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$

we have:

\(\ds \map {y_1} x\) \(=\) \(\ds \sin x\)
\(\ds \map {y_2} x\) \(=\) \(\ds \cos x\)
\(\ds \leadsto \ \ \) \(\ds \map {y_1'} x\) \(=\) \(\ds \cos x\) Derivative of Sine Function
\(\ds \map {y_2'} x\) \(=\) \(\ds -\sin x\) Derivative of Cosine Function


By the Method of Variation of Parameters, we have that:

$y_p = v_1 y_1 + v_2 y_2$

where:

\(\ds v_1\) \(=\) \(\ds -\int \frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds v_2\) \(=\) \(\ds \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x\)

where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.


We have that:

\(\ds \map W {y_1, y_2}\) \(=\) \(\ds y_1 {y_2}' - y_2 {y_1}'\) Definition of Wronskian
\(\ds \) \(=\) \(\ds \sin x \paren {- \sin x} - \cos x \cos x\)
\(\ds \) \(=\) \(\ds -\paren {\sin^2 x + \cos^2 x}\)
\(\ds \) \(=\) \(\ds -1\)


Hence:

\(\ds v_1\) \(=\) \(\ds \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds -\int \frac {\cos x \csc x} {-1} \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {\cos x} {\sin x} \rd x\) Definition of Cosecant
\(\ds \) \(=\) \(\ds \map \ln {\sin x}\) Primitive of $\cot x$


\(\ds v_2\) \(=\) \(\ds \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {\sin x \csc x} {-1} \rd x\)
\(\ds \) \(=\) \(\ds -\int \frac {\sin x} {\sin x} \rd x\) Definition of Cosecant
\(\ds \) \(=\) \(\ds -x\)


It follows that:

$y_p = \sin x \map \ln {\sin x} - x \cos x$


So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

$y = y_g + y_p = C_1 \sin x + C_2 \cos x + \sin x \map \ln {\sin x} - x \cos x$

is the general solution to $\paren 1$.

$\blacksquare$


Sources

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