91Ó°ÊÓ

The given expressions are $y^{\text{'}\text{'}}+y^{\text{'}}=sechx$.

In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

Given the following differential equation,

$y^{\text{'}\text{'}}−y=\mathrm{sech}\left(x\right)$

And, given that the homogeneous solution to such differential equation, is given by

$y_1=\sinh \left(x\right)\text{ â¶Ä„â¶Ä„}{y}_2=\cosh \left(x\right)$

Hence, we can find the particular solution to such differential equation, using the following equation,

$$\begin{gathered} y_p=y_2\left(x\right)∫\frac{y_1\left(x\right)f\left(x\right)}{W\left(x\right)}dx−y_1\left(x\right)∫\frac{y_2\left(x\right)f\left(x\right)}{W\left(x\right)}dx \\ W\left(x\right)=\left|\begin{array}{ll}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right| \end{gathered}$$

And, hence the Wronskian of the homogeneous equation, is thus

$W\left(x\right)=\left|\begin{array}{cc}\sinh \left(x\right)& \cosh \left(x\right)\\ \cosh \left(x\right)& \sinh \left(x\right)\end{array}\right|=\sinh {\left(x\right)}^2−\cosh {\left(x\right)}^2=−1$

And, knowing the Wronskian of the solutions to the homogeneous equation, and comparing the given differential equation with equation $12.19$, we find that

$f\left(x\right)=\mathrm{sech}\left(x\right)$

find the particular solution to the given differential equation,

$y_p=\cosh \left(x\right)∫\frac{\sinh \left(x\right)\mathrm{sech}\left(x\right)}{−1}dx−\sinh \left(x\right)∫\frac{\cosh \left(x\right)\mathrm{sech}\left(x\right)}{−1}dx$

Factorizing the negative sign out of the integration, we get

$y_p=−\cosh \left(x\right)∫\sinh \left(x\right)\mathrm{sech}\left(x\right)dx+\sinh \left(x\right)∫\cosh \left(x\right)\mathrm{sech}\left(x\right)dx$

And, knowing that

$\mathrm{sech}\left(x\right)=\frac{1}{\cosh \left(x\right)}$

Thus, we have

$y_p=−\cosh \left(x\right)∫\frac{\sinh \left(x\right)}{\cosh \left(x\right)}dx+\sinh \left(x\right)∫\frac{\cosh \left(x\right)}{\cosh \left(x\right)}dx$

And, we have

$\sinh \left(x\right)dx=d\cosh \left(x\right)$

Thus, simplifying the integration we get

$y_p=−\cosh \left(x\right)∫\frac{1}{\cosh \left(x\right)}d\cosh \left(x\right)+\sinh \left(x\right)∫1dx$

And, hence evaluating the integral to find the particular solution, thus we get

$y_p=−\cosh \left(x\right)\left(\ln \right(\cosh \left(x\right)\left)\right)+\sinh \left(x\right)x$

Hence, the particular solution is given by

$y_p=x\sinh \left(x\right)−\cosh \left(x\right)\left(\ln \right(\cosh \left(x\right)\left)\right)$