Answer to Question #222700 in Differential Equations for gavi

Question

Find the general solution of each of the following

1) y''+y=Tanx

Answer 1

$We \space have \space (D^2 \space + \space 1)y \space = \space tanx \space \\ A.E. \space is \space m^2 \space + \space 1 \space = \space 0 \space \\ i.e., \space m^2 \space = \space – \space 1\\ \space i.e., \space m \space = \space ± \space i \space \\ C.F. \space is \space C.F. \space = \space y_c \space = \space \space C_1cos \space x \space + \space \space C_2sinx \space \\ ∴ \space y \space = \space A \space cos \space x \space + \space B \space sinx \space ...(1) \space \\ be \space the \space complete \space solution \space of \space the \space given \space equation \space where \space A \space and \space B \\ \space are \space to \space be \space found \space \\ We \space have \space y_1 \space = \space cos \space x \space and \space y_2 \space = \space sin \space x \space \\ y′_1 \space = \space – \space sinx \space y′_2 \space = \space cos \space x\\ \space W \space = \space y_1y'_2 \space - \space y_2y'_1 \space = \space cos \space x \space . \space cos \space x \space + \space sin \space x \space . \space sin \space x \space = \space cos2x \space + \space sin2x \space = \space 1 \space \\ also \space \phi \space (x)= \space tan \space x \space \\ A' \space = \space \frac{-y_2 \space \space \space \phi \space (x)}{W} \space \\ \space \space \space \space \space = \space \frac{-sin \space x \space \space \space tan \space x \space }{1} \space \\ A' \space = \space \frac{-sin \space ^2 \space x}{cos \space x} \space \\ A \space = \space - \space \int \space \frac{sin \space ^2 \space x}{cos \space x} \space dx \space + \space C_1 \space \space \\ \space \space \space \space = \space - \space \int \space \frac{1-cos \space ^2 \space x}{cos \space x} \space dx \space + \space C_1 \space \space \\ \space \space \space \space = \space - \space \int \space {(sec \space x \space - \space cos \space x)}dx \space + \space C_1 \space \space \\ A= \space - \space [log \space (sec \space x \space + \space tan \space x \space )- \space sin \space x \space ]+ \space C_1 \space \\ B' \space = \space \frac{y_1 \space \phi \space (x)}{W} \space \\ B' \space = \space \frac{cos \space x \space tan \space x \space }{1} \space \\ B'=sin \space x \space \\ B= \space \int \space sin \space x \space dx \space + \space C_2 \space \\ B= \space -cos \space x \space + \space C_2 \space \\ Substitute \space these \space values \space of \space A \space and \space B \space in \space Eqn. \space (1), \space we \space get\\ \space y \space = \space [– \space log(secx \space + \space tanx) \space + \space sinx \space + \space \space C_1] \space cosx \space + \space [– \space cos \space x \space + \space \space C_2] \space sinx \space \\ y= \space C_1cos \space x \space + \space \space C_2sin \space x \space – \space cosx \space log \space (sec \space x \space + \space tan \space x) \space \\ This \space is \space the \space complete \space solution. \space \space$