y-y-cot-x-

Question Number 103788 by bemath last updated on 17/Jul/20

y^{″} - y = \cot x

Answered by mathmax by abdo last updated on 18/Jul/20

y^{^{″}} - y = \frac{cosx}{sinx}

h \to r^{2} - 1 = 0 \Rightarrow r = \bar{+} 1 \Rightarrow y_{h} = a e^{x} + {be}^{- x} = {au}_{1} + {bu}_{2}

W (u_{1}, u_{2}) = | \begin{matrix} e^{x} e^{- x} \\ e^{x} - e^{- x} \end{matrix} | = - 2 \neq 0

W_{1} = | \begin{matrix} o e^{- x} \\ \frac{cosx}{sinx} - e^{- x} \end{matrix} | = - \frac{e^{- x} cosx}{sinx}

W_{2} = | \begin{matrix} e^{x} 0 \\ e^{x} \frac{cosx}{sinx} \end{matrix} | = \frac{e^{x} cosx}{sinx}

v_{1} = \int \frac{w_{1}}{w} dx = \int \frac{e^{- x} cosx}{2 sinx} dx

v_{2} = \int \frac{w_{2}}{w} dx = - \frac{1}{2} \int \frac{e^{x} cosx}{sinx} dx

\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} = e^{x} \int \frac{e^{- x} cosx}{2 sinx} dx - e^{- x} \int \frac{e^{x} cosx}{2 sinx} dx

the general solution is y = y_{p} + y_{h}