find-solution-d-2-y-dx-2-4-dy-dx-3-cosec-

Question Number 125691 by fajri last updated on 13/Dec/20

f i n d s o l u t i o n :

\frac{d^{2} y}{{d x}^{2}} + 4 \frac{d y}{d x} = 3 c o s e c θ

Answered by mathmax by abdo last updated on 14/Dec/20

if cosec θ = \frac{1}{\sin θ} e \to y^{^{″}} + {4 y}^{^{'}} = \frac{3}{\sin θ} (the variable is x not θ!)

let y^{^{'}} = z \Rightarrow z^{^{'}} + z = \frac{3}{\sin θ}

h \to z^{^{'}} = - z \Rightarrow \frac{z^{^{'}}}{z} = - 1 \Rightarrow lnz = - x + c \Rightarrow z = k e^{- x}

lagrange method \to z^{^{'}} = k^{^{'}} e^{- x} - {ke}^{- x}

e \to k^{^{'}} e^{- x} - {ke}^{- x} + {ke}^{- x} = \frac{3}{\sin θ} \Rightarrow k^{^{'}} = \frac{3}{\sin θ} e^{x} \Rightarrow

k = \frac{3}{\sin θ} \int e^{x} + c = \frac{{3 e}^{x}}{\sin θ} + c \Rightarrow z = (\frac{{3 e}^{x}}{\sin θ} + c) e^{- x}

= \frac{3}{\sin θ} + {ce}^{- x}

y^{^{'}} = z \Rightarrow y^{^{'}} = \frac{3}{\sin θ} + {ce}^{- x} \Rightarrow y = \int (\frac{3}{\sin θ} + {ce}^{- x}) dx \Rightarrow

y (x) = \frac{3 x}{\sin θ} - c e^{- x} + λ