solve-x-y-e-x-y-x-sinx-

Question Number 46609 by maxmathsup by imad last updated on 29/Oct/18

s o l v e x y^{^{″}} - e^{- x} y^{^{'}} = x s i n x

Commented by maxmathsup by imad last updated on 11/Nov/18

h a n g e m e n t y^{^{'}} = z g i v e {x z}^{^{'}} - e^{- x} z = x s i n x

(h e) \Rightarrow {x z}^{^{'}} - e^{- x} z = 0 \Rightarrow {x z}^{^{'}} = e^{- x} z \Rightarrow \frac{z^{^{'}}}{z} = \frac{e^{- x}}{x} \Rightarrow l n ∣ z ∣= \int \frac{e^{- x}}{x} d x + α \Rightarrow

z = K e^{\int \frac{e^{- x}}{x} d x} m v c m e t h o d g v e z^{^{'}} = K^{^{'}} e^{\int \frac{e^{- x}}{x} d x} + K \frac{e^{- x}}{x} e^{\int \frac{e^{- x}}{x} d x}

(e) \Rightarrow {x K}^{^{'}} e^{\int \frac{e^{- x}}{x} d x} + {K e}^{- x} e^{\int \frac{e^{- x}}{x} d x} - e^{- x} K e^{\int \frac{e^{- x}}{x} d x} = x s i n x \Rightarrow

{x K}^{^{'}} e^{\int \frac{e^{- x}}{x} d x} = x s i n x \Rightarrow K^{^{'}} = s i n x e^{- \int \frac{e^{- x}}{x} d x} \Rightarrow

K (x) = \int_{.}^{x} (s i n t e^{- \int \frac{e^{- t}}{t} d t}) d t + λ \Rightarrow z = (\int_{.}^{x} (s i n t e^{- \int \frac{e^{- t}}{t} d t}) d t + λ) e^{\int \frac{e^{- x}}{x}} d x

b u t y^{^{'}} (x) = z (x) \Rightarrow y (x) = \int z (x) d x + β a n d t h e f u n c t i o n z i s d e t e r m i n e d .