solve-x-2-y-x-1-y-x-2-sinx-

Question Number 97625 by mathmax by abdo last updated on 08/Jun/20

solve x^{2} y^{^{″}} - (x + 1) y^{^{'}} = x^{2} sinx

Answered by mathmax by abdo last updated on 09/Jun/20

let y^{^{'}} = z so (e) \Rightarrow x^{2} z^{^{'}} - (x + 1) z = x^{2} sinx

(he) \to x^{2} z^{^{'}} - (x + 1) z = 0 \Rightarrow x^{2} z^{^{'}} = (x + 1) z \Rightarrow \frac{z^{^{'}}}{z} = \frac{x + 1}{x^{2}} = \frac{1}{x} + \frac{1}{x^{2}} \Rightarrow

\ln ∣ z ∣ = \ln ∣ x ∣ - \frac{1}{x} + c \Rightarrow z = k ∣ x ∣ e^{- \frac{1}{x}} let find solution on] 0, + \infty [\Rightarrow

z = {kxe}^{- \frac{1}{x}} mvc method \to z^{^{'}} = k^{^{'}} x e^{- \frac{1}{x}} + k (e^{- \frac{1}{x}} + x \times \frac{1}{x^{2}} e^{- \frac{1}{x}})

= k^{^{'}} x e^{- \frac{1}{x}} + k e^{- \frac{1}{x}} + \frac{k}{x} e^{- \frac{1}{x}}

(e) \Rightarrow k^{^{'}} x^{3} e^{- \frac{1}{x}} + {kx}^{2} e^{- \frac{1}{x}} + kx e^{- \frac{1}{x}} - kx (x + 1) e^{- \frac{1}{x}} = x^{2} sinx \Rightarrow

(k^{^{'}} x^{3} + {kx}^{2} + k - {kx}^{2} - kx) e^{- \frac{1}{x}} = x^{2} sinx \Rightarrow k^{^{'}} x^{3} e^{- \frac{1}{x}} = x^{2} sinx \Rightarrow

k^{^{'}} = \frac{sinx}{x} e^{\frac{1}{x}} \Rightarrow k (x) = \int^{x} \frac{sint}{t} e^{\frac{1}{t}} dt + c \Rightarrow

z (x) = x e^{- \frac{1}{x}} {\int^{x} \frac{sint}{t} e^{\frac{1}{t}} dt + c} = x e^{- \frac{1}{x}} \int^{x} \frac{sint}{t} e^{\frac{1}{t}} dt + cx e^{- \frac{1}{x}}

y^{^{'}} = z \Rightarrow y (x) = \int^{x} z (u) du + λ

= \int^{x} {u e^{- \frac{1}{u}} \int^{u} \frac{sint}{t} e^{\frac{1}{t}} dt + cu e^{- \frac{1}{u}}} du + λ