solve y^(′′) −3y^′ +2y =((sinx)/x)


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Question Number 98657 by mathmax by abdo last updated on 15/Jun/20

$solve y^{^{″}} - {3 y}^{^{'}} + 2 y = \frac{sinx}{x}$
	Answered by maths mind last updated on 15/Jun/20
	$y′′−3y′+2y=0 ⇒y=ae^x +be^(2x) let y=ke^x particular sokution ⇒y′=(k′+k)e^x ,y′′=(k′′+2k′+k)e^x ⇒ (k′′−k′)e^x =((sin(x))/x) ⇒k′′−k′=((sin(x)e^(−x) )/x)=Im{(e^((i−1)x) /x)} k′−k=∫(e^((i−1)x) /x)dx=∫(e^u /u)du=Ei(u)=Ei((i−1)x) ⇒k′−k=E_i ((i−1)x) k=se^x ⇒s′=∫e^(−x) E_i ((i−1)x) =[−e^(−x) E_i (i−1)x)]+∫e^(−x) .(e^((i−1)x) /x)dx =E_i ((i−2)x) k=e^x E_i ((i−2)x)) Y_p =Im{e^(2x) E_i ((i−2)x))} e^(2x) Im{∫_(−∞) ^x (e^((i−2)x) /x)dx} =e^(2x) Im{∫((e^(−2x) (cos(x)+isin(x)))/x)dx} =e^(2x) [((E_i ((i−2)x)−E_i ((−i−2)x))/(2i))] S=ae^x +be^(2x) +e^(2x) [((E_i ((i−2)x)−E_i ((−i−2)x))/(2i))]$
	$y^{″} - 3 y^{'} + 2 y = 0$ $\Rightarrow y = {a e}^{x} + {b e}^{2 x}$ $l e t y = {k e}^{x} p a r t i c u l a r s o k u t i o n$ $\Rightarrow y^{'} = (k^{'} + k) e^{x}, y^{″} = (k^{″} + 2 k^{'} + k) e^{x} \Rightarrow$ $(k^{″} - k^{'}) e^{x} = \frac{s i n (x)}{x}$ $\Rightarrow k^{″} - k^{'} = \frac{s i n (x) e^{- x}}{x} = I m {\frac{e^{(i - 1) x}}{x}}$ $k^{'} - k = \int \frac{e^{(i - 1) x}}{x} d x = \int \frac{e^{u}}{u} d u = E i (u) = E i ((i - 1) x)$ $\Rightarrow k^{'} - k = E_{i} ((i - 1) x)$ $k = {s e}^{x} \Rightarrow s^{'} = \int e^{- x} E_{i} ((i - 1) x)$ $= [- e^{- x} E_{i} (i - 1) x)] + \int e^{- x} . \frac{e^{(i - 1) x}}{x} d x$ $= E_{i} ((i - 2) x)$ $k = e^{x} E_{i} ((i - 2) x))$ $Y_{p} = I m {e^{2 x} E_{i} ((i - 2) x))}$ $e^{2 x} I m {\int_{- \infty}^{x} \frac{e^{(i - 2) x}}{x} d x}$ $= e^{2 x} I m {\int \frac{e^{- 2 x} (c o s (x) + i s i n (x))}{x} d x}$ $= e^{2 x} [\frac{E_{i} ((i - 2) x) - E_{i} ((- i - 2) x)}{2 i}]$ $S = {a e}^{x} + {b e}^{2 x} + e^{2 x} [\frac{E_{i} ((i - 2) x) - E_{i} ((- i - 2) x)}{2 i}]$
	Commented by mathmax by abdo last updated on 15/Jun/20

	$thank you sir .$
	Answered by mathmax by abdo last updated on 15/Jun/20

	$(he) \to y^{^{″}} - {3 y}^{^{'}} + 2 y = 0 \to r^{2} - 3 r + 2 = 0 \Rightarrow r^{2} - 1 - 3 r + 3 = 0 \Rightarrow$ $(r - 1) (r + 1) - 3 (r - 1) = 0 \Rightarrow (r - 1) (r - 2) = 0 \Rightarrow r = 1 or r = 2$ $\Rightarrow y_{h} = a e^{x} + b e^{2 x} = {au}_{1} + b u_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{x} e^{2 x} \\ e^{x} {2 e}^{2 x} \end{matrix} \| = {2 e}^{3 x} - e^{3 x} = e^{3 x} \neq 0$ $W_{1} = \| \begin{matrix} 0 e^{2 x} \\ \frac{sinx}{x} {2 e}^{2 x} \end{matrix} \| = - e^{2 x} \frac{sinx}{x}$ $W_{2} = \| \begin{matrix} e^{x} 0 \\ e^{x} \frac{sinx}{x} \end{matrix} \| = e^{x} \frac{sinx}{x}$ $v_{1} = \int \frac{W_{1}}{W} dx = \int \frac{- e^{2 x} \frac{sinx}{x}}{e^{3 x}} dx = - \int e^{- 5 x} \frac{sinx}{x} dx$ $v_{2} = \int \frac{W_{2}}{W} dx = \int \frac{e^{x} \frac{sinx}{x}}{e^{3 x}} = \int e^{- 2 x} \frac{sinx}{x} dx \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = - e^{x} \int^{x} e^{- 5 t} \frac{sint}{t} dt + e^{2 x} \int^{x} e^{- 2 t} \frac{sint}{t} dt$ $the general solution is y = y_{h} + y_{p}$ $= {ae}^{x} + {be}^{2 x} - e^{x} \int^{x} e^{- 5 t} \frac{sint}{t} dt + e^{2 x} \int^{x} e^{- 2 t} \frac{sint}{t} dt$
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