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Question Number 105163 by bemath last updated on 26/Jul/20

x^2 y′′+xy′−y=(1/(1+x^2 ))

$${x}^{\mathrm{2}} {y}''+{xy}'−{y}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Answered by bramlex last updated on 27/Jul/20

we solve homogenous equation  This is an Euler−Cauchy   y=x^n → { ((y′=nx^(n−1) )),((y′′=(n^2 −n)x^(n−2) )) :}  ⇒x^2 (n^2 −n)x^(n−2) +x(nx^(n−1) )−x^n =0  n^2 −n+n−1=0 →n = ±1  y_h  = C_1 x+(C_2 /x)  let y_1  = x ; y_2 = x^(−1)   → { ((y_1 ^′  = 1)),((y_2 ′=−x^(−2) )) :} ⇒ W(y_1 ,y_2 )= determinant (((x       x^(−1) )),((1    −x^(−2) )))=−2x^(−1)   u_1 = −∫((x^(−1) .(1/(x^2 (x^2 +1))))/(−2x^(−1) )) dx   u_1 =∫(1/2)((1/x^2 )−(1/(1+x^2 ))) dx  u_1 = −(1/(2x))−(1/2)tan^(−1) (x)  u_2 = ∫((x.(1/(x^2 (1+x^2 ))))/(−2x^(−1) )) dx  u_2 = ∫(( −1)/(2(1+x^2 ))) dx = −(1/2)tan^(−1) (x)  particular solution   y_p  = y_1 u_1 +y_2 u_2   y_p  = x(−(1/(2x))−(1/2)tan^(−1) (x))+(1/x)(−(1/2)tan^(−1) (x))  y_p = −(1/2)−(x/2) tan^(−1) (x) −(1/(2x))tan^(−1) (x)  General solution   y_G  = C_1 x+(C_2 /x)−(x/2) tan^(−1) (x)−(1/(2x))tan^(−1) (x)−(1/2)  ★▼◊

$${we}\:{solve}\:{homogenous}\:{equation} \\ $$$${This}\:{is}\:{an}\:{Euler}−{Cauchy}\: \\ $$$${y}={x}^{{n}} \rightarrow\begin{cases}{{y}'={nx}^{{n}−\mathrm{1}} }\\{{y}''=\left({n}^{\mathrm{2}} −{n}\right){x}^{{n}−\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left({n}^{\mathrm{2}} −{n}\right){x}^{{n}−\mathrm{2}} +{x}\left({nx}^{{n}−\mathrm{1}} \right)−{x}^{{n}} =\mathrm{0} \\ $$$${n}^{\mathrm{2}} −{n}+{n}−\mathrm{1}=\mathrm{0}\:\rightarrow{n}\:=\:\pm\mathrm{1} \\ $$$${y}_{{h}} \:=\:{C}_{\mathrm{1}} {x}+\frac{{C}_{\mathrm{2}} }{{x}} \\ $$$${let}\:{y}_{\mathrm{1}} \:=\:{x}\:;\:{y}_{\mathrm{2}} =\:{x}^{−\mathrm{1}} \\ $$$$\rightarrow\begin{cases}{{y}_{\mathrm{1}} ^{'} \:=\:\mathrm{1}}\\{{y}_{\mathrm{2}} '=−{x}^{−\mathrm{2}} }\end{cases}\:\Rightarrow\:{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)=\begin{vmatrix}{{x}\:\:\:\:\:\:\:{x}^{−\mathrm{1}} }\\{\mathrm{1}\:\:\:\:−{x}^{−\mathrm{2}} }\end{vmatrix}=−\mathrm{2}{x}^{−\mathrm{1}} \\ $$$${u}_{\mathrm{1}} =\:−\int\frac{{x}^{−\mathrm{1}} .\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{−\mathrm{2}{x}^{−\mathrm{1}} }\:{dx}\: \\ $$$${u}_{\mathrm{1}} =\int\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:{dx} \\ $$$${u}_{\mathrm{1}} =\:−\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right) \\ $$$${u}_{\mathrm{2}} =\:\int\frac{{x}.\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}}{−\mathrm{2}{x}^{−\mathrm{1}} }\:{dx} \\ $$$${u}_{\mathrm{2}} =\:\int\frac{\:−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right) \\ $$$${particular}\:{solution}\: \\ $$$${y}_{{p}} \:=\:{y}_{\mathrm{1}} {u}_{\mathrm{1}} +{y}_{\mathrm{2}} {u}_{\mathrm{2}} \\ $$$${y}_{{p}} \:=\:{x}\left(−\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)+\frac{\mathrm{1}}{{x}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$${y}_{{p}} =\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:−\frac{\mathrm{1}}{\mathrm{2}{x}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right) \\ $$$$\mathcal{G}{eneral}\:{solution}\: \\ $$$${y}_{\mathcal{G}} \:=\:{C}_{\mathrm{1}} {x}+\frac{{C}_{\mathrm{2}} }{{x}}−\frac{{x}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}{x}}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\bigstar\blacktrinagledown\lozenge \\ $$

Answered by mathmax by abdo last updated on 26/Jul/20

x^2 y^(′′)  +xy^′ −y =(1/(1+x^2 ))  h→x^2 y^((2))  +xy^((1)) −y =0 let y =x^m  ⇒y^′  =mx^(m−1)  and y^((2))  =m(m−1)x^(m−2)   e⇒m(m−1)x^m +mx^m −x^m  =0 ⇒(m^2 −m+m−1)x^m  =0 ⇒  m^2 −1=0 ⇒m =+^− 1 ⇒y =αx +(β/x) =αu_1  +β u_2   W(u_1  ,u_2 ) = determinant (((x          (1/x))),((1          −(1/x^2 ))))=−(1/x)−(1/x) =−(2/x) ≠0  W_1 = determinant (((o            (1/x))),(((1/(1+x^2 ))      −(1/x^2 ))))=−(1/(x(1+x^2 )))  W_2 = determinant (((x            0)),((1             (1/(1+x^2 )))))=(x/(1+x^2 ))  v_1 =∫ (w_1 /w)dx =−∫  (1/(x(1+x^2 )))×(((−x)/2))dx =(1/2) ∫  (dx/((1+x^2 )))=(1/2)arctanx  =∫ (w_2 /w)dx =∫  (x/(1+x^2 ))×(−(x/2))dx =−(1/2)∫ (x^2 /(1+x^2 ))dx  =−(1/2)∫((1+x^2 −1)/(1+x^2 ))dx =−(x/2) +(1/2)arctanx ⇒  y_p =u_1 v_1  +u_2 v_2 =(x/2)arctanx+(1/x)(−(x/2) +(1/2) arctanx)  =(x/2) arctanx−(1/2) +(1/(2x)) arctanx =(1/2)(x+(1/x))arctanx−(1/2)  the general solution is   y =αx +βx^(−1)  +(1/2)(x+(1/x))arctanx −(1/2)

$$\mathrm{x}^{\mathrm{2}} \mathrm{y}^{''} \:+\mathrm{xy}^{'} −\mathrm{y}\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{h}\rightarrow\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\left(\mathrm{2}\right)} \:+\mathrm{xy}^{\left(\mathrm{1}\right)} −\mathrm{y}\:=\mathrm{0}\:\mathrm{let}\:\mathrm{y}\:=\mathrm{x}^{\mathrm{m}} \:\Rightarrow\mathrm{y}^{'} \:=\mathrm{mx}^{\mathrm{m}−\mathrm{1}} \:\mathrm{and}\:\mathrm{y}^{\left(\mathrm{2}\right)} \:=\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)\mathrm{x}^{\mathrm{m}−\mathrm{2}} \\ $$$$\mathrm{e}\Rightarrow\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)\mathrm{x}^{\mathrm{m}} +\mathrm{mx}^{\mathrm{m}} −\mathrm{x}^{\mathrm{m}} \:=\mathrm{0}\:\Rightarrow\left(\mathrm{m}^{\mathrm{2}} −\mathrm{m}+\mathrm{m}−\mathrm{1}\right)\mathrm{x}^{\mathrm{m}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{m}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{m}\:=\overset{−} {+}\mathrm{1}\:\Rightarrow\mathrm{y}\:=\alpha\mathrm{x}\:+\frac{\beta}{\mathrm{x}}\:=\alpha\mathrm{u}_{\mathrm{1}} \:+\beta\:\mathrm{u}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} \:,\mathrm{u}_{\mathrm{2}} \right)\:=\begin{vmatrix}{\mathrm{x}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\end{vmatrix}=−\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{x}}\:=−\frac{\mathrm{2}}{\mathrm{x}}\:\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{o}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}}}\\{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\end{vmatrix}=−\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\end{vmatrix}=\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{w}_{\mathrm{1}} }{\mathrm{w}}\mathrm{dx}\:=−\int\:\:\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}×\left(\frac{−\mathrm{x}}{\mathrm{2}}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctanx} \\ $$$$=\int\:\frac{\mathrm{w}_{\mathrm{2}} }{\mathrm{w}}\mathrm{dx}\:=\int\:\:\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }×\left(−\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=−\frac{\mathrm{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctanx}\:\Rightarrow \\ $$$$\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} =\frac{\mathrm{x}}{\mathrm{2}}\mathrm{arctanx}+\frac{\mathrm{1}}{\mathrm{x}}\left(−\frac{\mathrm{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{arctanx}\right) \\ $$$$=\frac{\mathrm{x}}{\mathrm{2}}\:\mathrm{arctanx}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2x}}\:\mathrm{arctanx}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{arctanx}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{is}\: \\ $$$$\mathrm{y}\:=\alpha\mathrm{x}\:+\beta\mathrm{x}^{−\mathrm{1}} \:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{arctanx}\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

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