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Question Number 135061 by liberty last updated on 09/Mar/21

$$ \\ $$ How can I solve the differential equation (1+x^2)^2y′′+2x(1+x^2)y′+4y=0\\n

Answered by EDWIN88 last updated on 10/Mar/21

let u = arctan x ⇒(dy/dx) = (dy/du). (du/dx) = (1/(1+x^2 )). (dy/du)  then (d^2 y/dx^2 ) = −((2x)/((1+x^2 )^2 )) (dy/du) + (1/(1+x^2 )) . (1/(1+x^2 )) (d^2 y/du^2 )  substuting to original DE  (1+x^2 )^2 [ ((−2x)/((1+x^2 )^2 )) (dy/du) + (1/((1+x^2 )^2 )) (d^2 y/du^2 ) ]+2x(1+x^2 )[ (1/(1+x^2 )) (dy/du) ]+4y = 0  ⇔ −2x (dy/dx) + (d^2 y/du^2 ) + 2x (dy/du) + 4y = 0  ⇔ (d^2 y/du^2 ) + 4y = 0   the characteristic equation λ^2 +4 = 0  has roots λ = ± 2i   General solution   y= C_1 cos 2u + C_2  sin 2u   y= C_1  cos (2arctan x) + C_2  sin (2arctan x)

$$\mathrm{let}\:\mathrm{u}\:=\:\mathrm{arctan}\:\mathrm{x}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{dy}}{\mathrm{du}}.\:\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }.\:\frac{\mathrm{dy}}{\mathrm{du}} \\ $$ $$\mathrm{then}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:−\frac{\mathrm{2x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{\mathrm{dy}}{\mathrm{du}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:.\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} } \\ $$ $$\mathrm{substuting}\:\mathrm{to}\:\mathrm{original}\:\mathrm{DE} \\ $$ $$\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \left[\:\frac{−\mathrm{2x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{\mathrm{dy}}{\mathrm{du}}\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} }\:\right]+\mathrm{2x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left[\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\frac{\mathrm{dy}}{\mathrm{du}}\:\right]+\mathrm{4y}\:=\:\mathrm{0} \\ $$ $$\Leftrightarrow\:−\mathrm{2x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} }\:+\:\mathrm{2x}\:\frac{\mathrm{dy}}{\mathrm{du}}\:+\:\mathrm{4y}\:=\:\mathrm{0} \\ $$ $$\Leftrightarrow\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} }\:+\:\mathrm{4y}\:=\:\mathrm{0}\: \\ $$ $$\mathrm{the}\:\mathrm{characteristic}\:\mathrm{equation}\:\lambda^{\mathrm{2}} +\mathrm{4}\:=\:\mathrm{0} \\ $$ $$\mathrm{has}\:\mathrm{roots}\:\lambda\:=\:\pm\:\mathrm{2}{i}\: \\ $$ $$\mathbb{\color{mathred}{G}}\mathrm{\color{mathred}{e}\color{mathred}{n}\color{mathred}{e}\color{mathred}{r}\color{mathred}{a}\color{mathred}{l}}\color{mathred}{\:}\mathrm{\color{mathred}{s}\color{mathred}{o}\color{mathred}{l}\color{mathred}{u}\color{mathred}{t}\color{mathred}{i}\color{mathred}{o}\color{mathred}{n}}\color{mathred}{\:} \\ $$ $$\mathrm{\color{mathred}{y}}\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{C}}_{\mathrm{\color{mathred}{1}}} \mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\mathrm{\color{mathred}{2}\color{mathred}{u}}\color{mathred}{\:}\color{mathred}{+}\color{mathred}{\:}\mathrm{\color{mathred}{C}}_{\mathrm{\color{mathred}{2}}} \color{mathred}{\:}\mathrm{\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}\mathrm{\color{mathred}{2}\color{mathred}{u}}\color{mathred}{\:} \\ $$ $$\mathrm{\color{mathred}{y}}\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{C}}_{\mathrm{\color{mathred}{1}}} \color{mathred}{\:}\mathrm{\color{mathred}{c}\color{mathred}{o}\color{mathred}{s}}\color{mathred}{\:}\color{mathred}{\left(}\mathrm{\color{mathred}{2}\color{mathred}{a}\color{mathred}{r}\color{mathred}{c}\color{mathred}{t}\color{mathred}{a}\color{mathred}{n}}\color{mathred}{\:}\mathrm{\color{mathred}{x}}\color{mathred}{\right)}\color{mathred}{\:}\color{mathred}{+}\color{mathred}{\:}\mathrm{\color{mathred}{C}}_{\mathrm{\color{mathred}{2}}} \color{mathred}{\:}\mathrm{\color{mathred}{s}\color{mathred}{i}\color{mathred}{n}}\color{mathred}{\:}\color{mathred}{\left(}\mathrm{\color{mathred}{2}\color{mathred}{a}\color{mathred}{r}\color{mathred}{c}\color{mathred}{t}\color{mathred}{a}\color{mathred}{n}}\color{mathred}{\:}\mathrm{\color{mathred}{x}}\color{mathred}{\right)}\color{mathred}{\:} \\ $$

Commented byliberty last updated on 10/Mar/21

nice

$$\mathrm{nice} \\ $$

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