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
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)](https://www.tinkutara.com/question/Q135064.png)
$$\mathrm{let}\:\mathrm{u}\:=\:\mathrm{arctan}\:\mathrm{x}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{dy}}{\mathrm{du}}.\:\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }.\:\frac{\mathrm{dy}}{\mathrm{du}} \\ $$$$\mathrm{then}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:−\frac{\mathrm{2x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{\mathrm{dy}}{\mathrm{du}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:.\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} } \\ $$$$\mathrm{substuting}\:\mathrm{to}\:\mathrm{original}\:\mathrm{DE} \\ $$$$\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \left[\:\frac{−\mathrm{2x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{\mathrm{dy}}{\mathrm{du}}\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} }\:\right]+\mathrm{2x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left[\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\frac{\mathrm{dy}}{\mathrm{du}}\:\right]+\mathrm{4y}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:−\mathrm{2x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} }\:+\:\mathrm{2x}\:\frac{\mathrm{dy}}{\mathrm{du}}\:+\:\mathrm{4y}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{du}^{\mathrm{2}} }\:+\:\mathrm{4y}\:=\:\mathrm{0}\: \\ $$$$\mathrm{the}\:\mathrm{characteristic}\:\mathrm{equation}\:\lambda^{\mathrm{2}} +\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{roots}\:\lambda\:=\:\pm\:\mathrm{2}{i}\: \\ $$$$\mathbb{G}\mathrm{eneral}\:\mathrm{solution}\: \\ $$$$\mathrm{y}=\:\mathrm{C}_{\mathrm{1}} \mathrm{cos}\:\mathrm{2u}\:+\:\mathrm{C}_{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2u}\: \\ $$$$\mathrm{y}=\:\mathrm{C}_{\mathrm{1}} \:\mathrm{cos}\:\left(\mathrm{2arctan}\:\mathrm{x}\right)\:+\:\mathrm{C}_{\mathrm{2}} \:\mathrm{sin}\:\left(\mathrm{2arctan}\:\mathrm{x}\right)\: \\ $$
Commented by liberty last updated on 10/Mar/21
$$\mathrm{nice} \\ $$