How-can-I-solve-the-differential-equation-1-x-2-2y-2x-1-x-2-y-4y-0-

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 \Rightarrow \frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx} = \frac{1}{1 + x^{2}} . \frac{dy}{du}

then \frac{d^{2} y}{{dx}^{2}} = - \frac{2 x}{{(1 + x^{2})}^{2}} \frac{dy}{du} + \frac{1}{1 + x^{2}} . \frac{1}{1 + x^{2}} \frac{d^{2} y}{{du}^{2}}

substuting to original DE

{(1 + x^{2})}^{2} [\frac{- 2 x}{{(1 + x^{2})}^{2}} \frac{dy}{du} + \frac{1}{{(1 + x^{2})}^{2}} \frac{d^{2} y}{{du}^{2}}] + 2 x (1 + x^{2}) [\frac{1}{1 + x^{2}} \frac{dy}{du}] + 4 y = 0

\Leftrightarrow - 2 x \frac{dy}{dx} + \frac{d^{2} y}{{du}^{2}} + 2 x \frac{dy}{du} + 4 y = 0

\Leftrightarrow \frac{d^{2} y}{{du}^{2}} + 4 y = 0

the characteristic equation λ^{2} + 4 = 0

has roots λ = \pm 2 i

G eneral solution

y = C_{1} \cos 2 u + C_{2} \sin 2 u

y = C_{1} \cos (2 \arctan x) + C_{2} \sin (2 \arctan x)

Commented by liberty last updated on 10/Mar/21

nice

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