Question

PROBLEMA DIFÍCIL CÁLCULO (50 puntos)

Sea f(x) la función de la foto.

Demuestre que dicha función es solución de la ecuación diferencial:

$y'' + y = arctan({x}^{3} )$
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Answer 1

$\displaystyle\\f(x)=\int_{0}^{x}\arctan(t^3)(\sin x\cos t-\cos x\sin t)~dt\\ \\f(x)=\int_{0}^{x}\arctan(t^3)\sin x\cos t~dt -\int_{0}^{x}\arctan(t^3)\cos x\sin t~dt \\ \\f(x)=\sin x\int_{0}^{x}\arctan(t^3)\cos t~dt -\cos x\int_{0}^{x}\arctan(t^3)\sin t~dt$

$\displaystyle\\f'(x)=\left[\cos x\int_{0}^{x}\arctan(t^3)\cos t~dt+\sin x  \arctan x^3 \cos x\right]-\\ \\- \left[-\sin x \int_{0}^{x}\arctan(t^3)\sin t~dt+\cos x \arctan(x^3)\sin x\right]\\ \\\\f'(x)=\cos x\int_{0}^{x}\arctan(t^3)\cos t~dt+\sin x \int_{0}^{x}\arctan(t^3)\sin t~dt\\ \\f''(x)=\left[-\sin x \int_{0}^{x}\arctan(t^3)\cos t~dt+\cos^2 x \arctan(x^3)\right]+\\\\+\left[\cos x \int_{0}^{x}\arctan(t^3)\sin t~dt + \sin^2x \arctan(x^3)\right]$

$f''(x)=-f(x)+\arctan(x^3)\\ \\\boxed{f''(x)+f(x)=\arctan(x^3)}$