Verified answer

Resposta:

Encontremos [tex]f(x),[/tex] sabendo que [tex]\frac{d^2f}{dx^2} = x + cos(x).[/tex]

Temos:

[tex]\dfrac{df}{dx} = \int {x + cos(x)}\, dx\\\\\\\Longleftrightarrow \dfrac{df}{dx} = \dfrac{1}{2}x^2 + sin(x) + C_1\\\\\\\Longrightarrow \int df = \int {\dfrac{1}{2}x^2 + sin(x) + C_1}\, dx\\\\\\\Longleftrightarrow f = \dfrac{1}{6}x^3 - cos(x) + C_1x + C_2[/tex]

Apliquemos as condições de contorno para descobrirmos as constantes de integração:

[tex]f'(0) = 0\\\\\Longleftrightarrow \dfrac{1}{2} \cdot 0 + sin(0) + C_1 = 2\\\\\Longleftrightarrow C_1 = 2[/tex]

[tex]f(0) = 1\\\\\Longleftrightarrow \dfrac{1}{6} \cdot 0^3 - cos(0) + 2 \cdot 0 + C_2 = 1\\\\\Longleftrightarrow C_2 = 2[/tex]

Assim:

[tex]\boxed{f(x) = \dfrac{1}{6}x^3 - cos(x) + 2x + 2}[/tex]