Question

Resolva a equação diferencial sujeita às condições iniciais dadas. F"(x) = 4x - 1 f'(2) = -2; f (1) = 3

Answer 1

Resposta:

$\sf f''(x)=4x-1$

$\sf \int\sf f''(x)\,dx=\int\sf(4x-1)dx$

$\sf f'(x)=\int\sf4x\,dx-\int\sf1\,dx$

$\sf f'(x)=2x^2-x+\mathnormal{C}$

Se f'(2) = - 2:

$\sf f'(2)=2(2)^2-2+\mathnormal{C}=-\,2$

$\sf 2(4)-2+\mathnormal{C}=-\,2$

$\sf 8-2+\mathnormal{C}=-\,2$

$\sf 6+\mathnormal{C}=-\,2$

$\sf \mathnormal{C}=-\,2-6$

$\sf \mathnormal{C}=-\,8$

Logo:

$\sf f'(x)=2x^2-x-8$

$\sf \int\sf f'(x)\,dx=\int\sf(2x^2-x-8)\,dx$

$\sf f(x)=\int\sf2x^2dx-\int\sf x\,dx-\int\sf8\,dx$

$\sf f(x)=\frac{2x^3}{3}-\frac{x^2}{2}-8x+\mathnormal{C}$

Se f(1) = 3:

$\sf f(1)=\frac{2(1)^3}{3}-\frac{1^2}{2}-8(1)+\mathnormal{C}=3$

$\sf \frac{2}{3}-\frac{1}{2}-8+\mathnormal{C}=3$

$\sf \mathnormal{C}=3-\frac{2}{3}+\frac{1}{2}+8$

$\sf \mathnormal{C}=\frac{18}{6}-\frac{4}{6}+\frac{3}{6}+\frac{48}{6}$

$\sf \mathnormal{C}=\frac{65}{6}$

Portanto:

$\red{\sf f(x)=\frac{2x^3}{3}-\frac{x^2}{2}-8x+\frac{65}{6}}$

Answer 2

Resposta:

f(x)=2/3x3-1/2x2-4x+41/6 correta

Explicação passo a passo:

AVA