Xét phương trình
$f'(x) . f(x) = x^4 + x^2$
$\Leftrightarrow \dfrac{df(x)}{dx} f(x) = x^4 + x^2$
$\Leftrightarrow f(x) df(x) = (x^4 + x^2)dx$
Lấy nguyên hàm 2 vế ta có
$\displaystyle\int f(x) df(x) =\displaystyle \int (x^4 + x^2) dx$
$\Leftrightarrow \dfrac{f^2(x)}{2} = \dfrac{x^5}{5} + \dfrac{x^3}{3} + c$
$\Leftrightarrow f^2(x) = \dfrac{2x^5}{5} + \dfrac{2x^3}{3} + c$
$\Leftrightarrow f(x) = \sqrt{\dfrac{2x^5}{5} + \dfrac{2x^3}{3} + c}$
Do $f(0) = 2$ nên ta có
$f(0) = \sqrt{0 + 0 + c}$
$\Leftrightarrow 2 = \sqrt{c}$
Vậy $c = 4$
Do đó
$f(x) = \sqrt{\dfrac{2x^5}{5} + \dfrac{2x^3}{3} + 4}$
Do đó
$f(2) = \sqrt{\dfrac{64}{5} + \dfrac{16}{3} + 4}$
$\Leftrightarrow f(2) = \sqrt{\dfrac{332}{15}}$
$\Leftrightarrow f^2(2) = \dfrac{332}{15}$
