$f'(x)-2xf(x) = 2x^3 f^2(x)\\ \displaystyle \Leftrightarrow \frac{2xf(x)-f'(x)}{f^2(x)}=-2x^3\\ \displaystyle \Leftrightarrow \frac{2x.e^{x^2}.f(x)-e^{x^2}f'(x)}{f^2(x)}=-2x^3.e^{x^2}\\ \displaystyle \Leftrightarrow (\frac{e^{x^2}}{f(x)})'=-2x^3.e^{x^2}\\ \displaystyle \Leftrightarrow \displaystyle \int (\frac{e^{x^2}}{f(x)})'dx=- \int 2x^3.e^{x^2}dx$
Đặt $t=x^2$ thì $dt=2xdx$
$ \displaystyle \Leftrightarrow \frac{e^{x^2}}{f(x)}=-\int t.e^t dt\\\displaystyle \Leftrightarrow \frac{e^{x^2}}{f(x)}=-(t.e^t-\int e^t dt)\\\displaystyle \Leftrightarrow \frac{e^{x^2}}{f(x)}=(1-t)e^t+C\\\Leftrightarrow \displaystyle \frac{e^{x^2}}{f(x)}=(1-x^2)e^{x^2}+C$
Do $f(0)=1$ nên $C=0$
Do đó $ \displaystyle \frac{e^{x^2}}{f(x)}=(1-x^2)e^{x^2}$ hay $\displaystyle f(x)=\frac{1}{1-x^2}$
Vậy [tex]\displaystyle f(\frac{1}{2})=\frac{4}{3}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
