Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function with $f(0)=0$. If $y=f(x)$ s

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function with $f(0)=0$. If $y=f(x)$ satisfies the differential equation $\frac{d y}{d x}=(2+5 y)(5 y-2)$, then the value of $10\lim _{x \rightarrow-\infty} f(x)$ is

Professor PC Thomas Classes · Answer

(0.4) : $\frac{d y}{d x}=(5 y+2)(5 y-2)$\ $\Rightarrow \frac{d y}{25 y^{2}-4}=d x$\ On integrating, we get\ $\frac{1}{25} \cdot \frac{5}{2 \cdot 2} \log \left|\frac{y-2 / 5}{y+2 / 5}ight|=x+\lambda$\ Then, $\log \left(\frac{5 y-2}{5 y+2}ight)=20(x+\lambda)$.\ Again, $f(0)=0$, so $\lambda=0$. Then,\ $\left(\frac{5 y-2}{5 y+2}ight)=e^{20 x}$\ Now, $\lim _{x ightarrow-\infty}\left|\frac{5 f(x)-2}{5 f(x)+2}ight|=\lim _{x ightarrow-\infty} e^{20 x}=0$\ $	herefore \quad \lim _{x ightarrow-\infty} f(x)=\frac{2}{5}=0.4$\