Question

Let $y_{c} : R \to (0, \infty)$ be the solution of the Bernoulli’s equation
$\frac{d y}{d x} - y + y^{3} = 0, y (0) = c > 0 \cdot$
Then, for every $c > 0$ , which one of the following is true?

Answer: $lim_{x \to \infty} y_{c} (x) = 1$

Lets first find solution of given Bernoulli’s equation.

We have,

\begin{aligned} \frac{d y}{d x} - y + y^{3} & = & 0 \\ \frac{d y}{d x} - y & = & - y^{3} \\ \frac{1}{- y^{3}} \frac{d y}{d x} - \frac{y}{- y^{3}} & = & \frac{- y^{3}}{- y^{3}} (dividing both sides with - y^{3}) \\ - y^{- 3} \frac{d y}{d x} + y^{- 2} & = & 1 \dots \dots (1) \end{aligned}

Now, let $y^{- 2} = t$ , Then differentiate w. r. to $x$ , we get

- 2 y^{- 3} \frac{d y}{d x} = \frac{d t}{d x} ⟹ - y^{- 3} \frac{d y}{d x} = \frac{1}{2} \cdot \frac{d t}{d x}

from (1), we get

\frac{1}{2} \cdot \frac{d t}{d x} + t = 1 ⟹ \frac{d t}{d x} + 2 t = 2

Now, I.F. = $e^{\int 2 d x} = e^{2 x}$
So, the solution is

\begin{aligned} t \cdot e^{2 x} & = & \int 2 \cdot e^{2 x} d x \\ y^{- 2} \cdot e^{2 x} & = & 2 \cdot \frac{e^{2 x}}{2} + k \\ \frac{e^{2 x}}{y^{2}} & = & e^{2 x} + k \dots \dots (2) \end{aligned}

Now, given that $y (0) = c > 0$

\begin{aligned} So, \frac{e^{2 \cdot 0}}{c^{2}} & = & e^{2 \cdot 0} + k \\ \frac{1}{c^{2}} & = & 1 + k \\ ∴ k & = & \frac{1}{c^{2}} - 1 \end{aligned}

Now, put the value of k in eq(2), we get

\begin{aligned} \frac{e^{2 x}}{y^{2}} & = & e^{2 x} + \frac{1}{c^{2}} - 1 \\ \frac{1}{y^{2}} & = & 1 + (\frac{1}{c^{2}} - 1) \frac{1}{e^{2 x}} (dividing both sides with e^{2 x}) \\ y^{2} & = & \frac{1}{1 + (\frac{1}{c^{2}} - 1) \frac{1}{e^{2 x}}} \\ ∴ y & = & \frac{1}{\sqrt{1 + (\frac{1}{c^{2}} - 1) \frac{1}{e^{2 x}}}} (∵ y_{c} : R \to (0, \infty)) \end{aligned}

Now, we know that if $x \to \infty$ then all of these $\frac{1}{x}, \frac{1}{e^{x}}, \frac{c}{e^{2 x}}$ and so on goes to zero.

Hence, term $(\frac{1}{c^{2}} - 1) \frac{1}{e^{2 x}}$ become zero when $x \to \infty$ , given c > 0.

So, $lim_{x \to \infty} y_{c} (x) = \frac{1}{\sqrt{1 + 0}} = \frac{1}{\sqrt{1}} = 1 \cdot$

Below, you can see the graph of $y = \frac{1}{\sqrt{1 + (\frac{1}{c^{2}} - 1) \frac{1}{e^{2 x}}}} as x \to \infty and c > 0.$

So, correct answer is option-2, $\to lim_{x \to \infty} y_{c} (x) = 1$