Question

Let \(y_c:\R \rightarrow(0,\infin)\) be the solution of the Bernoulli’s equation
\(\frac{dy}{dx}-y+y^3=0,\ \ \ \ \ \ \ y(0)=c \gt 0.\)
Then, for every 𝑐 > 0, which one of the following is true ?

• A. \(\lim\limits_{x\rightarrow \infin}y_c(x)=0\)
• B. \(\lim\limits_{x\rightarrow \infin}y_c(x)=1\)
• C. \(\lim\limits_{x\rightarrow \infin}y_c(x)=e\)
• D. \(\lim\limits_{x\rightarrow \infin}y_c(x)\) does not exist

Answer 1

The Correct Option is B

To solve the problem, we need to analyze the differential equation given and understand its behavior as \( x \rightarrow \infty \).

The given equation is:

\(\frac{dy}{dx} - y + y^3 = 0\)

This can be rewritten as: 

\(\frac{dy}{dx} = y - y^3\)

The equation is a first-order non-linear differential equation. We can separate variables to solve it:

\(\frac{1}{y - y^3}\frac{dy}{dx} = 1\)

Separate the variables:

\(\int \frac{1}{y(1-y^2)} \, dy = \int 1 \, dx\)

We proceed by partial fraction decomposition:

\(\frac{1}{y(1-y^2)} = \frac{A}{y} + \frac{B}{1-y} + \frac{C}{1+y}\)

Solving for constants \(A\), \(B\), and \(C\), we find:

\(A = 1, \, B = -\frac{1}{2}, \, C = -\frac{1}{2}\)

This gives:

\(\int \left(\frac{1}{y} - \frac{1/2}{1-y} - \frac{1/2}{1+y}\right) \, dy = \int 1 \, dx\)

Integrate each term:

\(\ln|y| - \frac{1}{2}\ln|1-y| - \frac{1}{2}\ln|1+y| = x + C\)

As \( x \rightarrow \infty \), the competition between terms \( y - y^3 \) guides the behavior.

The equation can stabilize (reach a steady state) when \( y - y^3 = 0 \Rightarrow y(1-y^2) = 0 \).

This yields the potential steady-state solutions:

\(y = 0 \text{ or } y = \pm 1\)

Since \( y(x) \) is in \((0, \infty)\) (from the problem statement), only \( y = 1 \) is a valid solution. Therefore, \( \lim_{x \rightarrow \infty} y_c(x) = 1 \) for any initial condition \( y(0) = c > 0 \).

Conclusion: The correct answer, based on our reasoning and analysis, is:

\(\lim\limits_{x\rightarrow \infin}y_c(x)=1\)