Question

If y(x) is the solution of the differential equation $\frac{dy}{dx} = -2x(y-1)$ with y(0) = 1, then find the $\lim_{x \to \infty} y(x)$.

Answer 1

1

Answer 2

The given differential equation is $\frac{dy}{dx} = -2x(y-1)$. This is a first-order separable differential equation.

1. Separate variables: $\frac{dy}{y-1} = -2x dx$

2. Integrate both sides: $\int \frac{dy}{y-1} = \int -2x dx$ $\ln|y-1| = -x^2 + C$

3. Solve for y: $|y-1| = e^{-x^2+C} = e^C e^{-x^2}$ Let $A = \pm e^C$. Then, we have: $y-1 = A e^{-x^2}$ $y(x) = 1 + A e^{-x^2}$

4. Apply the initial condition $y(0) = 1$: Substitute $x=0$ and $y=1$: $1 = 1 + A e^{-0^2}$ $1 = 1 + A$ $A = 0$

5. Write the particular solution: Substituting $A=0$ into the general solution, we get: $y(x) = 1 + 0 \cdot e^{-x^2} = 1$

6. Find the limit as $x \to \infty$: $\lim_{x \to \infty} y(x) = \lim_{x \to \infty} 1 = 1$

The given differential equation is separable. After separating variables and integrating, we obtain $\ln|y-1| = -x^2 + C$. Exponentiating both sides yields $y-1 = A e^{-x^2}$. Using the initial condition $y(0)=1$, we find the constant $A=0$. This means the particular solution is $y(x)=1$. The limit of $y(x)$ as $x \to \infty$ is therefore $\lim_{x \to \infty} 1 = 1$.