Question

Let y(x) be the solution of the deferential equation \(\frac{dy}{dx}+3x^2y=x^2\), for xεR, 
satisfying the initial condition y(0)=4. 
Then \(lim \,\,y_{n→∞}\) y(x) is equal to ____ (Rounded off to decimal places)

Answer 1

Correct Answer: 0.32 - 0.34

The given differential equation can be solved using the method of integrating factors. The equation is: \(\frac{dy}{dx} + 3x^2y = x^2\). This is a first-order linear differential equation in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), with \(P(x) = 3x^2\) and \(Q(x) = x^2\).

First, we find the integrating factor \(\mu(x)\) given by \(\mu(x) = e^{\int P(x)dx} = e^{\int 3x^2dx} = e^{x^3}\).

Multiply the entire differential equation by the integrating factor: 
\(e^{x^3}\frac{dy}{dx} + 3x^2e^{x^3}y = x^2e^{x^3}\).

The left-hand side is the derivative of \((e^{x^3}y)\), simplifying to: 
\(\frac{d}{dx}(e^{x^3}y) = x^2e^{x^3}\).

Integrating both sides with respect to \(x\): 
\(\int \frac{d}{dx}(e^{x^3}y)dx = \int x^2e^{x^3}dx\).

The left side integrates to \(e^{x^3}y\). For the right side, use substitution: let \(u = x^3\), then \(du = 3x^2dx\) or \(x^2dx = \frac{1}{3}du\).

Thus, 
\(\int x^2e^{x^3}dx = \int \frac{1}{3}e^udu = \frac{1}{3}e^u + C = \frac{1}{3}e^{x^3} + C\).

Equating both results, we get: 
\(e^{x^3}y = \frac{1}{3}e^{x^3} + C\).

Solving for \(y\), we have: 
\(y = \frac{1}{3} + Ce^{-x^3}\).

Using the initial condition \(y(0) = 4\), find \(C\): 
\(4 = \frac{1}{3} + Ce^0\) which implies \(C = \frac{11}{3}\).

Thus, the solution is \(y(x) = \frac{1}{3} + \frac{11}{3}e^{-x^3}\).

Calculating \(\lim_{x \to \infty} y(x)\):
As \(x \rightarrow \infty\), \(e^{-x^3} \rightarrow 0\), so \(\lim_{x \to \infty} y(x) = \frac{1}{3}\).

The computed limit, \(\frac{1}{3} = 0.3333\).