Solving the Given Differential Equation

Solving differential equations is a fundamental topic in calculus. This problem involves finding the limit of the solution \(y(x)\) of a first-order linear differential equation as \(x \to 0\), given an initial condition.

The given differential equation is \(\frac{{dy}}{{dx}} + 4xy = {x^3}\). This is a first-order linear differential equation of the form \(\frac{{dy}}{{dx}} + P(x)y = Q(x)\), where \(P(x) = 4x\) and \(Q(x) = x^3\). To solve this type of differential equation, we first find the integrating factor.

Step 1: Find the Integrating Factor (IF)

The integrating factor for a first-order linear differential equation is given by \(IF = e^{\int P(x) dx}\).

Here, \(P(x) = 4x\).

\(IF = e^{\int 4x dx} = e^{4 \int x dx} = e^{4 \left(\frac{x^2}{2}\right)} = e^{2x^2}\).

So, the integrating factor is \(e^{2x^2}\). This factor helps us solve the differential equation.

Step 2: Find the General Solution

Multiply the entire differential equation by the integrating factor:

\(e^{2x^2} \left( \frac{{dy}}{{dx}} + 4xy \right) = e^{2x^2} x^3\)

The left side is the derivative of \(y \cdot IF\):

\(\frac{d}{dx} (y \cdot e^{2x^2}) = x^3 e^{2x^2}\)

Now, integrate both sides with respect to \(x\):

\(y \cdot e^{2x^2} = \int x^3 e^{2x^2} \, dx + C\)

To evaluate the integral \(\int x^3 e^{2x^2} \, dx\), we use substitution and integration by parts.

Let \(u = 2x^2\). Then \(du = 4x \, dx\), so \(x \, dx = \frac{1}{4} du\). Also, \(x^2 = \frac{u}{2}\).

The integral becomes \(\int x^2 \cdot x e^{2x^2} \, dx = \int \frac{u}{2} e^u \frac{1}{4} du = \frac{1}{8} \int u e^u \, du\).

Using integration by parts (\(\int v \, dw = vw - \int w \, dv\)) with \(v = u\) and \(dw = e^u \, du\), we get \(dv = du\) and \(w = e^u\).

\(\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u = e^u(u-1)\).

Substitute back \(u = 2x^2\): \(e^{2x^2}(2x^2 - 1)\).

So, \(\int x^3 e^{2x^2} \, dx = \frac{1}{8} e^{2x^2}(2x^2 - 1)\).

The general solution for the differential equation is:

\(y \cdot e^{2x^2} = \frac{1}{8} e^{2x^2}(2x^2 - 1) + C\)

\(y(x) = \frac{1}{8}(2x^2 - 1) + C e^{-2x^2}\)

This is the general solution involving the constant \(C\). We need to use the initial condition to find the specific solution.

Step 3: Use Initial Condition to Find Specific Solution

We are given the initial condition \(y(0) = 0\).

Substitute \(x = 0\) and \(y(0) = 0\) into the general solution:

\(y(0) = \frac{1}{8}(2(0)^2 - 1) + C e^{-2(0)^2}\)

\(0 = \frac{1}{8}(0 - 1) + C e^0\)

\(0 = -\frac{1}{8} + C \cdot 1\)

\(C = \frac{1}{8}\)

Substitute the value of \(C\) back into the general solution to get the specific solution of the differential equation:

\(y(x) = \frac{1}{8}(2x^2 - 1) + \frac{1}{8} e^{-2x^2}\)

Step 4: Evaluate the Limit as \(x \to 0\)

We need to find \(\mathop {\lim }\limits_{x \to 0} y(x)\) for the specific solution we found.

\(\mathop {\lim }\limits_{x \to 0} y(x) = \mathop {\lim }\limits_{x \to 0} \left( \frac{1}{8}(2x^2 - 1) + \frac{1}{8} e^{-2x^2} \right)\)

We can evaluate the limit of each term separately:

• \(\mathop {\lim }\limits_{x \to 0} \frac{1}{8}(2x^2 - 1) = \frac{1}{8}(2(0)^2 - 1) = \frac{1}{8}(0 - 1) = -\frac{1}{8}\)
• \(\mathop {\lim }\limits_{x \to 0} \frac{1}{8} e^{-2x^2} = \frac{1}{8} e^{-2(0)^2} = \frac{1}{8} e^0 = \frac{1}{8} \cdot 1 = \frac{1}{8}\)

Therefore, the limit is:

\(\mathop {\lim }\limits_{x \to 0} y(x) = -\frac{1}{8} + \frac{1}{8} = 0\)

The limit of the solution \(y(x)\) as \(x\) approaches 0 is 0.