Question

4.b.) solve \\( \frac { d y } { d x } + 3 x ^ { 2 } y = 6 x ^ { 2 } \\)

Explanation:

Step1: Identify the type of differential equation

This is a first - order linear ordinary differential equation of the form \(\frac{dy}{dx}+P(x)y = Q(x)\), where \(P(x)=3x^{2}\) and \(Q(x) = 6x^{2}\).

Step2: Find the integrating factor

The integrating factor \(\mu(x)=e^{\int P(x)dx}\).
Since \(\int3x^{2}dx=x^{3}\), then \(\mu(x)=e^{x^{3}}\).

Step3: Multiply the entire equation by the integrating factor

We get \(e^{x^{3}}\frac{dy}{dx}+3x^{2}e^{x^{3}}y = 6x^{2}e^{x^{3}}\).
The left - hand side is the derivative of the product \(y\mu(x)\) by the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (here \(u = y\) and \(v=e^{x^{3}}\), \(u^\prime=\frac{dy}{dx}\), \(v^\prime = 3x^{2}e^{x^{3}}\)). So \(\frac{d}{dx}(ye^{x^{3}})=6x^{2}e^{x^{3}}\).

Step4: Integrate both sides

\(\int\frac{d}{dx}(ye^{x^{3}})dx=\int6x^{2}e^{x^{3}}dx\).
Let \(t=x^{3}\), then \(dt = 3x^{2}dx\) and \(\int6x^{2}e^{x^{3}}dx = 2\int e^{t}dt=2e^{t}+C=2e^{x^{3}}+C\).
So \(ye^{x^{3}}=2e^{x^{3}}+C\).

Step5: Solve for \(y\)

Divide both sides by \(e^{x^{3}}\) (since \(e^{x^{3}}
eq0\) for all real \(x\)), we get \(y = 2 + Ce^{-x^{3}}\).

Answer:

\(y=2 + Ce^{-x^{3}}\)