Linear Ode 6Ade89

1. **State the problem:** Solve the first-order linear differential equation $$y' - 3x^2 y = \frac{6 e^{x^3} (\ln x)^2}{x}, \quad x > 0.$$\n\n2. **Identify the type and formula:** This is a linear ODE of the form $$y' + P(x) y = Q(x)$$ where $$P(x) = -3x^2$$ and $$Q(x) = \frac{6 e^{x^3} (\ln x)^2}{x}.$$\n\nThe integrating factor (IF) is given by $$\mu(x) = e^{\int P(x) dx}.$$\n\n3. **Calculate the integrating factor:**\n$$\int P(x) dx = \int -3x^2 dx = -x^3 + C,$$\nso we take $$\mu(x) = e^{-x^3}.$$\n\n4. **Multiply the entire ODE by the integrating factor:**\n$$e^{-x^3} y' - 3x^2 e^{-x^3} y = e^{-x^3} \cdot \frac{6 e^{x^3} (\ln x)^2}{x}.$$\nSimplify the right side: $$e^{-x^3} \cdot e^{x^3} = 1,$$ so\n$$e^{-x^3} y' - 3x^2 e^{-x^3} y = \frac{6 (\ln x)^2}{x}.$$\n\n5. **Recognize the left side as a derivative:**\nThe left side is the derivative of $$y e^{-x^3}$$ because\n$$\frac{d}{dx} \left(y e^{-x^3}\right) = y' e^{-x^3} + y \cdot \frac{d}{dx} e^{-x^3} = y' e^{-x^3} - 3x^2 y e^{-x^3}.$$\n\n6. **Rewrite the equation:**\n$$\frac{d}{dx} \left(y e^{-x^3}\right) = \frac{6 (\ln x)^2}{x}.$$\n\n7. **Integrate both sides with respect to $$x$$:**\n$$y e^{-x^3} = \int \frac{6 (\ln x)^2}{x} dx + C.$$\n\n8. **Evaluate the integral:** Let $$u = \ln x,$$ then $$du = \frac{1}{x} dx,$$ so\n$$\int \frac{6 (\ln x)^2}{x} dx = 6 \int u^2 du = 6 \cdot \frac{u^3}{3} + C = 2 (\ln x)^3 + C.$$\n\n9. **Substitute back:**\n$$y e^{-x^3} = 2 (\ln x)^3 + C,$$\nso\n$$y = e^{x^3} \left(2 (\ln x)^3 + C\right).$$\n\n**Final answer:**\n$$\boxed{y = e^{x^3} \left(2 (\ln x)^3 + C\right)}.$$