Initial Value 211244

1. **State the problem:** We need to solve the initial value problem given by the differential equation $$\frac{dy}{dx} = x^2 (y - 2)$$ with the initial condition $$y(0) = 5$$. 2. **Identify the type of differential equation:** This is a first-order separable differential equation because the right side can be written as a product of a function of $$x$$ and a function of $$y$$. 3. **Separate variables:** Rewrite the equation as: $$\frac{dy}{y - 2} = x^2 \, dx$$ 4. **Integrate both sides:** Integrate the left side with respect to $$y$$ and the right side with respect to $$x$$: $$\int \frac{1}{y - 2} \, dy = \int x^2 \, dx$$ 5. **Perform the integrations:** $$\int \frac{1}{y - 2} \, dy = \ln|y - 2| + C_1$$ $$\int x^2 \, dx = \frac{x^3}{3} + C_2$$ 6. **Combine constants:** Let $$C = C_2 - C_1$$, then: $$\ln|y - 2| = \frac{x^3}{3} + C$$ 7. **Solve for $$y$$:** Exponentiate both sides: $$|y - 2| = e^{\frac{x^3}{3} + C} = e^C e^{\frac{x^3}{3}}$$ Let $$A = e^C > 0$$, so: $$y - 2 = \pm A e^{\frac{x^3}{3}}$$ 8. **Use initial condition to find $$A$$:** At $$x=0$$, $$y=5$$: $$5 - 2 = \pm A e^0 = \pm A$$ So: $$3 = \pm A$$ Since $$A > 0$$, we take $$A = 3$$. 9. **Write the final solution:** $$y - 2 = 3 e^{\frac{x^3}{3}}$$ Or: $$\boxed{y = 2 + 3 e^{\frac{x^3}{3}}}$$