solve the initial - value problem $\frac{dy}{dx}=x^{2},y(1)=3$.

Explanation:

Step1: Integrate both sides

Integrate $\frac{dy}{dx}=x^{2}$ with respect to $x$. We know that $\int dy=\int x^{2}dx$. $y=\int x^{2}dx$

Step2: Calculate the integral

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n = 2$, we have $\int x^{2}dx=\frac{x^{3}}{3}+C$. So $y=\frac{x^{3}}{3}+C$.

Step3: Find the value of C

Substitute the initial condition $x = 1$ and $y=3$ into $y=\frac{x^{3}}{3}+C$. $3=\frac{1^{3}}{3}+C$ $3=\frac{1}{3}+C$ $C=3-\frac{1}{3}=\frac{9 - 1}{3}=\frac{8}{3}$

Step4: Write the final solution

Substitute $C=\frac{8}{3}$ back into $y=\frac{x^{3}}{3}+C$. $y=\frac{x^{3}}{3}+\frac{8}{3}$

Answer:

$y=\frac{x^{3}}{3}+\frac{8}{3}$