Answer:
[tex]y=x^2+x+3[/tex]
Step-by-step explanation:
Solve the given initial-value problem.
[tex]\dfrac{dy}{dx} =2x+1; \ y(0)=3[/tex]
This is a separable differential equation. We can solve these as follows.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Separable Differential Equation:}}\\\frac{dy}{dx} =f(x)g(y)\\\\\rightarrow\int\frac{dy}{g(y)}=\int f(x)dx \end{array}\right }[/tex]
[tex]\dfrac{dy}{dx} =2x+1\\\\\Longrightarrow dy=(2x+1)dx\\\\\Longrightarrow \int dy=\int(2x+1)dx\\\\\Longrightarrow \boxed{y=x^2+x+C}[/tex]
Use the initial condition to find the arbitrary constant "C."
[tex]y=x^2+x+C; \ \text{Recall} \ y(0)=3\\\\\Longrightarrow3=(0)^2+0+C\\\\\Longrightarrow \boxed{C=3}\\\\\therefore \boxed{\boxed{y=x^2+x+3}}[/tex]
Thus, the problem is solved.
