Solving a Differential Equation

Solve the differential equation \(\dfrac{dy}{dx}=x^{2}+2x+1\) with the boundary condition when \(x=3,\) then \(y={-}1.\)

Solution We begin by finding the general solution of the differential equation, namely \[ y=\frac{x^{3}}{3}+x^{2}+x+C \]

To determine the number \(C\), we use the boundary condition when \(x=3,\) then \(y={-}1\). \[ \begin{array}{rl@{\qquad}l} -1 &=\dfrac{3^{3}}{3}+3^{2}+3+C & \color{#0066A7}{{x=3,\quad y=-1}} \\[8pt] C&=-22 \end{array} \]

The particular solution of the differential equation with the boundary condition when \(x=3,\) then \(y=-1\), is \[ y=\frac{x^{3}}{3}+x^{2}+x-22 \]