2.1.1.2 Solved using first_order_ode_quadrature

Entering first order ode quadrature solver

\[\begin {aligned} x^{\prime }&=3 t^{2}+4 t\\ x \left (1\right ) &= 0\\ \end {aligned}\]

Because the ODE has the form \(x^{\prime }=f(t)\), the solution requires only integration. Therefore

\begin{align*} dx &= \left (3 t^{2}+4 t\right ) \, dt\\ x &= \int { \left (3 t^{2}+4 t\right ) \, dt}\\ &= t^{3}+2 t^{2}+c_1 \end{align*}

Applying the initial condition \(x \left (1\right ) = 0\), the solution becomes

\begin{align*} x &= t^{3}+2 t^{2}-3 \\ \end{align*}

Direction field with Isoclines	\( x = t^{3}+2 t^{2}-3 \)

Summary of solutions found

\[ x = t^{3}+2 t^{2}-3 \]