2.1.7.2 Solved using first_order_ode_autonomous

Entering first order ode autonomous solver

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

Since the ode has the form \(x^{\prime }=f(x)\) and initial conditions \(\left (t_0,x_0\right ) \) are given such that they satisfy the ode itself, then we can write

\begin{align*} 0 &= \left . f(x) \right |_{x=x_0}\\ 0 &= 0 \end{align*}

And the solution is immediately written as

\begin{align*}x&=x_0\\ x&=1 \end{align*}

Singular solutions are found by solving

\begin{align*} x^{2}-3 x +2&= 0 \end{align*}

for \(x\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} x&=1 \end{align*}

Direction field with Isoclines	\( x = 1 \)

Summary of solutions found

\[ x = 1 \]