2.2.8.2 second order linear constant coeff

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

Entering second order linear constant coefficient ode solverThis is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A x''(t) + B x'(t) + C x(t) = 0 \]

Where in the above \(A=1, B=-3, C=2\). Let the solution be \(x=e^{\lambda t}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\lambda t}-3 \lambda \,{\mathrm e}^{\lambda t}+2 \,{\mathrm e}^{\lambda t} = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda t}\) gives

\[ \lambda ^{2}-3 \lambda +2 = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form. Using the quadratic formula the roots are

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=-3, C=2\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {3}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {-3^2 - (4) \left (1\right )\left (2\right )}\\ &= {\frac {3}{2}} \pm {\frac {1}{2}} \end{align*}

Hence

\begin{gather*} \begin {aligned} \lambda _1 &= {\frac {3}{2}} + {\frac {1}{2}}\\ \lambda _2 &= {\frac {3}{2}} - {\frac {1}{2}} \end {aligned} \end{gather*}

Which simplifies to

\begin{gather*} \begin {aligned} \lambda _1 &= 2\\ \lambda _2 &= 1 \end {aligned} \end{gather*}

Since the roots are distinct, the solution is

\begin{gather*} \begin {aligned} x &= c_1 e^{\lambda _1 t} + c_2 e^{\lambda _2 t}\\ x &= c_1 e^{\left (2\right )t} +c_2 e^{\left (1\right )t} \end {aligned} \end{gather*}

Or

\begin{align*} x &=c_1 \,{\mathrm e}^{2 t}+c_2 \,{\mathrm e}^{t} \end{align*}

Applying the initial conditions \([x \left (0\right ) = 0, x^{\prime }\left (0\right ) = 1]\), the solution becomes

\begin{align*} x &= {\mathrm e}^{t} \left ({\mathrm e}^{t}-1\right ) \\ \end{align*}

\(x = {\mathrm e}^{t} \left ({\mathrm e}^{t}-1\right )\)	Direction field