second order linear constant coeﬀ

2.7.3.1 second order linear constant coeﬀ

\[\begin {aligned} y^{\prime \prime }+12 y&=7 y^{\prime } \end {aligned}\]

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

\[ A y''(x) + B y'(x) + C y(x) = 0 \]

Where in the above \(A=1, B=-7, C=12\). Let the solution be \(y=e^{\lambda x}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\lambda x}-7 \lambda \,{\mathrm e}^{\lambda x}+12 \,{\mathrm e}^{\lambda x} = 0 \tag {1} \]

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

\[ \lambda ^{2}-7 \lambda +12 = 0 \tag {2} \]

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 \(A=1, B=-7, C=12\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {7}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {-7^2 - (4) \left (1\right )\left (12\right )}\\ &= {\frac {7}{2}} \pm {\frac {1}{2}} \end{align*}

Hence

\begin{gather*} \begin {aligned} \lambda _1 &= {\frac {7}{2}} + {\frac {1}{2}}\\ \lambda _2 &= {\frac {7}{2}} - {\frac {1}{2}} \end {aligned} \end{gather*}

Which simpliﬁes to

\begin{gather*} \begin {aligned} \lambda _1 &= 4\\ \lambda _2 &= 3 \end {aligned} \end{gather*}

Since the roots are distinct, the solution is

\begin{gather*} \begin {aligned} y &= c_1 e^{\lambda _1 x} + c_2 e^{\lambda _2 x}\\ y &= c_1 e^{\left (4\right )x} +c_2 e^{\left (3\right )x} \end {aligned} \end{gather*}

\begin{align*} y &={\mathrm e}^{4 x} c_1 +c_2 \,{\mathrm e}^{3 x} \end{align*}

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