### Differential equations: Linear second-order differential equations

### Homogeneous linear 2nd-order ODEs with constant coefficients

Before, we saw that any homogeneous linear second-order ODE has a general solution with two free parameters. That does not mean that the solution is easy to write down. Think, for instance, of #y''=\e^{-t^2}#. The fact that the right-hand side has no antiderivative that can be written in terms of known standard functions, we see that there is no simple way to describe the solution #y# of the ODE explicitly. After all, if there were, then is derived function would be simple to express (thanks to the calculation rules for differentiation), but such a derivative would be an antiderivative of #\e^{-t^2}#.

In the special case of a homogeneous linear equation with constant coefficients, the solution can be found. Here we introduce the notions necessary for formulating the solution. *Later*, we will give the full solution.

A differential equation of the form \[a\cdot\frac{\dd^2y}{\dd t^2}+b\cdot\frac{\dd y}{\dd t}+c\cdot y=0\] where \(a\), \(b\), and \(c\) are arbitrary real constants with \(a\neq 0\), is called a **linear homogeneous differential equation of the second order with constant coefficients.**

If #a=0# and #b\ne0#, then the equation is linear of first order. At the *Solution of the exponential growth equation* we have seen that then the general solution is #y=C\cdot \e^{-\frac{c}{b}\cdot t}#.

If #a\ne0#, then we can divide all terms of the ODE by #a#. The result then is in **standard form**, where the coefficient of #\frac{\dd^2y}{\dd t^2}# is equal to #1#.

If #c=0# and #b\ne0#, then the equation can be solved by first solving the \(a\cdot\frac{\dd u}{\dd t}+b\cdot u=0\) by use of *Solution of the exponantial growth equation*: \( u = C\cdot \e^{-\frac{b}{a}\cdot t}\), and next solving #\frac{\dd y}{\dd t}=u#. The general solution then is #y =A\e^{-\frac{b}{a}\cdot t}+B#, where #A# and #B# are integration constants zijn.

The general solution of the differential equation of the form \[a\cdot\frac{\dd^2y}{\dd t^2}+b\cdot\frac{\dd y}{\dd t}+c\cdot y=0\] can be found by trying powers of \(\e\) as solutions.

Characteristic equation

Suppose that \[y(t)=\e^{\lambda t}\] is a solution of the linear second-order ODE \[a\cdot y''+b\cdot y'+c\cdot y=0\] where \(a\), \(b\), and \(c\) are given real constants with \(a\neq 0\). Then, substituting the function rule for #y(t)# in the ODE gives the equation \[a\cdot\lambda^2\cdot\e^{\lambda t}+b\cdot\lambda \cdot\e^{\lambda t}+c \cdot\e^{\lambda t}=0\]

and, after dividing by \(\e^{\lambda t}\) , \[a\cdot\lambda^2+b\cdot\lambda + c=0\] This equation is called the **characteristic equation** of the differential equation. The left-hand side is also called the **characteristic polynomial** of the ODE.

Every solution of this quadratic equation gives a solution \( y = \e^{\lambda t}\) of the differential equation. If #\lambda# is a double root of the characteristic equation, then \( t\cdot\e^{\lambda\cdot t}\) is also a solution of the ODE.

If #a=0# and #b\ne0#, then the characteristic equation is linear: \(b\cdot\lambda + c=0\), and there is always a solution: \(\lambda=-\frac{c}{b}\). The corresponding solution #y(t)=\e^{\lambda\cdot t}# is then a special case of the *solution of the exponential growth-equation*.

If #a\ne0# and #c=0#, then #\lambda=0# is a solution, which corresponds to the constant solution #y=1# of the ODE, \(a\cdot\frac{\dd^2y}{\dd t^2}+b\cdot\frac{\dd y}{\dd t}=0\).

The nature of the solutions of a linear homogeneous differential equation of second order with constant coefficients is determined by the sign of the *discriminant* \(D=b^2-4a\cdot c\). If \(D\gt0\), then there are two real solutions. If \(D=0\), then we have a unique solution. Finally, if \(D\lt0\), then there are no real solutions, but two complex solutions. *Later* we will look into the solutions in each of these three cases.

Similarly, repeatedly differentiating #y_2# \[y_2=\e^{-2 t},\qquad \frac{\dd y_2}{\dd t}= -2 \e^{-2 t},\qquad \frac{\dd^2y_2}{\dd t^2}=4\e^{-2 t}\] So \[\frac{\dd^2y_2}{\dd t^2}+\frac{\dd y_2}{\dd t}-2 y_2(t)= 4 \e^{-2 t} + \left(-2 \e^{-2 t} \right)-2 \e^{-2 t} = \e^{-2 t}\cdot (4-2-2)=0\]

*Uniqueness of specific solutions of linear second-order differential equations*, the general solution of the given GDV is #y = A\cdot\e^{t}+B\cdot\e^{-2 t}#, where #A# and #B# are integration constants.

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