Homogeneous linear 2nd-order ODEs with constant coefficients

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 the quadratic equation has two different solutions, say #\lambda_1# and #\lambda_2#, then the corresponding solutions of the ODE, #y_1=\e^{\lambda_1\cdot t}# and #y_2=\e^{\lambda_2\cdot t}#, are linearly independent. In particular, in view of Uniqueness of specific solutions of linear second-order differential equations, the general solution of the ODE is then #y = A\cdot y_1+B\cdot y_2#.

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, \(a\cdot\frac{\dd^2y}{\dd t^2}+b\cdot\frac{\dd y}{\dd t}=0\).

Attempting a solution of a special form is called the Ansatz method. Later we will come back to it.

The formulation of the definition makes it clear that, if #\lambda# is a solution of the characteristic equation, then #y=\e^{\lambda\cdot t}# is a solution of the ODE.

Suppose that #\lambda# is the only solution of the characteristic equation. This means that #\lambda=\frac{-b}{2a}#, and that the discriminant \(D=b^2-4a\cdot c\) is equal to #0#. We show that also #y = t\cdot \e^{\lambda\cdot t}# a solution of the ODE. To this end, we first calculate the first and second derivative of #y#: \[\frac{\dd y}{\dd t}=(1+\lambda \cdot t)\cdot\e^{\lambda\cdot t}\qquad\text{and}\qquad \frac{\dd^2y}{\dd t^2}=(\lambda\cdot t+2)\cdot\lambda \cdot\e^{\lambda \cdot t}\] We now prove #y# is a solution by filling the function rules for #y#, #\frac{\dd y}{\dd t}#, and #\frac{\dd^2y}{\dd t^2}# into the ODE: \[\begin{array}{rcl}\displaystyle a\cdot\frac{\dd^2y}{\dd t^2}+b\cdot\frac{\dd y}{\dd t}+c\cdot y(t) &=& a \cdot(\lambda\cdot t+2)\cdot\lambda \cdot\e^{\lambda \cdot t}+b\cdot (1+\lambda\cdot t)\cdot \e^{\lambda \cdot t}+c \cdot t\cdot\e^{\lambda \cdot t}\\ &&\phantom{x}\color{blue}{\frac{\dd y}{\dd t}=(1+\lambda \cdot t)\cdot\e^{\lambda t}\qquad\text{and}\qquad \frac{\dd^2y}{\dd t^2}=(\lambda t+2)\cdot\lambda \cdot\e^{\lambda \cdot t}} \\ &=& \bigl((a\cdot\lambda^2+b\cdot\lambda+c)\cdot t+(2a\cdot\lambda+b)\bigr) \cdot \e^{\lambda \cdot t}\\ &=& -\dfrac{b^2-4a\cdot c}{4a} \cdot t\cdot \e^{\lambda\cdot t}\\ &&\phantom{xx}\color{blue}{\lambda=\frac{-b}{2a}}\\ &=& 0\\ &&\phantom{xx}\color{blue}{b^2-4a\cdot c=D=0}\end{array}\]

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.

Show that \(y_1=\e^{-2 t}\) and \(y_2=\e^{-t}\) are solutions of the ODE \[\frac{\dd^2y}{\dd t^2}+3\frac{\dd y}{\dd t }+2 y(t)=0\]

Repeatedly differentiating #y_1# gives \[y_1=\e^{-2 t},\qquad \frac{\dd y_1}{\dd t}= -2 \e^{-2 t},\qquad \frac{\dd^2y_1}{\dd t^2}=4 \e^{-2 t}\] So \[\frac{\dd^2y_1}{\dd t^2}+3\frac{\dd y_1}{\dd t}+2 y_1(t)= 4 \e^{-2 t} + 3\cdot \left(-2 \e^{-2 t} \right)+ 2 \e^{-2 t} = \e^{-2 t}\cdot(4-6+2)=0\]
Similarly, repeatedly differentiating #y_2# \[y_2=\e^{-t},\qquad \frac{\dd y_2}{\dd t}= - \e^{-t},\qquad \frac{\dd^2y_2}{\dd t^2}=\e^{-t}\] So \[\frac{\dd^2y_2}{\dd t^2}+3\frac{\dd y_2}{\dd t}+2 y_2(t)= \e^{-t} + 3\cdot \left(- \e^{-t} \right)+ 2 \e^{-t} = \e^{-t}\cdot (1-3+2)=0\]

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

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