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 $y_1=\e^{t}$ and $y_2=\e^{2 t}$ 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 $y_1=\e^{t},\qquad \frac{\dd y_1}{\dd t}= \e^{t},\qquad \frac{\dd^2y_1}{\dd t^2}= \e^{t}$ So $\frac{\dd^2y_1}{\dd t^2}-3\frac{\dd y_1}{\dd t}+2 y_1(t)= \e^{t} -3\cdot \e^{t}+ 2 \e^{t} = \e^{t}\cdot(1-3+2)=0$
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 $\frac{\dd^2y_2}{\dd t^2}-3\frac{\dd y_2}{\dd t}+2 y_2(t)= 4 \e^{2 t} -3\cdot 2\e^{2 t}+ 2 \e^{2 t} = \e^{2 t}\cdot (4-6+2)=0$

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

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