Solving homogeneous linear ODEs with constant coefficients

Differential equations: Linear second-order differential equations

Solving homogeneous linear ODEs with constant coefficients

Homogeneous linear differential equations of second order with constant coefficients can be solved fully.

General solution of a homogeneous linear second-order ODE with constant coefficients

Consider the ODE \[a\cdot\frac{\dd^2y}{\dd t^2}+b\cdot\frac{\dd y}{\dd t}+c\cdot y(t)=0\] where #a#, #b#, and #c# are constants with #a\ne0#. The corresponding characteristic equation is \[a\cdot\lambda^2+b\cdot\lambda+c=0\] The discriminant of this quadratic equation in #\lambda# is \(D=b^2-4a\cdot c\). We distinguish three cases according to the sign of #D# and give the general solution with integration constants #A# and #B# in each case:

If #D# is positive, then the characteristic equation of the ABC-formula two real solutions, say \[\lambda_1=\frac{-b+\sqrt{D}}{2a}\qquad\text{and}\qquad \lambda_2=\frac{-b-\sqrt{D}}{2a}\] The general solution of the differential equation is then \[ y(t)=A\cdot\e^{\lambda_1\cdot t}+B\cdot\e^{\lambda_2\cdot t}\]
If \(D\) is equal to zero, then, in accordance with the abc-formula, the characteristic equation has exactly one solution, namely \[\lambda=\frac{-b}{2a}\] The general solution is then \[y(t)=(A+B\cdot t)\cdot\e^{\lambda\cdot t}\]
If \(D\) is negative, then the characteristic equation has no real solutions, but two complex conjugate solutions #\alpha\pm\beta\, \ii#, where \[\alpha=-\frac{b}{2a}\text{ and } \beta=\frac{\sqrt{-D}}{2a}\] The general solution of the differential equation is then \[y(t)=\e^{\alpha\cdot t}\cdot\bigl(A\cdot\cos(\beta \cdot t)+B\cdot\sin(\beta \cdot t)\bigr)\]

It follows from the Uniqueness of specific solutions of linear second-order differential equations that the general solution is a linear combination #y(t)=A\cdot y_1(t)+B\cdot y_2(t)# of the two basic solutions #y_1# and #y_2#, where #A# and #B# are real numbers (the integration constants). In each of the three cases for the sign of the discriminant \(D=b^2-4ac\), it is easy to verify that the solutions #y_1# and #y_2# indicated in the table below satisfy the ODE.

discriminant	\(y_1\)	\(y_2\)
\(D\gt0\)	\(\e^{\lambda_1\cdot t}\)	\( \e^{\lambda_2\cdot t}\)
\(D=0\)	\(\e^{\lambda\cdot t}\)	\( t \cdot\e^{\lambda t}\)
\(D\lt0\)	\( \e^{\alpha t}\cdot \cos(\beta t)\)	\( \e^{\alpha t}\cdot\sin(\beta t)\)

For cases #D\gt 0# and #D=0#, we proved this earlier. Below we provide a proof for the case #D\lt0#. We treat only #y_2(t) =\e^{\alpha t}\cdot\sin(\beta t)#; the proof for #y_1(t)# is very similar. We start with the calculation of the first and second derivative of #y_2#:

\[\begin{array}{rcl}\dfrac{\dd y_2}{\dd t}&=&\left(\alpha\cdot\sin(\beta t)+\beta\cdot\cos(\beta t)\right)\cdot\e^{\alpha t}\\ &&\text{en}\\ \dfrac{\dd^2y_2}{\dd t^2}&=&\left(2\alpha\beta\cdot\cos(\beta t)+(\alpha^2-\beta^2)\sin(\beta t)\right)\cdot\e^{\alpha t}\end{array} \] This allows us to evaluate the left side of the ODE: \[\begin{array}{rcl}\displaystyle a\frac{\dd^2y_2}{\dd t^2}+b\cdot\frac{\dd y_2}{\dd t}+c\cdot y_2(t) &=& a\left(2\alpha\beta\cdot\cos(\beta t)+(\alpha^2-\beta^2)\sin(\beta t)\right)\cdot\e^{\alpha t}\\ &&+b\cdot(\alpha\cdot\sin(\beta t)+\beta\cdot\cos(\beta t) )\e^{\alpha t}\\ &&+c \cdot \sin(\beta t)\cdot\e^{\alpha t}\\ &&\phantom{x}\color{blue}{\text{function rules of }\frac{\dd y_2}{\dd t}\text{ and }\frac{\dd^2y_2}{\dd t^2}\text{ filled in}} \\ &=& \left(2a\alpha+b\right) \cdot\beta\cos(\beta t)\cdot \e^{\alpha \cdot t}\\ &&+\left(a(\alpha^2-\beta^2)+b\alpha+c\right) \cdot\sin(\beta t) \cdot \e^{\alpha \cdot t}\\ &&\phantom{xx}\color{blue}{\text{terms with }\cos\text{ and }\sin\text{ collected}}\\ &=&\left(\dfrac{b^2}{2a}+b\alpha\right) \cdot\sin(\beta t) \cdot \e^{\alpha \cdot t}\\ &&\phantom{xx}\color{blue}{\alpha=\frac{-b}{2a}\text{ en }a(\alpha^2-\beta^2)=\frac{b^2+D}{4a}=\frac{b^2}{2a}-c}\\&=&0\\ &&\phantom{xx}\color{blue}{\alpha=\frac{-b}{2a}}\end{array}\]

If #D\lt0#, then, according to the ABC-formula, there are two imaginary zeros, say \[\lambda_1=\frac{-b+\ii\,\sqrt{-D}}{2a}\qquad\text{and}\qquad \lambda_2=\frac{-b-\ii\,\sqrt{-D}}{2a}\] The following two functions are solutions of the differential equation \[y_1(t)=\e^{\lambda_1\cdot t}\qquad\text{and}\qquad y_2(t)=\e^{\lambda_2\cdot t}\] In fact, any solution can be written as a linear combination of these two "basic solutions". In other words, the general solution of the differential equation can be written as \[y(t)=c_1\cdot y_1(t)+c_2\cdot y_2(t)=c_1\cdot\e^{\lambda_1t}+c_2\cdot\e^{\lambda_2\cdot t}\] for certain constants \(c_1\) and \(c_2\). These constants can be determined by boundary conditions.

You might feel a little easier if we do not work with complex solutions, but only with real numbers and functions. This can be arranged by working with special combinations of the complex solutions. Suppose that the standard form of the complex number \(\lambda_1\) is equal to \(\alpha+{\ii}\,\beta\). Then the complex number \(\lambda_2\) is equal to \(\alpha-{\ii}\,\beta\) and \[\alpha=-\frac{b}{2a}\qquad\text{and}\qquad \beta=\frac{\sqrt{4a\cdot c-b^2}}{2a}\] The two complex solutions are complex conjugates of each other. Then: \[\begin{aligned}\frac{\e^{\lambda_1t} +\e^{\lambda_2t}}{2} &= \frac{\e^{(\alpha+{\ii}\,\beta)t} +\e^{(\alpha-{\ii}\,\beta)t}}{2}\\ \\ &= \e^{\alpha t}\cdot \frac{\e^{\ii\,\beta\, t} +\e^{-{\ii}\,\beta\, t}}{2}\\ \\ &= \e^{\alpha t}\cdot \cos(\beta t)\end{aligned}\] This is the first real solution of the differential equation. Another combination gives a second real solution \[ \begin{aligned}\frac{\e^{\lambda_1t} -\e^{\lambda_2t}}{2{\ii}} &= \frac{\e^{(\alpha+{\ii}\,\beta)t} -\e^{(\alpha-{\ii}\,\beta)t}}{2{\ii}}\\ \\ &= \e^{\alpha t}\cdot \frac{\e^{{\ii}\,\beta\, t} +\e^{-{\ii}\,\beta\, t}}{2{\ii}}\\ \\ &= \e^{\alpha t}\cdot \sin(\beta t)\end{aligned}\] A linear combination of the two real solutions, with integration constants \(A\) and \(B\), makes up the general solution: \[y(t)=\e^{\alpha t}\cdot \bigl(A\cos(\beta t)+B\sin(\beta t)\bigr)\]

Solve the following differential equation: \[\frac{\dd^2y}{\dd t^2}+5\frac{\dd y}{\dd t}+6\, y(t)=0\]
Use #c# and #d# as constants.

#y(t)=# #c\cdot\e^{-3t}+d\cdot\e^{-2t}#

The characteristic equation of the given homogeneous linear differential equation of second order is \[ \lambda^2+5\lambda+6=0\]
The discriminant \(D\) of the quadratic polynomial is equal to \[D = 5 ^ 2-4 \cdot 6 = 1 = 1 ^ 2 \] Since #D\gt0#, the characteristic equation has two real solutions:\[\lambda_{1,2}= \frac{-5\pm 1}{2}\] that is, \[\lambda_1=-3 \qquad\text{and}\qquad\lambda_2=-2\]

According to theorem General solution of the homogeneous linear second-order ODE with constant coefficients, the differential equation has the following general solution: \[y(t)=c\cdot\e^{-3t}+d\cdot\e^{-2t}\] with constants \(c\) and \(c\).

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