# 2nd order homogenous constant coefficient ODE question

• quasar987
In summary, the conversation discusses solving a second order homogeneous ODE by supposing a solution of exponential type and using Euler's formula to convert it to a second order polynomial equation. The two forms of the solution, c1e^{wx}+c2e^{\overline{w}x} and e^x(Acos(bx)+Bsin(bx)), are equivalent but may lead to different results when solving an initial value problem. This can be due to working with complex numbers or the assumption that c1 and c2 are real.
quasar987
Homework Helper
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If we have a constant coefficient second order homogeneous ODE, the way to solve this is to suppose a solution of the exponential type. This yields a second order polynomial equation (the "characteristic equation") that the exponent must satisfy. In case the solutions of the characteristic equation are complex, we get a solution of the form $c1e^{wx}+c2e^{\overline{w}x}$ (call this form 1). The general solution of the ODE is usually presented in the form e^x(Acos(bx)+Bsin(bx)) (call this form 2) by using Euler's formula for e^{a+ib}.

To my eyes, forms 1 and 2 are equivalent. So then, how come trying to solve an initial value problem with form 1 does not work?!?

Consider for instance y''-2y+2=0 with y(0)=1, y'(0)=0. The roots of the char. equ. are 1+i and 1-i. If we try to find c1 and c2 using the initial values given we are lead to the impossible system c1 + c2 = 1, c1 + c2=0, c1 - c2=0. With form 2 however, we find A=1, B=-1.

Thanks, I'm confused!

quasar987 said:
If we have a constant coefficient second order homogeneous ODE, the way to solve this is to suppose a solution of the exponential type. This yields a second order polynomial equation (the "characteristic equation") that the exponent must satisfy. In case the solutions of the characteristic equation are complex, we get a solution of the form $c1e^{wx}+c2e^{\overline{w}x}$ (call this form 1). The general solution of the ODE is usually presented in the form e^x(Acos(bx)+Bsin(bx)) (call this form 2) by using Euler's formula for e^{a+ib}.

To my eyes, forms 1 and 2 are equivalent. So then, how come trying to solve an initial value problem with form 1 does not work?!?

Consider for instance y''-2y+2=0 with y(0)=1, y'(0)=0.
Surely you meant y'' - 2y' + 2y = 0, based on the characteristic equation you have.
quasar987 said:
The roots of the char. equ. are 1+i and 1-i. If we try to find c1 and c2 using the initial values given we are lead to the impossible system c1 + c2 = 1, c1 + c2=0, c1 - c2=0. With form 2 however, we find A=1, B=-1.

Thanks, I'm confused!

If you checked your solution against what you wrote -- y'' - 2y + 2 = 0 -- that's a problem right there.

Either pair of functions e(1 + i)x, e(1 - i)x and excos(x), exsin(x), is a basis for the solution space of your differential equation. Due to your typo, you might have been trying to calculate the constants that go with your initial conditions using the wrong DE.

So there are three things we can do.
They are all equivalent.
1)Work freely with complex numbers
2)Use complex numbers but convert back to real numbers at the end.
3)Work only with real numbers.

As far as your example your equations seem to result from a false assumption that c1 and c2 are real.
The initial conditions give the system (or some variation of it)
c1+c2=0
c1-c2=i

which is easily solved

quasar987 said:
If we try to find c1 and c2 using the initial values given we are lead to the impossible system c1 + c2 = 1, c1 + c2=0, c1 - c2=0. With form 2 however, we find A=1, B=-1.

Thanks, I'm confused!

Regardless of what equation you were trying to solve, I'm confused about how you got 3 equations from 2 initial conditions.

So sorry, the correct DE is y'' - 2y' + 2y = 0.

But anyhow, I see what I did wrong there. My mistake was indeed that I assumed for a second that c1 and c2 are real! Thanks for the help.

## 1. What is a second order homogenous constant coefficient ODE?

A second order homogenous constant coefficient ODE (ordinary differential equation) is a mathematical equation that involves a function, its first derivative, and its second derivative, where all the coefficients are constant. The term "homogenous" means that all the terms in the equation have the same degree, and the term "constant coefficient" means that the coefficients do not depend on the independent variable.

## 2. What is the general form of a second order homogenous constant coefficient ODE?

The general form of a second order homogenous constant coefficient ODE is: a*y'' + b*y' + c*y = 0, where a, b, and c are constants and y is the dependent variable.

## 3. How do you solve a second order homogenous constant coefficient ODE?

To solve a second order homogenous constant coefficient ODE, you can use the characteristic equation method. This involves finding the roots of the characteristic equation a*r^2 + b*r + c = 0, where r is the independent variable. The roots of this equation will determine the form of the solution, which can be either real or complex.

## 4. What are the boundary conditions for a second order homogenous constant coefficient ODE?

The boundary conditions for a second order homogenous constant coefficient ODE are the initial values or values at specific points that must be provided in order to find a unique solution. These conditions can be specified as y(x0) = y0 and y'(x0) = y'0, where x0 is the initial value of the independent variable and y0 and y'0 are the corresponding initial values of the function and its derivative.

## 5. What are the applications of second order homogenous constant coefficient ODEs?

Second order homogenous constant coefficient ODEs are widely used in physics, engineering, and other fields to model various physical phenomena such as motion, oscillations, and vibrations. They are also used in control systems, signal processing, and other areas of mathematics and science.

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