Hi,

If i have two conjugate roots as solution to 4th order ode, such that the two roots are proportional, how to find the four independent solutions of the equation?

Thanks

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- June 22nd 2013, 03:59 PMmopen80basis of 4th order ode
Hi,

If i have two conjugate roots as solution to 4th order ode, such that the two roots are proportional, how to find the four independent solutions of the equation?

Thanks - June 22nd 2013, 07:19 PMHallsofIvyRe: basis of 4th order ode
I'm not sure what you mean by "two conjugate roots as solution to 4th order ode". Perhaps you mean two conjugate roots of the

**characteristic equation**of the ode? But then I don't know what you mean by "the two roots are proportional". The characteristic equation of a 4th order ode is a fourth degree polynomial equation which generally has**four**roots. Just knowing two of the roots is not enough to tell you all four independent solutions. Or do you mean that two solutions are conjugates and there exist a second pair that are proportional to the first? - June 22nd 2013, 07:28 PMmopen80Re: basis of 4th order ode
- July 11th 2013, 01:45 PMHallsofIvyRe: basis of 4th order ode
So two solutions are of the form a+ bi and a- bi and the other two are ak+ bki and ak- bki for some constant k?

If two solutions to the characteristic equation are a+ bi and a- bi then two independent solutions are and . The other solutions to the characteristic, ak+ bki and ak- bki, give and .

(The fact that the second pair of solutions to the characteristic equation are proportional to the first pair is not really of importance. If one pair of solutions is , and the other is l and then the four independent solutions to the differential equation are , , , and . - July 11th 2013, 04:59 PMmopen80Re: basis of 4th order ode