# Thread: basis of 4th order ode

1. ## basis 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

2. ## Re: 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?

3. ## Re: basis of 4th order ode

Sorry for inaccuracy, yes i mean that "two solutions are conjugates and there exist a second pair that are proportional to the first"; how to find the four independent solutions?
Originally Posted by HallsofIvy
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?

4. ## Re: 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 $\displaystyle e^{ax}cos(bx)$ and $\displaystyle ^{ax}sin(bx)$. The other solutions to the characteristic, ak+ bki and ak- bki, give $\displaystyle e^{akx}cos(bkx)$ and $\displaystyle e^{akx}sin(bkx)$.

(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 $\displaystyle a_1+b_1i$, $\displaystyle a_1- b_1i$ and the other is l$\displaystyle a_2+ b_2i$ and $\displaystyle a_2- b_2i$ then the four independent solutions to the differential equation are $\displaystyle e^{a_1x}cos(b_1x)$, $\displaystyle e^{a_1x}sin(b_1x)$, $\displaystyle e^{a_2x}cos(b_2x)$, and $\displaystyle e^{a_2x}sin(b_2x)$.

5. ## Re: basis of 4th order ode

Originally Posted by HallsofIvy
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 $\displaystyle e^{ax}cos(bx)$ and $\displaystyle ^{ax}sin(bx)$. The other solutions to the characteristic, ak+ bki and ak- bki, give $\displaystyle e^{akx}cos(bkx)$ and $\displaystyle e^{akx}sin(bkx)$.

(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 $\displaystyle a_1+b_1i$, $\displaystyle a_1- b_1i$ and the other is l$\displaystyle a_2+ b_2i$ and $\displaystyle a_2- b_2i$ then the four independent solutions to the differential equation are $\displaystyle e^{a_1x}cos(b_1x)$, $\displaystyle e^{a_1x}sin(b_1x)$, $\displaystyle e^{a_2x}cos(b_2x)$, and $\displaystyle e^{a_2x}sin(b_2x)$.
Ok, but when i calculated the the Wronskian of the 4 solutions i found it equals to zero. This means that they are not linearly independent! in this case how to find the 4 linearly independent solutions?

Thanks