# Third order linear ODE Wolfram Math. problem

1. Nov 8, 2009

### dperkovic

I have a problem with this homogenous ODE:
$$\frac{d^3x}{dt^3} + 2\frac{dx}{dt} = 0$$

With substitution $$x=e^\lambda t$$, I got characteristic polinom: $$\lambda^3+2\lambda=0$$.
So,
$$\lambda_1 = 0$$, $$\lambda_{1,2} = \pm\sqrt{2}i$$

Because we have one real,and one pair of complex roots of characteristic polinom, general solution of ODE is:

$$x(t) =C1e^{0\cdot t} + C2\cos(\sqrt{2}t) +C3\sin(\sqrt{2}t)$$,
or:
$$x(t) =C1 + C2\cos(\sqrt{2}t) +C3\sin(\sqrt{2}t)$$

But, Wolfram Mathematica's command DSolve[x'''[t] + 2*x'[t] == 0, x[t], t],give to me this:

$$x(t) =C1 - \frac{C2\cos(\sqrt{2}t)}{\sqrt{2}} +\frac{C3\sin(\sqrt{2}t)}{\sqrt{2}}$$.

As we can see, there is difference between my and wolfram's solution. Why ?!?! Where is a mistake in my procedure ?

Last edited: Nov 8, 2009
2. Nov 8, 2009

### Staff: Mentor

There is no difference in the solutions, remember the constants of integration are arbitrary.

3. Nov 8, 2009

### dperkovic

Thank's. I thought that. But still, I can't figure out why constants in Mathematica's solution is different than constants i my solution. From which algorithm comming this "complicated" constants ?

4. Nov 8, 2009

### Staff: Mentor

The Mathematica DSolve usually looks for general solutions that match your particular equation and then it uses the form of the general solution to solve your specific problem. Often the general solution has such "complicated" constants.

5. Nov 8, 2009

### Staff: Mentor

In this case the general form is probably something like:
$$a x'''(t)+b x'(t)=0$$

Which has the general solution:
$$\frac{\text{C1} \sqrt{a} \sin \left(\frac{\sqrt{b} t}{\sqrt{a}}\right)}{\sqrt{b}}-\frac{\text{C2} \sqrt{a} \cos \left(\frac{\sqrt{b} t}{\sqrt{a}}\right)}{\sqrt{b}}+\text{C3}$$

Which turns into exactly the expression you have above for a=1 and b=2

6. Nov 9, 2009

### dperkovic

This is the explanation which I looking for . Thank you !!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook