- #1
dperkovic
- 17
- 0
I have a problem with this homogenous ODE:
[tex]
\frac{d^3x}{dt^3} + 2\frac{dx}{dt} = 0$
[/tex]
With substitution [tex] x=e^\lambda t[/tex], I got characteristic polinom: [tex]\lambda^3+2\lambda=0[/tex].
So,
[tex]\lambda_1 = 0[/tex], [tex]\lambda_{1,2} = \pm\sqrt{2}i[/tex]
Because we have one real,and one pair of complex roots of characteristic polinom, general solution of ODE is:
[tex]x(t) =C1e^{0\cdot t} + C2\cos(\sqrt{2}t) +C3\sin(\sqrt{2}t)[/tex],
or:
[tex]x(t) =C1 + C2\cos(\sqrt{2}t) +C3\sin(\sqrt{2}t)[/tex]
But, Wolfram Mathematica's command DSolve[x'''[t] + 2*x'[t] == 0, x[t], t],give to me this:
[tex]x(t) =C1 - \frac{C2\cos(\sqrt{2}t)}{\sqrt{2}} +\frac{C3\sin(\sqrt{2}t)}{\sqrt{2}}[/tex].
As we can see, there is difference between my and wolfram's solution. Why ?? Where is a mistake in my procedure ?
[tex]
\frac{d^3x}{dt^3} + 2\frac{dx}{dt} = 0$
[/tex]
With substitution [tex] x=e^\lambda t[/tex], I got characteristic polinom: [tex]\lambda^3+2\lambda=0[/tex].
So,
[tex]\lambda_1 = 0[/tex], [tex]\lambda_{1,2} = \pm\sqrt{2}i[/tex]
Because we have one real,and one pair of complex roots of characteristic polinom, general solution of ODE is:
[tex]x(t) =C1e^{0\cdot t} + C2\cos(\sqrt{2}t) +C3\sin(\sqrt{2}t)[/tex],
or:
[tex]x(t) =C1 + C2\cos(\sqrt{2}t) +C3\sin(\sqrt{2}t)[/tex]
But, Wolfram Mathematica's command DSolve[x'''[t] + 2*x'[t] == 0, x[t], t],give to me this:
[tex]x(t) =C1 - \frac{C2\cos(\sqrt{2}t)}{\sqrt{2}} +\frac{C3\sin(\sqrt{2}t)}{\sqrt{2}}[/tex].
As we can see, there is difference between my and wolfram's solution. Why ?? Where is a mistake in my procedure ?
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