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I'm self teaching myself about ODE's and have come across the following question which I'm a bit unsure of:

Find the general solution of the ODE: $ \frac {d^2x} {dt^2} − 6x = 0.$ Find a solution of this ODE that satisfies the two conditions: $ x(0) = 2; x(t) → 0 \text { as }t → +∞.$

Any help would be really appreciated

Unblazon
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1 Answers1

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This is a second order linear differential equation with constant coefficients. In that case, one may use the following ansatz: $$x=e^{\lambda t}$$ After substitution into the differential equation, this gives the following characteristic equation: $$\lambda^2-6=0 \tag{1}$$ Evaluating for $\lambda$ gives: $$\lambda=\pm \sqrt{6}$$ Note that if $x_1(t)$ and $x_2(t)$ are solutions to the differential equation, then so is: $$x(t)=c_1\cdot x_1(t)+c_2\cdot x_2(t)$$ For any constants $c_1$ and $c_2$. Using the roots of the characteristic equation on $(1)$, we obtain: $$x(t)=c_1e^{\lambda_1 t}+c_2 e^{\lambda_2 t}$$ Therefore, our general solution is given by: $$x(t)=c_1e^{-\sqrt{6} t}+c_2 e^{\sqrt{6} t} \tag{2}$$ Can you now use the two conditions to determine the appropriate values of $c_1$ and $c_2$? If not, feel free to ask.

projectilemotion
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  • Thanks very much for your answer. First condition is fine, second I'm a bit unsure would I use log? thanks – Unblazon Apr 06 '17 at 15:29
  • You're welcome. No need to use logarithms. Notice that $e^{-\sqrt{6}t}\to 0$ and $e^{\sqrt{6}t}\to \infty$ as $t\to \infty$. Therefore, since you require $x(t)\to 0$, you know that $c_2=0$ is the only possible way this will occur. Now, you should be able to find $c_1$. (Substitute $t=0$ and $x=2$ in the solution given on $(2)$) – projectilemotion Apr 06 '17 at 15:31
  • @projectilemotion Why we have to stop after we have find two particular solutions? How do we know that there is no other function e.g. (a trigonometric) which could also satisfy the above differential equation? It seems that for an "nth" order the general solution is the linear combination of "n" linear indepedent functions but the above doesn't prove this. How can we prove it? – Antonios Sarikas Nov 16 '20 at 20:16
  • @user599310 See The number of solutions to an $n^{th}$ order differential equation for an explanation on why an $n$-th order linear homogeneous differential equation should have $n$ linearly independent solutions. – projectilemotion Nov 16 '20 at 20:42