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I have a second order differential equation:

$$y''=-a^2y(t)$$

I search some possible solution such as:

$$y(t)=\sin(at)$$ or $$y(t)=\cos(at)$$

Is $y(t)= 0 $ also a possible solution or not?

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Definitely $y(t)=0$ is a solution as it satisfies the given equation. –  Tapu Feb 3 '13 at 17:01
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Plug in and observe that $y=0$ is a solution. Now for further insight: this is a second order homogeneous linear ODE. So the solution set is a 2-dimensional vector space. You have found a basis in the case $a\neq 0$. Now $y$ is a solution if and only if it is a linear combination of your two solutions. Of course, this is the case of $y=0$. –  1015 Feb 3 '13 at 17:15
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2 Answers 2

up vote 2 down vote accepted

$y(t)=0$ is called the trivial solution. So the easy solution which doesn't really represent anything interesting. But since it solves the differential equation it is a solution. Also, when you would find possible solutions involving $\sin$ and $\cos$ (or exponentials) there will be arbitrary constants $c_1$ and $c_2$ multiplying them. You can view the trivial solution as just setting these constants equal to 0.

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So, $D^2y+a^2y=0\implies D=\pm i\cdot a$

So if $a\ne0,y=Ae^{i\cdot ax}+Be^{-i\cdot ax}$ where $A,B$ are arbitrary constants.

So, $y=A(\cos ax+i\sin ax)+B(\cos ax-i\sin ax)=(A+B)\cos ax+i(A-B)\sin ax$

If $a=0,D=0\implies y=(A+Bx)e^{ix\cdot0}=A+Bx$ as both the are equal.

Any constant set of values of $A,B$ this will satisfy $y''+a^2y=0$

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Why are you assuming $a\ne0$? :) –  Tapu Feb 3 '13 at 17:03
    
@Tapu, conventionally $a$ is assumed non-zero. Anyway, the answer has been edited to consider that case. –  lab bhattacharjee Feb 3 '13 at 17:05
    
Please do not mind, but I have never seen an $i$ , where conventionally $y$ is assumed to be real :) I mean it is better to write the solution as $y(t)=A\cos ax+B\sin ax$. –  Tapu Feb 3 '13 at 17:08
    
@Tapu, why can't $A,B$ be complex, to make $A+B, i(A-B)$ real/complex as we like? –  lab bhattacharjee Feb 3 '13 at 17:26
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