- #1

jellicorse

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## Homework Statement

I have been trying to follow a solution to a problem I had but do not quite understand the whole thing. I wondered if anybody could clear it up for me.

Let [tex]a_0[/tex] be the initial value of 'a' for which the transition from one type of behaviour to another occurs.

The general solution to the ODE is [tex]y=(a+3)e^{\frac{t}{2}}-3e^{\frac{t}{3}}[/tex]

So to get the critical value between a divergence in behaviour for this equation is when [tex]\frac{dy}{dt}=0[/tex]

Differentiating the general solution gives:

[tex]\frac{dy}{dt}= \frac{1}{2}(a+3)e^{\frac{t}{2}}-e^{\frac{t}{3}}[/tex]

[tex] 0 =\frac{1}{2}(a+3)e^{\frac{t}{2}}-e^{\frac{t}{3}}[/tex]

Up to here I understand the working. But the next step doesn't make any sense to me where they say

__"The functions of [tex]e^{\frac{t}{3}}[/tex] and [tex]e^{\frac{t}{2}}[/tex] are not equal to zero..."__That's OK but I don't see how it leads to this:

__"...so each term on the RHS is independently zero"__

According to their solution, this leads to [tex]\frac{1}{2}(a+3)=0[/tex], and so on.

There are two things I am having difficulty with here:

- I don't understand what is meant by "each term on the RHS is independently zero". Surely if [tex]e^{\frac{t}{3}}\neq 0[/tex] then [tex](\frac{a+3}{2})e^{\frac{t}{2}}=e^{\frac{t}{3}}[/tex] to make the entire expression zero?

- This is just a general example and it's not stated that it models anything in reality. But if this was a ODE to represent reality and 't' represented time, wouldn't it make sense to find the critical initial value by differentiating wrt 't' and then find the value of y(0), when t=0?

I would be very grateful if anyone could give me some pointers!