Initial Value problem 1st order ODE

[jonnburton]

I was wondering whether anyone could explain a step in the working i have been lookoing at for this problem.

The differential equation has the general solution \(\displaystyle y=\frac{4}{5}(2sin(t)-cos(t))+(a+\frac{4}{5})e^{\frac{t}{2}}\)


Let \(\displaystyle a_0\) be the value of the initial value 'a' for which a transition from one type of behaviour to another occurs.


so, to get this value, differentiate both sides:

\(\displaystyle \frac{dy}{dt}=\frac{4}{5}(2cos(t)+sin(t))+\frac{1}{2}(a+\frac{4}{5})e^{\frac{t}{2}}\)


The critical value occurs when \(\displaystyle \frac{dy}{dt} =0\)




But I am not sure how they have got to this next step:

\(\displaystyle 0=\frac{1}{2}(a+\frac{4}{5})e^{\frac{t}{2}}\)

In the notes to the working it says "each term on the RHS is zero, since both are independent factors". I am not sure what they mean by this and how it is the case that each term necessarily equals zero. Why couldn't one term equal a negative value of a certain magnitude and another equal a positive value of the same magnitude? And I can't see how any value of 't' would allow both sin(t) and cos(t) to be zero.

I'd be very grateful if anybody could tell me how this works.

 

[Deleted member 4993]

I was wondering whether anyone could explain a step in the working i have been lookoing at for this problem.

The differential equation has the general solution \(\displaystyle y=\frac{4}{5}(2sin(t)-cos(t))+(a+\frac{4}{5})e^{\frac{t}{2}}\)


Let \(\displaystyle a_0\) be the value of the initial value 'a' for which a transition from one type of behaviour to another occurs.


so, to get this value, differentiate both sides:

\(\displaystyle \frac{dy}{dt}=\frac{4}{5}(2cos(t)+sin(t))+\frac{1}{2}(a+\frac{4}{5})e^{\frac{t}{2}}\)


The critical value occurs when \(\displaystyle \frac{dy}{dt} =0\)




But I am not sure how they have got to this next step:

\(\displaystyle 0=\frac{1}{2}(a+\frac{4}{5})e^{\frac{t}{2}}\)

In the notes to the working it says "each term on the RHS is zero, since both are independent factors". I am not sure what they mean by this and how it is the case that each term necessarily equals zero. Why couldn't one term equal a negative value of a certain magnitude and another equal a positive value of the same magnitude? And I can't see how any value of 't' would allow both sin(t) and cos(t) to be zero.

I'd be very grateful if anybody could tell me how this works.

think about this:

2cos(t) + sin(t) = √5*[2/√5 * cos(t) + 1/√5 * sin(t)] = √5 * cos(Θ - t)

where: cos(Θ) = 2/√5 and sin(Θ) = 1/√5

Now you can see cos(Θ - t) = 0 when (Θ - t) = ±(2n+1)*π/2

 

[jonnburton]

Thank you Subhotosh. I will have to refresh my memory on that aspect of trigonometry again. I can certainly see it is possible to that term to equal zero now.

I don't think this means that the term \(\displaystyle (2sin(t)-cos(t)\) necessarily equals zero though, as the working in the book seems to state.

 

[Deleted member 4993]

Thank you Subhotosh. I will have to refresh my memory on that aspect of trigonometry again. I can certainly see it is possible to that term to equal zero now.

I don't think this means that the term \(\displaystyle (2sin(t)-cos(t)\) necessarily equals zero though, as the working in the book seems to state.

It does mean that at a certain value of 't' we get \(\displaystyle (2sin(t)-cos(t) = 0\)

 

[jonnburton]

I think I've got it now, thanks Subhotosh.

I had to have a look through the earlier trig to refresh my memory and then think about the problem in question.