ODE with constant coefficient ##b##

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Second order ODE using ansatz $$u = e^{λt}$$

For this problem,

I can confused why they don't include the case where $b = 0$ since $b < 0$.

That is, why don't they include $λ^2e^{λt} + e^{λt} = 0$ when solving the associated homogenous equation? This gives the commentary solution $u_h = \cos t + \sin t$ which is not included in their complex case, even thought it accounts for $-2 < b < 1$.

Thanks for any help!

 

[Orodruin]

It is included in $-2 < b < 1$. There is no need to create a separate case for it.

There are however numerous typos in the solution.

 

[ChiralSuperfields]

It is included in $-2 < b < 1$. There is no need to create a separate case for it.

There are however numerous typos in the solution.

Thank you for your reply @Orodruin!

Sorry I am still confused. The two complementary solutions are different.

When $$b = 0$$ for $$-2 < b < 1$$, then $$u_h = c_1\cos(2t) + c_2\sin(2t)$$ from the complex case in the solution.

However, from my expression is $$u_h = c_1\cos(t) + c_2\sin(t)$$ from setting the b coefficient zero in the second order ODE.

Do you please know why there is a difference here in the solutions?

Thanks!

 

[Orodruin]

Do you please know why there is a difference here in the solutions?

That would be some of the many many many many typos in the provided solution. The complex root solution contains several of them.

It should be:
$$
u_h = e^{b\color{Red}{t}/2} \left[ c_1 \cos\left( \frac{t}{\color{Red}{2}} \sqrt{4 - b^2}\right)
+ c_2 \sin\left( \frac{t}{\color{Red}{2}} \sqrt{4 - b^2}\right)\right]
$$

 

[ChiralSuperfields]

That would be some of the many many many many typos in the provided solution. The complex root solution contains several of them.

It should be:
$$
u_h = e^{b\color{Red}{t}/2} \left[ c_1 \cos\left( \frac{t}{\color{Red}{2}} \sqrt{4 - b^2}\right)
+ c_2 \sin\left( \frac{t}{\color{Red}{2}} \sqrt{4 - b^2}\right)\right]
$$

Thank you for your reply @Orodruin!

You said there are many typos in this solution. Do you please know what the other typos are?

Thanks!

 

[Orodruin]

It is symptomatic of many of the solutions posted with your threads and therefore seemingly of your learning material. Unfortunately, I don’t feel like keeping pointing out those typos is a very fruitful endeavour. I would suggest looking for better learning material.

 

[ChiralSuperfields]

It is symptomatic of many of the solutions posted with your threads and therefore seemingly of your learning material. Unfortunately, I don’t feel like keeping pointing out those typos is a very fruitful endeavour. I would suggest looking for better learning material.

Thank you for your reply @Orodruin!

I understand. I would look for better material, however, I still have to use this material since it is from the university course I am taking in ODEs :(

Otherwise I would just read a textbook

 

[ChiralSuperfields]

That would be some of the many many many many typos in the provided solution. The complex root solution contains several of them.

It should be:
$$
u_h = e^{b\color{Red}{t}/2} \left[ c_1 \cos\left( \frac{t}{\color{Red}{2}} \sqrt{4 - b^2}\right)
+ c_2 \sin\left( \frac{t}{\color{Red}{2}} \sqrt{4 - b^2}\right)\right]
$$

Thank you for your reply @Orodruin!

I've been in touch with the Prof and there is no typo in the solution (at least the complex case part you said there was). Do you please know why you think this is a typo (I am getting the same solution as you by the way)?

Thanks!

 

[Orodruin]

Thank you for your reply @Orodruin!

I've been in touch with the Prof and there is no typo in the solution (at least the complex case part you said there was). Do you please know why you think this is a typo (I am getting the same solution as you by the way)?

Thanks!

Yes there are. There are obvious typos in the complex solution as pointed out in red in my previous post.

Just try to insert the given solution and you will see that it is, in fact, not a solution. Insert the corrected solution from post #4 and you will see that it is a solution.

Failure to recognise this is a worrying sign. Did you point out exactly where the typos were?

A similar issue, by the way, holds for the real solution, where the $b/2$ is not multiplied by $t$ as it should be. It is furthermore unclear from that solution whether the $t$ after the square root is a multiplication or a division. Your professor should use parentheses to make this clear. The easiest remedy is to put parentheses around the full root expression:
$$
\exp\left([b/2 + \sqrt{b^2 - 4}/2] t\right)
$$
Also note how the root was messed up: Using $4 - b^2$ instead of $b^2 - 4$ in the real ($b < -2$) case would make the square root argument negative and therefore make the square root imaginary.

The correct solutions are:
Complex case:
$$
u_h = e^{b\color{Red}{t}/2} \left[c_1 \cos\left(\frac{t\sqrt{4-b^2}}{\color{Red}{2}}\right) + c_2 \sin\left(\frac{t\sqrt{4-b^2}}{\color{Red}{2}}\right)\right]
$$
Real case:
$$
u_h = c_1 \exp\left( \frac{\color{Red}{t}}{2}[b + \sqrt{\color{Red}{+}b^2 \color{Red}{-} 4}] \right) + c_2 \exp\left( \frac{\color{Red}{t}}{2}[b - \sqrt{\color{Red}{+}b^2 \color{Red}{-} 4}] \right)
$$

I have marked in red where typos of the provided solution have been corrected. The repeated root solutions is free of typos.

Your professor is welcome here to try to argue their piece if they still think that they do not have typos (they would be wrong, but they are free to present it and be corrected).

 

[Orodruin]

By the way. If your professor does not trust someone else's computations, here are the computations performed by Mathematica:

Here, uh[t_] is the expression provided by the solution and uH[t_] is the corrected expression. Inserted into the LHS of the differential equation, it is clear that the corrected expression satisfies the differential equation whereas the provided expression does not.

 

[ChiralSuperfields]

Yes there are. There are obvious typos in the complex solution as pointed out in red in my previous post.

Just try to insert the given solution and you will see that it is, in fact, not a solution. Insert the corrected solution from post #4 and you will see that it is a solution.

Failure to recognise this is a worrying sign. Did you point out exactly where the typos were?

A similar issue, by the way, holds for the real solution, where the $b/2$ is not multiplied by $t$ as it should be. It is furthermore unclear from that solution whether the $t$ after the square root is a multiplication or a division. Your professor should use parentheses to make this clear. The easiest remedy is to put parentheses around the full root expression:
$$
\exp\left([b/2 + \sqrt{b^2 - 4}/2] t\right)
$$
Also note how the root was messed up: Using $4 - b^2$ instead of $b^2 - 4$ in the real ($b < -2$) case would make the square root argument negative and therefore make the square root imaginary.

The correct solutions are:
Complex case:
$$
u_h = e^{b\color{Red}{t}/2} \left[c_1 \cos\left(\frac{t\sqrt{4-b^2}}{\color{Red}{2}}\right) + c_2 \sin\left(\frac{t\sqrt{4-b^2}}{\color{Red}{2}}\right)\right]
$$
Real case:
$$
u_h = c_1 \exp\left( \frac{\color{Red}{t}}{2}[b + \sqrt{\color{Red}{+}b^2 \color{Red}{-} 4}] \right) + c_2 \exp\left( \frac{\color{Red}{t}}{2}[b - \sqrt{\color{Red}{+}b^2 \color{Red}{-} 4}] \right)
$$

I have marked in red where typos of the provided solution have been corrected. The repeated root solutions is free of typos.

Your professor is welcome here to try to argue their piece if they still think that they do not have typos (they would be wrong, but they are free to present it and be corrected).

By the way. If your professor does not trust someone else's computations, here are the computations performed by Mathematica:
View attachment 343870
Here, uh[t_] is the expression provided by the solution and uH[t_] is the corrected expression. Inserted into the LHS of the differential equation, it is clear that the corrected expression satisfies the differential equation whereas the provided expression does not.

Thank you very much @Orodruin! That is a good idea to consult Mathematica so your solution is definitely the correct one!

 

[Orodruin]

That is a good idea to consult Mathematica so your solution is definitely the correct one!

Oh, I had no doubts. It is just additional weight in the argument. It really shouldn’t be necessary to convince someone who teach the subject for a living that there are obvious typos in the solution.

Sorry to come off as aggressive. It just really rubs me the wrong way. A couple of typos anyone can make, but when they are this abundant and on top you don’t even own up to them when they are pointed out … sorry, but then one is simply not putting any effort into teaching.