# Thread: teacher couldn't help me understand: y'-2y=sin(x) to -sin x(B + 2A) + cos x(A-2B)=...

1. ## teacher couldn't help me understand: y'-2y=sin(x) to -sin x(B + 2A) + cos x(A-2B)=...

Hello

I'm posting this after an extra math lesson where teacher tries to help us who need it. She couldn't break it to me though so I'm asking if there's a person who can use words that I understand and make me learn this one equation.

We calculated this:
y' - 2y = sin x to the point where we got:
-sin x(B + 2A) + cos x(A-2B) = sin x
I just can get my head around the next part even with the help of my teacher. I wish there was a way to make me understand how to solve that and why is the solution like it is. I don't want a straight answer but to understand.

I would start by moving all the sin f(x):s to the right and cos to the left but teacher said it would "break the idea". She then started to explain something about multipliers the sin functions have, which are -1 and 1. After that I have no idea how her thought progressed.

I'm in distress for not understanding something that seems so obvious to the other people!

Joona

2. Hi Joona,
could you clarify what the question is? Maybe take a picture of the problem? What is y'? A? B?

Regards,
Lev

3. Originally Posted by Saraghtera
Hello

I'm posting this after an extra math lesson where teacher tries to help us who need it. She couldn't break it to me though so I'm asking if there's a person who can use words that I understand and make me learn this one equation.

We calculated this:
y' - 2y = sin x to the point where we got:
-sin x(B + 2A) + cos x(A-2B) = sin x
I just can get my head around the next part even with the help of my teacher. I wish there was a way to make me understand how to solve that and why is the solution like it is. I don't want a straight answer but to understand.

I would start by moving all the sin f(x):s to the right and cos to the left but teacher said it would "break the idea". She then started to explain something about multipliers the sin functions have, which are -1 and 1. After that I have no idea how her thought progressed.

I'm in distress for not understanding something that seems so obvious to the other people!

Joona
It isn't obvious to me (yet)!

It's not generally a good idea to drop helpers in the middle of a problem with no context. As I see it, you have probably assumed that y = A cos(x) + B sin(x), so that y' = -A sin(x) + B cos(x), and the differential equation yields -(B + 2A)sin(x) + (A - 2B)cos(x) = sin x. That part makes sense, now that I've gone through it. (Note that the way I wrote it makes it easier to be sure what it means.)

But you haven't said what the next step is, that you need to understand. Rather than just assume we know the entire solution process, can you show us what you were given, and ask specific questions about the reason for each step? We can't tell you "why is the solution like it is" if we don't know what the solution is.

Presumably, you are trying to solve for A and B, given that this equation is true for all x. (In fact, that may be the point you are missing -- it is not uncommon to have trouble with this idea.) My own inclination would be, as you seem to be saying, to put it in the form ___cos(x) = ___sin(x), and then make a conclusion. What did your teacher do instead?

4. Originally Posted by lev888
Hi Joona,
could you clarify what the question is? Maybe take a picture of the problem? What is y'? A? B?

Regards,
Lev

Okay I'm leaving a picture that should clarify what I've done so far. I was so into the task I forgot to explain everything.
-The last line is where I can't proceed.
-"Kaava" is Formula in English. I used it to solve the first part of the task.

Forgive me for being so vague. I'm not used to posting math problems in forums and the problem itself is hard for me so explaining it in English added some extra challenge for me.

Joona

5. Originally Posted by Dr.Peterson
It isn't obvious to me (yet)!

It's not generally a good idea to drop helpers in the middle of a problem with no context. As I see it, you have probably assumed that y = A cos(x) + B sin(x), so that y' = -A sin(x) + B cos(x), and the differential equation yields -(B + 2A)sin(x) + (A - 2B)cos(x) = sin x. That part makes sense, now that I've gone through it. (Note that the way I wrote it makes it easier to be sure what it means.)

But you haven't said what the next step is, that you need to understand. Rather than just assume we know the entire solution process, can you show us what you were given, and ask specific questions about the reason for each step? We can't tell you "why is the solution like it is" if we don't know what the solution is.

Presumably, you are trying to solve for A and B, given that this equation is true for all x. (In fact, that may be the point you are missing -- it is not uncommon to have trouble with this idea.) My own inclination would be, as you seem to be saying, to put it in the form ___cos(x) = ___sin(x), and then make a conclusion. What did your teacher do instead?

Hello Dr. Peterson

I attached a picture of the page I've calculated on. Now you can see the thought process throughout the problem. I'm stuck at the last line. It might be a bit inconvenient for you that there are some words in Finnish but I guess its the numbers that count!

Forgive me for being vague. It's my first post here and the problem itself was hard for me which didn't make it easy to explain.

As for the question "What did your teacher do instead?": We ran out of time and she left me to this problem haha. She's a nice lady but we sure don't speak the same language sometimes.

Joona

6. Originally Posted by Saraghtera
Hello Dr. Peterson

I attached a picture of the page I've calculated on. Now you can see the thought process throughout the problem. I'm stuck at the last line. It might be a bit inconvenient for you that there are some words in Finnish but I guess its the numbers that count!

Forgive me for being vague. It's my first post here and the problem itself was hard for me which didn't make it easy to explain.

As for the question "What did your teacher do instead?": We ran out of time and she left me to this problem haha. She's a nice lady but we sure don't speak the same language sometimes.

Joona
This confirms my guess about what you are doing, but doesn't make it clear where your difficulty is. I wanted you to also confirm what the goal of the next step is.

Can you show us what you tried doing next? I told you what I would do, which seems to be what you said you tried, but I can't be sure without seeing it.

As I said, you currently have -(B + 2A)sin(x) + (A - 2B)cos(x) = sin x. I would now add (B + 2A)sin(x) to each side, so that it would be in the form C cos(x) = D sin(x). Now, this must be true for ALL values of x -- this, as I mentioned, is the key to the problem. The only way this can happen is if C and D are both zero; otherwise it would be true for only certain values of x. (You could solve for x!)

So you can write a system of two equations in the variable A and B, and solve for them.

Try that, and show what you get.

7. Originally Posted by Dr.Peterson
This confirms my guess about what you are doing, but doesn't make it clear where your difficulty is. I wanted you to also confirm what the goal of the next step is.

Can you show us what you tried doing next? I told you what I would do, which seems to be what you said you tried, but I can't be sure without seeing it.

As I said, you currently have -(B + 2A)sin(x) + (A - 2B)cos(x) = sin x. I would now add (B + 2A)sin(x) to each side, so that it would be in the form C cos(x) = D sin(x). Now, this must be true for ALL values of x -- this, as I mentioned, is the key to the problem. The only way this can happen is if C and D are both zero; otherwise it would be true for only certain values of x. (You could solve for x!)

So you can write a system of two equations in the variable A and B, and solve for them.

Try that, and show what you get.

Firstly, I really appreciate your dedication towards helping me. Secondly I want to refer back to the original problem given by the teacher as an answer to your question "I wanted you to also confirm what the goal of the next step is":
Solve Differential Equations: y'-2y = sin x
I hope this clarifies the result I'm aiming for because that's all we know. I've been trying to fiddle around with the numbers and see what I get but I feel like I'm just losing my direction doing that.

I could simplify the equation to: Acos(x) - 2Bcos(x) = sin(x) + Bsin(x) + 2Asin(x) but not really know what to do with that info.

I did solve it for a and got a really horrific looking answer:
A = (sin(x) + Bsin(x) + 2Bcos(x)) / (cos(x) - 2sin(x))
where
(cos(x) - 2sin(x)) is not zero.Other than that I really don't know what is the expected answer for the question at hand. The things I tried really don't get me anywhere too.

To add an answer to your suggestion to create a system of two equations. I wasn't able to form that one as it has been 8 years since I had my matriculation examination, when I last used them.

Joona

8. Originally Posted by Saraghtera
Firstly, I really appreciate your dedication towards helping me. Secondly I want to refer back to the original problem given by the teacher as an answer to your question "I wanted you to also confirm what the goal of the next step is":
Solve Differential Equations: y'-2y = sin x
I hope this clarifies the result I'm aiming for because that's all we know. I've been trying to fiddle around with the numbers and see what I get but I feel like I'm just losing my direction doing that.

I could simplify the equation to: Acos(x) - 2Bcos(x) = sin(x) + Bsin(x) + 2Asin(x) but not really know what to do with that info.

I did solve it for a and got a really horrific looking answer:
A = (sin(x) + Bsin(x) + 2Bcos(x)) / (cos(x) - 2sin(x))
where
(cos(x) - 2sin(x)) is not zero.Other than that I really don't know what is the expected answer for the question at hand. The things I tried really don't get me anywhere too.

To add an answer to your suggestion to create a system of two equations. I wasn't able to form that one as it has been 8 years since I had my matriculation examination, when I last used them.

Joona
The "goal of the next step" is to solve for the constants A and B, right? That's what I hoped you'd tell me, because your teacher should have told you what you were trying to do. The goal of the whole problem is important, too, and we'll get to that, but it's not the immediate goal.

But you seem to have missed my point, so I'll say it a third time, with more context. In solving a differential equation, you are trying to find a function, y = f(x) that makes the differential equation true. This means you are not just talking about equations that are true for certain values of x, but rather about things that are true for ALL x. That's what a function is.

You assumed that f(x) = A cos(x) + B sin(x), where A and B are some as yet unknown constants. Right now, you need to find their values. Keep in mind that they must be constants. You're not solving for A as a function of x.

So you want -(B + 2A)sin(x) + (A - 2B)cos(x) = sin x to be true for ALL x. Simplify it to (A - 2B) cos(x) = sin x + (B + 2A) sin(x), and further to (A - 2B) cos(x) = (1 + B + 2A) sin(x). Don't expand it, as you did; that doesn't help. My point was that the only way a multiple of cos(x) can ALWAYS equal a multiple of sin(x), is if both multiples are ZERO. Think about that, until you see WHY it has to be true. This is, as I've said, a key idea here, and probably the one idea that is most preventing you from following what your teacher says.

Now, this means that

A - 2B = 0
1 + B + 2A = 0

Can you see that? When this is true, then the equation is 0 cos(x) = 0 sin(x), which is indeed true for all x.

Do you know how to solve a system of linear equations? If not, look it up. That is part of what it takes to learn a new subject: when something is required that you are assumed to know, you have to refresh your knowledge so that you meet the prerequisite. Don't just wait for someone else to do it for you, or you'll be forever dependent on someone else. The alternative is to go back and take the course in which you learn to solve systems of equations, so that you meet the prerequisites for the course you are taking now.

9. Originally Posted by Dr.Peterson
The "goal of the next step" is to solve for the constants A and B, right? That's what I hoped you'd tell me, because your teacher should have told you what you were trying to do. The goal of the whole problem is important, too, and we'll get to that, but it's not the immediate goal.

But you seem to have missed my point, so I'll say it a third time, with more context. In solving a differential equation, you are trying to find a function, y = f(x) that makes the differential equation true. This means you are not just talking about equations that are true for certain values of x, but rather about things that are true for ALL x. That's what a function is.

You assumed that f(x) = A cos(x) + B sin(x), where A and B are some as yet unknown constants. Right now, you need to find their values. Keep in mind that they must be constants. You're not solving for A as a function of x.

So you want -(B + 2A)sin(x) + (A - 2B)cos(x) = sin x to be true for ALL x. Simplify it to (A - 2B) cos(x) = sin x + (B + 2A) sin(x), and further to (A - 2B) cos(x) = (1 + B + 2A) sin(x). Don't expand it, as you did; that doesn't help. My point was that the only way a multiple of cos(x) can ALWAYS equal a multiple of sin(x), is if both multiples are ZERO. Think about that, until you see WHY it has to be true. This is, as I've said, a key idea here, and probably the one idea that is most preventing you from following what your teacher says.

Now, this means that

A - 2B = 0
1 + B + 2A = 0

Can you see that? When this is true, then the equation is 0 cos(x) = 0 sin(x), which is indeed true for all x.
Hi

(A - 2B) cos(x) = (1 + B + 2A) sin(x) this form makes it a tad bit easier to understand what I'm looking at. If I were to simplify it even more, for the sake of argument, I wouldn't be completely wrong to put it down as A cos(x) = B sin(x)? And to build up on that though, the only possible way to have that hold true is to have A = 0 and B = 0 for all x values.

TOP 3 from your explanation that helped me to learn
-Unknown constants
-For all x values
-In solving a differential equation, you are trying to find a function, y = f(x) that makes the differential equation true
I really like the way you found out the things what I was unsure about and emphasized them to me.
Had I come to this conclusion on my own, I would've doubted myself with a question "How can I be sure there's no other possible constants that would hold true for the function?. Maybe there are two random constant numbers that still hold true for the function."

The doubt cleared after some graph drawing. I plotted two graphs: 1 * sin(x) and 1 * cos(x) and it became very apparent that the two can never be the same with all x values other than zero as the graphs are not in line with each other and the multiplier only grows the wave height (so to speak?). Without the experience and visualization, though, I find it really hard to come to that conclusion. Most of the time I start the calculation process by drawing as it makes the problem visual, hence easier to understand to me. Before the simplification I really didn't know how to visualize though.

Thank you, Dr. Peterson, for your patience with me and for that you took your time to see this through with me. I got a great deal of new keywords from you for my future studies. I'm going to mention you to my teacher next Thursday when we go through the problems together with her.

Lastly I want to stick with the part "That is part of what it takes to learn a new subject: when something is required that you are assumed to know, you have to refresh your knowledge so that you meet the prerequisite. Don't just wait for someone else to do it for you, or you'll be forever dependent on someone else." for a moment here.

I'm coming from a background studying a short curriculum of maths in high school and we're now building up on information acquired from the long curriculum in university. My problem lies very much in the lack of cumulative knowledge of the long curriculum maths. My metacognition for maths is in child's boots what comes to finding new solutions and having learning mechanisms. So I came to the forums in search of keywords. Someone who has the knowledge could here possibly offer some seeds for me to build upon. If I compare this to my skill in programming for example, I find it very easy to find new information and solutions based on my already achieved knowledge.

So I'm now still a bit dependent on others with my learning but once I receive the knowledge to learn I'm all set. I just need enough of those "seeds" to start building on.

You've definitely offered me plenty of concrete information but also broader knowledge on the subject and I'm going to put in good use!

Joona

10. Originally Posted by Saraghtera
(A - 2B) cos(x) = (1 + B + 2A) sin(x) this form makes it a tad bit easier to understand what I'm looking at. If I were to simplify it even more, for the sake of argument, I wouldn't be completely wrong to put it down as A cos(x) = B sin(x)? And to build up on that though, the only possible way to have that hold true is to have A = 0 and B = 0 for all x values.
I said the same thing, but used C and D to avoid confusion with the A and B used in the problem.

Originally Posted by Saraghtera
TOP 3 from your explanation that helped me to learn
-Unknown constants
-For all x values
-In solving a differential equation, you are trying to find a function, y = f(x) that makes the differential equation true
I really like the way you found out the things what I was unsure about and emphasized them to me.
Had I come to this conclusion on my own, I would've doubted myself with a question "How can I be sure there's no other possible constants that would hold true for the function?. Maybe there are two random constant numbers that still hold true for the function."
At one point I considered explicitly saying something about this, but then I realized that it doesn't really matter. You defined f(x) as you did because you are looking for a solution -- not necessarily all solutions, which comes later. So all you really need is to find a pair of values that work, not to be sure there are no other pairs of values. But it's good that you went on to explore this idea, because asking yourself such questions is the way to really come to understand.

Originally Posted by Saraghtera
The doubt cleared after some graph drawing. I plotted two graphs: 1 * sin(x) and 1 * cos(x) and it became very apparent that the two can never be the same with all x values other than zero as the graphs are not in line with each other and the multiplier only grows the wave height (so to speak?). Without the experience and visualization, though, I find it really hard to come to that conclusion. Most of the time I start the calculation process by drawing as it makes the problem visual, hence easier to understand to me. Before the simplification I really didn't know how to visualize though.

Thank you, Dr. Peterson, for your patience with me and for that you took your time to see this through with me. I got a great deal of new keywords from you for my future studies. I'm going to mention you to my teacher next Thursday when we go through the problems together with her.
Excellent. You learned just as I hoped you would.

If you check my profile, you will find two other sites worth mentioning to your teacher, in which I participate.

Originally Posted by Saraghtera
Lastly I want to stick with the part "That is part of what it takes to learn a new subject: when something is required that you are assumed to know, you have to refresh your knowledge so that you meet the prerequisite. Don't just wait for someone else to do it for you, or you'll be forever dependent on someone else." for a moment here.

I'm coming from a background studying a short curriculum of maths in high school and we're now building up on information acquired from the long curriculum in university. My problem lies very much in the lack of cumulative knowledge of the long curriculum maths. My metacognition for maths is in child's boots what comes to finding new solutions and having learning mechanisms. So I came to the forums in search of keywords. Someone who has the knowledge could here possibly offer some seeds for me to build upon. If I compare this to my skill in programming for example, I find it very easy to find new information and solutions based on my already achieved knowledge.

So I'm now still a bit dependent on others with my learning but once I receive the knowledge to learn I'm all set. I just need enough of those "seeds" to start building on.

You've definitely offered me plenty of concrete information but also broader knowledge on the subject and I'm going to put in good use!

Joona
Wonderful. My comments were meant to encourage you to be an independent learner, and clearly I judged you correctly, that this is something you can do well. You can find the help you need (which is part of that independence!), and grow from there.

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