# learning differentual equations

#### PG1995

Joined Apr 15, 2011
818
Hi

Please help me with this problem. I'm really stuck on this one. I'm learning to work with differential equations.

Page #1: http://img850.imageshack.us/img850/4451/page1st.jpghttp://img707.imageshack.us/img707/2342/page1oip.jpg
Page #2: http://img59.imageshack.us/img59/943/page2tv.jpg
Page #3: http://img80.imageshack.us/img80/2509/page3qz.jpghttp://img228.imageshack.us/img228/7319/page3q.jpg

Thanks a lot for your help.

Regards
PG

PS: I have modified the linked pages to make my queries more understandable. Thanks.

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#### steveb

Joined Jul 3, 2008
2,436
PS: I have modified the linked pages to make my queries more understandable. Thanks.
For question 1, the author seems to be saying that you can't get the solution y=-2 from that particular equation directly. You have to go back to the original differential equation and note that this is a trivial solution to the problem.

He also mentions that you can modify the constant to get an equation that does reduce to y=-2, which leads to your Q2.

For Q2, there are a number of places you could make the constant change between equation 10 and equation 11. However, don't even do that because it is confusing. Instead take your first solution

$$y=2{{1+c e^{4x}}\over{1-ce^{4x}}$$

and divide the numerator and denominator by c to get the following

$$y=2{{1/c+e^{4x}}\over{1/c-e^{4x}}$$

Now, redefine the constant, and let c_3=1/c, to get the following.

$$y=2{{c_3+e^{4x}}\over{c_3-e^{4x}}$$

Now when c_3 is zero, you get y=-2.

You can also get that directly from the first equation, although mathematicians sometimes don't like this shortcut approach because you have to be careful. Since c_3 is zero, then c is infinity. If you plug c=infinity into your first solution you get the following.

$$y=2{{1+\infty e^{4x}}\over{1-\infty e^{4x}}$$

Then the ones are insignificant which leaves.

$$y=2{{\infty e^{4x}}\over{-\infty e^{4x}}} =-2$$

There is a more formal way to do this writing limits, but it amounts to the same thing.

If i get some time later I'll make an attempt at Q3 and Q4. My first reading of them left me confused, but a second reading in a more relaxed setting should make it clearer.

#### steveb

Joined Jul 3, 2008
2,436
For questions Q3 and Q4, I'm not sure I understand exactly what you are asking. You mentioned something about the y=0 solution being excluded by the method, but I wouldn't worry about a comment like that from the author. It's clear that y=0 is a trivial solution and it's clear that the casewise function provides an infinite number of valid solutions, even when y=0. Are you doubting that these are valid solutions? Or, are you worried that there are even more solutions not mentioned? Please clarify.

#### PG1995

Joined Apr 15, 2011
818
For questions Q3 and Q4, I'm not sure I understand exactly what you are asking. You mentioned something about the y=0 solution being excluded by the method, but I wouldn't worry about a comment like that from the author. It's clear that y=0 is a trivial solution and it's clear that the casewise function provides an infinite number of valid solutions, even when y=0. Are you doubting that these are valid solutions? Or, are you worried that there are even more solutions not mentioned? Please clarify.
Thanks a lot. I understand that it could be quite confusing for you to understand my queries. I'm doing my best to make it easy for you.

I have rephrased my Q3 and Q4. Please have a look on the attachment. You can find Page #3 useful from the my first post above:
Page #3: http://img228.imageshack.us/img228/7319/page3q.jpg

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#### steveb

Joined Jul 3, 2008
2,436
Thanks a lot. I understand that it could be quite confusing for you to understand my queries. I'm doing my best to make it easy for you.

I have rephrase my Q3 and Q4. Please have a look on the attachment. You can find Page #3 useful from the my first post above:
Page #3: http://img228.imageshack.us/img228/7319/page3q.jpg

OK, I think I understand you now.

For Q3, he is defining what is called a piecewise (or case dependent) function. The reason why y=0 for x<a is because that is how the function is defined. The author is pointing out that the function with a>0 does not match the initial condition, but if you define a piecewise function that forces y=0 for x<a, then the initial condition is met and each piece of the function still obeys the differential equation.

For Q4, you could write x<=a, if you want, because the two pieces of the function are equal when x=a, but traditionally, the piecewise function is defined in a way that the regions don't overlap, even when they are equal to each other at a point or over a region.

• PG1995

#### PG1995

Joined Apr 15, 2011
818
Thank you, Steve. I'm happy I was able to get understood.

QUESTION 1:
Fortunately enough, the Wikipedia page (at present) on piece-wise defined function was understandable for me.

[tex.]$f(x)=|x|=\dbinom{-x}{x}\QDATOPD. . {x<0}{x\geq 0}$[/tex]

(The above Latex code is not compiling for me on this site so please check the attachment. Could you please try it on your computer. Notice the period "." I have used in [tex.])

See, y is not equal to zero for x<a (for the function given in Latex code, a=0, and y=f(x)). So, in my opinion for y to be always "0" is not really necessary for a piece-wise function. Even in this graph of the function given above (in Latex code) y is not equal to zero when x<0, so it makes sense and I have no objections. So, I still don't understand why the author says here y=0 for x<a when y is just equal to zero over a small portion.

And you were right about Q4 in your last post that it could be written x<=a but it would be wrong in strict mathematics terms because the subdomains of a piece-wise function are disjoint sets.

QUESTION 2:

I have yet another question about the Page #3 from my first post. Please have a look on the yellow highlight.

The author says:
In addition, the initial-value problem (13) possesses infinitely more solutions, since for any choice of the parameter a > 0 the piecewise-defined function satisfies both the differential equation and the initial condition.
I don't see how when a>0 satisfies the initial value condition (in my opinion, initial condition is satisfied only when a=0). Initial condition is when x=0, y is also zero. Say, a = 3.

y = (x^2 - a^2)^2 / 16.

When you put x=0, then y=81/16. So, you see my problem.

Best regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
(The above Latex code is not compiling for me on this site so please check the attachment. Could you please try it on your computer. Notice the period "." I have used in [tex.])

See , y is not equal to zero for x<a. So, in my opinion for y to be always "0" is not defined. Even in this graph of the function given above y is not equal to zero when x<0, so it makes sense and I have no objections. So, I still don't understand why the author says here y=0 for x<a when y is just equal to zero over a small portion.
OK, let's look at question 1 first.

First, I could not get your TeX code to work. Also, I normally use the \cases command to get a piecewise function form, but I could not get this to work here.

The graph you showed does not have y=0 for x<0. You are correct about that. The plot is for the function without modification. As far as I can tell, the author then goes on to define a new function in which y does equal zero for x<0. He is free to define anything he likes, and if he says this new function is zero for x<0, then it is true, by definition. The original function does not meet the initial conditions, while the new modified function does meet the initial conditions.

I hope I'm understanding your question correctly.

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• PG1995

#### steveb

Joined Jul 3, 2008
2,436

QUESTION 2:

I have yet another question about the Page #3 from my first post. Please have a look on the yellow highlight.

The author says: I don't see how when a>0 satisfies the initial value condition. Initial condition is when x=0, y is also zero. Say, a = 3.

y = (x^2 - a^2)^2 / 16.

When you put x=0, then y=81/16. So, you see my problem.

Best regards
PG
OK, the second question seems related to the first question. Again, if the author wants to define a new function for which y=0 for x<a, then he is free to do so. The piecewise definition of a function gives a person great flexibility to define a wide range of unusual functions. The author sees that the basic function does not meet the initial conditions, so he modifies the problem area and manages to find a weird function that meets the initial conditions and also is a solution to the differential equation.

• PG1995

#### steveb

Joined Jul 3, 2008
2,436
[tex.]$f(x)=|x|=\dbinom{-x}{x}\QDATOPD. . {x<0}{x\geq 0}$[/tex]

(The above Latex code is not compiling for me on this site so please check the attachment. Could you please try it on your computer. Notice the period "." I have used in [tex.])
Here is a way to write that in tex here.

$$f(x) = \{ \matrix{-x,\;\;\;{\rm if}\;\;\;x \lt 0}\cr{\ \;x,\;\;\;{\rm if}\;\;\;x\ge 0 }$$

[tex.]f(x) = \{ \matrix{-x,\;\;\;{\rm if}\;\;\;x \lt 0}\cr{\ \;x,\;\;\;{\rm if}\;\;\;x\ge 0 }[/tex]

• PG1995

#### PG1995

Joined Apr 15, 2011
818

QUESTION 2:

I have yet another question about the Page #3 from my first post. Please have a look on the yellow highlight.

The author says:
In addition, the initial-value problem (13) possesses infinitely more solutions, since for any choice of the parameter a > 0 the piecewise-defined function satisfies both the differential equation and the initial condition.
I don't see how when a>0 satisfies the initial value condition (in my opinion, initial condition is satisfied only when a=0). Initial condition is when x=0, y is also zero. Say, a = 3.

y = (x^2 - a^2)^2 / 16.

When you put x=0, then y=81/16. So, you see my problem.
Thank you very much, Steve.

I was wrong about QUESTION 2 above. The author is talking about problem (13) which is actually differential equation we are seeking solution of, dy/dx = x*y^(1/2), y(0)=0. The author is correct.

As I have hard time compiling my LaTex code here, therefore please check the attachment to see why author was correct.

I do have some queries on the same problems. But I think it would be easy for you to help me if I start everything afresh in a new post. So, my next post is on the same topic but won't have anything to do with the rest of the thread above it. Please help me. Thanks.

Regards
PG

PS: At the end of the attachment, it reads: "...x needs to be >=a". The bold part is missing text.

#### steveb

Joined Jul 3, 2008
2,436
Hi

Page #1: http://img839.imageshack.us/img839/5540/ppage1.jpg
Page #2: http://img684.imageshack.us/img684/7079/ppage2.jpg
Page #3: http://img694.imageshack.us/img694/8255/ppage3.jpg

Regards
PG
Sorry, I missed these questions before. I'll try to answer them now. I'll handle each question in a separate post to provide focus.

Q1:

Personally, I think the author make much ado about nothing when he talks about there being a "dilemma". Mathematicians are much more afraid of infinity than engineers and physicists, it seems. The author is just trying to be careful when there is an infinity involved, but don't let that confuse you.

Take his line after equation 11. He says ...

$$-1+c=1+c$$ or $$-1=1$$

He has artificially created the dilemma by making this step, because this is not a valid step when c goes to infinity. In fact it is just the opposite. A proper limit would be as follows ...

$$\lim_{c\to \infty} (-1+c)=lim_{c\to \infty} (1+c)$$ results in $$c=c$$

Hence, there is really no dilemma, but rather this is a situation that requires caution. Previously, I showed you how to convert the problem by defining a new constant c3=1/c, which provides another viewpoint and avoids the problem with infinity.

You also compare different problems to each other in this question. I would recommend not doing this. Differential equations are a tricky subject and there are no rules that work in all cases. Just focus on each problem and learn the methods, solutions, tricks and pitfalls that each problem teaches.

You also seem bothered by the comment the author makes about losing the solution when y=0. Again, he is just being cautious with infinities. When you divide by y, then the validity of the operation breaks down when y=0. This does not mean that your resulting solution is not valid at y=0, it only means that your method of finding the solution is not valid at y=0. If you take your solution and plug it back into the original equation, and find that it works, then there is no problem because this method does not use division by zero. That method is the guess and check method and if you try a solution and it works, you have something valid.

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• PG1995

#### steveb

Joined Jul 3, 2008
2,436
For Q2, the solution y=2 is prohibited by the initial condition y=-2. There is just no way for this to work. However, the solution y=-2 obeys the initial condition and is a solution to the given differential equation. It's as simple as that.

Remember, when trying to find a solution, we sometimes have to use some mathematical tricks to search out and find the form of the basic solution. The tricks may induce some problem areas, (such as points that are undefined or infinities etc.). In such cases, one has to use logic. Are those points really problems or are the so-called "dilemmas" just artifacts of the method? The method is not the answer. Actually, the answer given by the method is a potential answer of the real problem. You then need to go back and check it, and look at the problem points. Also, you need to look for other pitfalls and other potential solutions that your method might have discarded by accident. This is very tricky stuff, but you are doing the correct thing by working on many examples and asking good questions.

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• PG1995

#### steveb

Joined Jul 3, 2008
2,436
For Q3, the author seems to be pointing out that sometimes teachers and students are content to find an answer, but he is warning that sometimes this is not the complete answer. He reminds you of an example that reveals a pitfall that students and even teachers might fall in.

By the way, I hope you realize that i'm not an expert in differential equations myself. I've learned about many equations and solutions through the years, but this subject is so vast and never-ending. I could easily fall into the mentioned pitfalls, and I have to think very hard when solving problems like the ones you are showing. When my work leads me to an obscure differential equation, I sometimes have to do extensive searching to see if that problem has been solved before. If it hasn't, then I take a crack at it myself. If I then fail in finding my own solution, I do a numerical solution by computer. Not all equations are solvable by mathematical analysis, and sometimes a computer (numerical analysis) is the only way. Real-world problems have a tendency to introduce complications that make the textbook answers useless. Anyway, just learn as much as you can, and realize that you will never know all about this subject, and luckily you don't need to know it all to be a good engineer or scientist.

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• PG1995

#### PG1995

Joined Apr 15, 2011
818
Q1:

Personally, I think the author make much ado about nothing when he talks about there being a "dilemma". Mathematicians are much more afraid of infinity than engineers and physicists, it seems. The author is just trying to be careful when there is an infinity involved, but don't let that confuse you.

Take his line after equation 11. He says ...

$$-1+c=1+c$$ or $$-1=1$$

He has artificially created the dilemma by making this step, because this is not a valid step when c goes to infinity. In fact it is just the opposite. A proper limit would be as follows ...

$$\lim_{c\to \infty} (-1+c)=lim_{c\to \infty} (1+c)$$ results in $$c=c$$

Hence, there is really no dilemma, but rather this is a situation that requires caution. Previously, I showed you how to convert the problem by defining a new constant c3=1/c, which provides another viewpoint and avoids the problem with infinity.

You also compare different problems to each other in this question. I would recommend not doing this. Differential equations are a tricky subject and there are no rules that work in all cases. Just focus on each problem and learn the methods, solutions, tricks and pitfalls that each problem teaches.

You also seem bothered by the comment the author makes about losing the solution when y=0. Again, he is just being cautious with infinities. When you divide by y, then the validity of the operation breaks down when y=0. This does not mean that your resulting solution is not valid at y=0, it only means that your method of finding the solution is not valid at y=0. If you take your solution and plug it back into the original equation, and find that it works, then there is no problem because this method does not use division by zero. That method is the guess and check method and if you try a solution and it works, you have something valid.
Thanks a lot, Steve. It is very kind of you for all the help and advice.

I need to ask some follow-on questions. I hope you don't mind.

Question 1:
To me, -1+c = 1+c, is a valid dilemma because we end up with -1=+1. The author did not do something out of the way; he went from equation (10) to equation (11) using a valid method.

-1+c = 1+c could be a valid expression when c goes to infinity but here we aren't dealing with limits.

What I say above is just my opinion to let you see what is troubling me. I don't intend to mean that what you have said is wrong. In a nutshell, I still don't understand why we end up with -1+c=1+c => -1=1.

Question 2:
If you proceed from step A to step B using some assumption such as that denominator is not to be zero, then wouldn't that assumption also hold for the step B. I think it should because step B is simply an outcome of preceding step A. Here, on this page in equations (8), (9), (10), you can see y cannot take values +/-2 because it makes denominator zero. But equation (11) can still take the value y=2 if you put c=0. The equation (11) is just an outcome of preceding steps, i.e. equations (8), (9), (10). Please help me.

Many thanks.

With best regards
PG

#### steveb

Joined Jul 3, 2008
2,436
Question 1:
To me, -1+c = 1+c, is a valid dilemma because we end up with -1=+1. The author did not do something out of the way; he went from equation (10) to equation (11) using a valid method.

-1+c = 1+c could be a valid expression when c goes to infinity but here we aren't dealing with limits.

What I say above is just my opinion to let you see what is troubling me. I don't intend to mean that what you have said is wrong. In a nutshell, I still don't understand why we end up with -1+c=1+c => -1=1.
I really feel the author is wrong here. What I suspect is that he is deliberately saying something wrong in order to simplify the situation and cause the least confusion to the student. Notice that you are confused by what he wrote, but also confused by what I wrote. I suspect that, in his experience, he finds that students struggle at this point and he wants to avoid it. However, I think it is a mistake for him to avoid it by making a mathematical mistake. This does the student more harm than good. Let's look at this carefully.

Lets's take the equation c-1=c+1 and assume that c equals either plus or minus infinity. Is this equation correct? Yes, sure it is. What is infinity plus one? It is infinity. What is infinity minus one? It is still infinity. Both sides are equal.

Let's look at what he did. Again, assume c is infinity and substitute it in to the equation.

$$\infty -1=\infty +1$$

Now let's try to subtract infinity from both sides, and see if we end up with a valid equation.

$$\infty -1 -\infty=\infty +1 -\infty$$

The author now claims that this means -1=1, and says there is a dilemma. Where did he go wrong? His mistake is that infinity minus infinity is not a valid operation in mathematics. It is similar to 0/0, and is indeterminate. Take a look at the following reference.

http://en.wikipedia.org/wiki/Indeterminate_form

Question 2:
If you proceed from step A to step B using some assumption such as that denominator is not to be zero, then wouldn't that assumption also hold for the step B. I think it should because step B is simply an outcome of preceding step A.
Yes, you are absolutely correct here, and I didn't mean to imply otherwise above.

Here, on this page in equations (8), (9), (10), you can see y cannot take values +/-2 because it makes denominator zero. But equation (11) can still take the value y=2 if you put c=0. The equation (11) is just an outcome of preceding steps, i.e. equations (8), (9), (10). Please help me.
Ask yourself a question. Is equations 8, 9 and 10 the same equation you started with? Didn't you change the problem when you rearranged the equation? The original equation had nothing in the denominator that could go to zero, while the rearrangement induced problem points. So, these so called problem points are only an issue with the method. As the author says, the problem solutions are subsequently recovered, if you think carefully. You can then verify these recovered solutions by plugging them in to the original problem and making sure they work (y=-2 works and y=2 doesn't work). Once you find these solutions, forget about the method. Pretend that your friend suggested y=-2 and y=2 to you as possible answers and pretend you just decided to try them. You then discover that one of them works. Now show me the dilemma in this method. Remember, guess and check is one of the accepted methods of getting to the solution of differential equations. You can see that you can't do this stuff blindly by method only. You really have to think carefully.

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• PG1995

#### PG1995

Joined Apr 15, 2011
818
I really feel the author is wrong here. What I suspect is that he is deliberately saying something wrong in order to simplify the situation and cause the least confusion to the student. Notice that you are confused by what he wrote, but also confused by what I wrote. I suspect that, in his experience, he finds that students struggle at this point and he wants to avoid it. However, I think it is a mistake for him to avoid it by making a mathematical mistake. This does the student more harm than good. Let's look at this carefully.

Lets's take the equation c-1=c+1 and assume that c equals either plus or minus infinity. Is this equation correct? Yes, sure it is. What is infinity plus one? It is infinity. What is infinity minus one? It is still infinity. Both sides are equal.

Let's look at what he did. Again, assume c is infinity and substitute it in to the equation.

$$\infty -1=\infty +1$$

Now let's try to subtract infinity from both sides, and see if we end up with a valid equation.

$$\infty -1 -\infty=\infty +1 -\infty$$

The author now claims that this means -1=1, and says there is a dilemma. Where did he go wrong? His mistake is that infinity minus infinity is not a valid operation in mathematics. It is similar to 0/0, and is indeterminate. Take a look at the following reference.

http://en.wikipedia.org/wiki/Indeterminate_form

Yes, you are absolutely correct here, and I didn't mean to imply otherwise above.

Ask yourself a question. Is equations 8, 9 and 10 the same equation you started with? Didn't you change the problem when you rearranged the equation? The original equation had nothing in the denominator that could go to zero, while the rearrangement induced problem points. So, these so called problem points are only an issue with the method. As the author says, the problem solutions are subsequently recovered, if you think carefully. You can then verify these recovered solutions by plugging them in to the original problem and making sure they work (y=-2 works and y=2 doesn't work). Once you find these solutions, forget about the method. Pretend that your friend suggested y=-2 and y=2 to you as possible answers and pretend you just decided to try them. You then discover that one of them works. Now show me the dilemma in this method. Remember, guess and check is one of the accepted methods of getting to the solution of differential equations. You can see that you can't do this stuff blindly by method only. You really have to think carefully.
Hi Steve

Q1:
First of all, let me offer my thanks for your help and above all your patience! I can understand probably this thread is rather getting frustrating for you because it can become hard to keep track of things. I'm sorry. I too want to be done with it soon.

Sorry, but I don't understand why you are taking infinity into this. I think somewhere in my mind I understand what you are trying to tell but I can't make a right connections in my mind. You also said something similar here. Let's have a look on this equation: x+1 = x+1 => 1=1 (after we subtract 'x' from both sides). On Page #2 the author ends up with -1+c=1+c => -1=1 (after subtracting 'c' from both sides). See, no discussion about infinity. Even if you prove that -1+c=1+c is correct when c-->inf then it becomes a limit problem which we are not discussing. At the end, I have pasted the links to the original problem being discussed.

Q2:
general solution:
solution set or family of solution
particular solution: solution found for specific value of constant
singular solution: when constant functions such as y=2 or y=-4 directly satisfy the equation
trivial solution: when the equation is satisfied by y=0.

[Note: The above definitions aren't precise and might be incorrect]

What is a complementary solution?

Page #1:
http://img839.imageshack.us/img839/5540/ppage1.jpg
Page #2: http://img684.imageshack.us/img684/7079/ppage2.jpg
Page #3: http://img694.imageshack.us/img694/8255/ppage3.jpg

Best regards
PG

#### steveb

Joined Jul 3, 2008
2,436
Q1:
First of all, let me offer my thanks for your help and above all your patience! I can understand probably this thread is rather getting frustrating for you because it can become hard to keep track of things. I'm sorry. I too want to be done with it soon.
You are welcome, and this is not frustrating at all. It is good for me to review this material, as it has been decades since I studied it formally. I expect it is more frustrating for you because you need to learn it quickly and efficiently to balance your study time and other obligations. So, I'm sorry I haven't found the right viewpoint to explain this in a way that makes it clear.

Sorry, but I don't understand why you are taking infinity into this. ... Even if you prove that -1+c=1+c is correct when c-->inf then it becomes a limit problem which we are not discussing. At the end, I have pasted the links to the original problem being discussed.
The only reason for bringing infinity into the discussion is that the constant c needs to be infinity to let the solution give the y=-2 form. So, I'm just trying to show that the solution obtained is more general than the author gives it credit for. The fact that you are not at the point of discussing limits is probably the stumbling point here. So, you have a choice to either try to understand limits now, or hold off and don't worry too much about my explanations. I'm not really sure how else to answer your question. The author takes a very confusing approach in my opinion. I looked at it and just said, why bother with that. Just convert the equation to an alternate form by taking the original equation.

$$y=2{{1+c e^{4x}}\over{1-ce^{4x}}$$

and divide the numerator and denominator by c to get the following

$$y=2{{1/c+e^{4x}}\over{1/c-e^{4x}}$$

Perhaps this step caused you confusion. To clarify, when you divide the numerator and denominator by c, you don't change the answer because c/c equals one, and (1/c)/(1/c) equals one. This is true even in the limits as c goes to infinity or 1/c goes to zero. If you are uncomfortable with this idea, then just accept it for now and know that later you will understand it when you study limits more formally.

Now, redefine the constant, and let c_3=1/c, to get the following.

$$y=2{{c_3+e^{4x}}\over{c_3-e^{4x}}$$

Now when c_3 is zero, you get y=-2.

Q2:
general solution:
solution set or family of solution
particular solution: solution found for specific value of constant
singular solution: when constant functions such as y=2 or y=-4 directly satisfy the equation
trivial solution: when the equation is satisfied by y=0.

[Note: The above definitions aren't precise and might be incorrect]

What is a complementary solution?

I'm not really familiar with that term, but my best guess is that the complementary solution is the general homogeneous solution, or the solution when the forcing term is zero.

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• PG1995

#### Tesla23

Joined May 10, 2009
406
My 2 cents worth is that you shouldn't stress about the published solution, it falls down at the first step:

$$\frac{dy}{dx}=y^2-4 , y(0) = -2$$

SOLUTION

we put the equation in the form

$$\frac{dy}{y^2-4}=dx$$ but we note that this is only valid for $$y\neq2 , -2$$

Now we see that the initial condition is $$y(0) = -2$$, so this approach WON"T WORK! If it can't give us the derivative at the starting point, what hope has it of integrating the equation.

Investigating closely we see that for $$y = -2$$, $$\frac{dy}{dx}=0$$, so once $$y = -2$$ the derivative vanishes and $$y$$ stays constant.

So $$y = -2$$ is the answer.

• PG1995

#### PG1995

Joined Apr 15, 2011
818
$$y=2{{1+c e^{4x}}\over{1-ce^{4x}}$$ (1)

and divide the numerator and denominator by c to get the following

$$y=2{{1/c+e^{4x}}\over{1/c-e^{4x}}$$ (2)

Now, redefine the constant, and let c_3=1/c, to get the following.

$$y=2{{c_3+e^{4x}}\over{c_3-e^{4x}}$$ (3)

Now when c_3 is zero, you get y=-2.
Hi again, The entries in blue in the quoted post are my additions.

c_3 = 1/c

When c_3=0 in (3) , this means c is infinity which implies for (2) y=-2 when x=0, and for (1) the expression is undefined because y=infinity/infinity.

This means (1) isn't equivalent to (2) and (3). We also do the same thing while finding limits. When the original function has singularity at a point for which we want to find the limit, we transform the expression into a new one (which I have read isn't considered equivalent to the original expression) which gets rid of that singularity and gives us the limit which was our objective. The same things happens here. We are getting rid of problem points to reach the solution for differential equation and aren't really concerned about equivalency of the steps in the method. I hope it makes some sense.

Regards
PG