Where did this substitution technique go wrong?

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In summary, I tried an alternative technique to see if it would work, but I'm still not sure how I went wrong. I'm exploring an alternative approach.
  • #1
askmathquestions
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We can solve ##y'(x) = (ax+b)y(x)## by rearranging to obtain ##\frac{y'}{y} = ax +b## and solving in terms of an exponential.

I tried an alternative technique to see if it would work, and somewhere I went wrong. The point of the technique is that a slightly simpler version of the problem should be "easier" to solve than the original version.

We start with ##y'(x) = (ax + b)y(x)## then make the substitution ##u = ax + b \implies \frac{u-b}{a} = x##. This is the tricky part now I think I am having trouble with. Is the left-hand side ##y'( \frac{u-b}{a} )## or is the left-hand side ##\frac{1}{a} y'( \frac{u-b}{a})## by the chain rule? I'm not entirely sure any more.

In either case, the next step is to relabel ##y( \frac{u-b}{a})## as ##z(u)##, so it's important that we know what is happening with the original left-hand side because we need to find a relationship for ##\frac{dz}{du}## by differentiating both sides of ##y( \frac{u-b}{a}) = z(u)##.

So the expression, after those couple substitutions, we should end up with an expression like ##z'(u) = \frac{u}{a}z(u)##, which may be incorrect, but then we could explicitly solve for ##z## and then back-substitute to obtain our original function ##y(x)## once we solve for ##z## in terms of ##u## explicitly.

In principle this is slightly easier than the original problem because there is no ##+b## term in the coefficient of ##z##.
 
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  • #2
You can treat ##\frac{dy}{y}## with no regard to the substitutions as
[tex]\frac{dy}{y}=(ax+b)dx=u \frac{1}{a}du[/tex]
[tex]\ln y = \frac{u^2}{2a}+C[/tex]
 
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  • #3
anuttarasammyak said:
You can treat ##\frac{dy}{y}## with no regard to the substitutions as
[tex]\frac{dy}{y}=(ax+b)dx=u \frac{1}{a}du[/tex]
[tex]\ln y = \frac{u^2}{2a}+C[/tex]
That's interesting, I'll have to play around with that. Although, I'm still looking to know specifically where the substitution I tried went wrong.
 
  • #4
[tex]\frac{dy}{dx}=(ax+b)y[/tex]
By the transformation
[tex]\frac{dy}{du}\frac{du}{dx}=uy[/tex]
[tex]a \frac{dy}{du}=uy[/tex]
[tex]\frac{dy}{y}=\frac{u}{a}du[/tex]

However I am afraid I do not catch fully what you are doing,
[tex]y'(x)=y'(\frac{u-b}{a})=\frac{d y(\frac{u-b}{a})}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{1}{a}\frac{dy}{du}[/tex] or so.
 
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  • #5
anuttarasammyak said:
[tex]\frac{dy}{dx}=(ax+b)y[/tex]
By the transformation
[tex]\frac{dy}{du}\frac{du}{dx}=uy[/tex]
[tex]a \frac{dy}{du}=uy[/tex]
[tex]\frac{dy}{y}=\frac{u}{a}du[/tex]

However I am afraid I do not catch fully what you are doing,
[tex]y'(x)=y'(\frac{u-b}{a})=\frac{d y(\frac{u-b}{a})}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{1}{a}\frac{dy}{du}[/tex] or so.
I don't quite catch your transformation. In my original problem, we make a substitution ##u = ax + b##, then we substitute ##y(\frac{u-b}{a}) = z(u)## to simplify. Somewhere along the way my differentiation assumption was incorrect.
 
  • #6
askmathquestions said:
We start with ##y'(x) = (ax + b)y(x)## then make the substitution ##u = ax + b \implies \frac{u-b}{a} = x##.

So the expression, after those couple substitutions, we should end up with an expression like ##z'(u) = \frac{u}{a}z(u)##, which may be incorrect,
It looks correct. This gives a general solution of$$z(u) = Ae^{\frac {u^2}{2a}}$$And$$y(x) = z(u) = z(ax + b) = Ae^{\frac{ax^2}{2} + bx + \frac{b^2}{2}}$$We can absorb the term ##e^{\frac{b^2}{2}}## into ##A## to give:
$$y(x) = Ae^{\frac{ax^2}{2} + bx}$$Note that it was simpler to solve the original equation directly:
$$\frac d{dx}\big (\ln y(x) \big ) = \frac{y'(x)}{y(x)} = ax + b$$$$\ln y(x) = \frac{ax^2}{2} + bx + C$$$$y(x) = Ae^{\frac{ax^2}{2} + bx}$$The substitution was unnecessary.
 
  • #7
PeroK said:
It looks correct. This gives a general solution of$$z(u) = Ae^{\frac {u^2}{2a}}$$And$$y(x) = z(u) = z(ax + b) = Ae^{\frac{ax^2}{2} + bx + \frac{b^2}{2}}$$We can absorb the term ##e^{\frac{b^2}{2}}## into ##A## to give:
$$y(x) = Ae^{\frac{ax^2}{2} + bx}$$Note that it was simpler to solve the original equation directly:
$$\frac d{dx}\big (\ln y(x) \big ) = \frac{y'(x)}{y(x)} = ax + b$$$$\ln y(x) = \frac{ax^2}{2} + bx + C$$$$y(x) = Ae^{\frac{ax^2}{2} + bx}$$The substitution was unnecessary.
I'm not saying the substitution was necessary, I'm exploring an alternative approach.

I'm still hung up on the ##u## substitution.

We have ##y'(x) = (ax+b)y(x)##, our original equation.

We make a substitution ##ax + b = u \implies \frac{u-b}{a} = x## giving us what exactly? I'm confused because I could interpret this as either

##\frac{d}{d\frac{u-b}{a}}y(\frac{u-b}{a}) = a \frac{d}{du}y(\frac{u-b}{a}) = y'(\frac{u-b}{a})## where the ##a## coefficients cancel due to the chain rule, which seems like it could be correct but somehow isn't,

or I could interpret this as ##\frac{d}{du}y( \frac{u-b}{a} ) = \frac{1}{a} y'(\frac{u-b}{a})## which also seems incorrect.
 
  • #8
I think you are better off introducing a new function ##z(u) = z(ax+b) = y(x)##.

If you do that, what could possibly go wrong?
 
  • #9
PeroK said:
I think you are better off introducing a new function ##z(u) = z(ax+b) = y(x)##.

If you do that, what could possibly go wrong?
I understand you think that, but my goal of this question is to understand how to utilize this substitution process. It seems helpful to understand the ##\frac{dy}{du} \frac{du}{dx}## argument too, though I don't know if that is directly analogous.
 
  • #10
askmathquestions said:
I understand you think that, but the goal of this question is to understand how to utilize this substitution process.
That is the mathematically correct way to do a substitution.
 
  • #11
PeroK said:
That is the mathematically correct way to do a substitution.
Okay, maybe it's not clear to me because there are many steps being left out.

We start with

## dy/dx = (ax+b)y(x) ## .

Okay, now what are you saying the next step should be?
 
  • #12
I have a maths tutorial on the multi-variable chain rule in the Insights section of this forum.

It might be a bit advanced for you at this stage, but it highlights the mathematical details behind the scenes.
 
  • #13
PeroK said:
I have a maths tutorial on the multi-variable chain rule in the Insights section of this forum.

It might be a bit advanced for you at this stage, but it highlights the mathematical details behind the scenes.
Sure, I can brush up on multivariable calculus at some point.
But, this still leaves me wondering about the original method. Can the original substitution I used work? Or is it intrisically flawed? What went wrong in my original approach? I still don't see concise answers to those questions.
 
  • #14
askmathquestions said:
Okay, maybe it's not clear to me because there are many steps being left out.

We start with

##y(x) = (ax+b)y(x)##.

Okay, now what are you saying the next step should be?
One of the things you are missing is that a derivative is a function. When you write ##y'(x)## that ##y'## is a function.

Similarly, if we have ##y(x) = z(ax +b)##, then that defines a new function ##z## and a new function ##z'##, which is the derivative of ##z##.

The chain rule says that$$y'(x) = az'(ax +b) = az'(u)$$You should check that out by using an example.

That then gives you the basis for your substitution.

Note that we define ##\frac{dz}{du} \equiv z'(u)##.

That insight justifies everything. Whereas, you got into a mess with:

askmathquestions said:
##\frac{d}{d\frac{u-b}{a}}y(\frac{u-b}{a}) = a \frac{d}{du}y(\frac{u-b}{a}) = y'(\frac{u-b}{a})##
Which is a real mess!
 
  • #15
PS in general the chain rule says that if$$h(x) = f(g(x))$$then$$h'(x) = f'(g(x))g'(x)$$That's one of the most important results in all mathematics and a full and precise understanding of this will help you enormously.
 
  • #16
PeroK said:
PS in general the chain rule says that if$$h(x) = f(g(x))$$then$$h'(x) = f'(g(x))g'(x)$$That's one of the most important results in all mathematics and a full and precise understanding of this will help you enormously.
Right, I'm aware of the chain rule, but it's ambiguous if I'm applying it correctly, and I also don't understand the intuition behind saying ##y(x) = z(ax+b)##. If ##y(x) = z(ax+b)##, what happens to the ##ax+b## coefficient of ##y## in the original equation?
 
  • #17
askmathquestions said:
Right, I'm aware of the chain rule, but it's ambiguous
No it isn't. If you think that, you haven't understood it.

askmathquestions said:
and I also don't understand the intuition behind saying ##y(x) = z(ax+b)##.
that's part of the substitution process.
askmathquestions said:
If ##y(x) = z(ax+b)##, what happens to the ##ax+b## coefficient of ##y## in the original equation?
##ax +b## gets replaced by ##u##. That's part of the substitution process.
 
  • #18
PeroK said:
No it isn't. If you think that, you haven't understood it.that's called a substitution!

##ax +b## isn't a coefficient. It's a function. A variable substitution leads to a composition of functions.
##d/dx f(g(x)) = f'(g(x)) \cdot g'(x)##

That's the chain rule, that's what I'm familiar with, no problem there.

Going back to the original problem, we have

##\frac{d}{dx} y(x) = (ax+b)y(x)## a non-autonomous differential equation.
Now I say ## u = ax + b## which also implies ##\frac{u-b}{a} = x## and so we have

##? = u y(\frac{u-b}{a})##

What is the left-side based on your understanding?
 
  • #19
The left hand side is ##az'(u)##, which can also be written as ##a\frac{dz}{du}##.

That's the chain rule that you are so sure you understand and yet unable to apply correctly!
 
  • #20
askmathquestions said:
##d/dx f(g(x)) = f'(g(x)) \cdot g'(x)##

That's the chain rule, that's what I'm familiar with, no problem there.

Going back to the original problem, we have

##\frac{d}{dx} y(x) = (ax+b)y(x)## a non-autonomous differential equation.
Now I say ## u = ax + b## which also implies ##\frac{u-b}{a} = x## and so we have

##? = u y(\frac{u-b}{a})##

What is the left-side based on your understanding?
Also, ##y(\frac{u-b}{a}) \equiv z(u)##.

Some people then get lazy and write ##y(u)## instead of ##z(u)##.

My argument against that is that it may confuse the inexperienced student. And that it's better to use the mathematically correct ##z(u)##.
 
  • #21
PeroK said:
The left hand side is ##az'(u)##, which can also be written as ##a\frac{dz}{du}##.

That's the chain rule that you are so sure you understand and yet unable to apply correctly!
There's intermediate steps that you're leaving out. My substitution appears to be different than yours, so keep that in mind.

So, we start with the equation

##\frac{d}{dx}y(x) = (ax+b)y(x)##. Then I make the substitution ##u = ax+b## which implies ##x = \frac{u-b}{a}##.

our equation becomes

##? = u y(\frac{u-b}{a})##

What is ##?## That's really the core of this question, is simply what that left-hand side should be after this substitution.

After we confirm what ##?## is, then I'll go back to your personal method and check if that works out too as a comparison.
 
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  • #22
I'm out!
 
  • #23
PeroK said:
I'm out!
Alright well I guess we have an unsolved mystery that stumps even the experienced members: what is ##?##?
 
  • #24
askmathquestions said:
Alright well I guess we have an unsolved mystery that stumps even the experienced members:
That's certainly true! An inability to learn is nothing to be proud of.
 
  • #25
PeroK said:
That's certainly true! An inability to learn is nothing to be proud of.
The ability to learn is equally as important as the ability to communicate, and communicate respectfully at that.

The derivative of ##\frac{d}{dx} log(x)^2## is ##\frac{2log(x)}{x}##.
The derivative of ##e^{\cos(x)}## is ##-e^{\cos(x)} \cdot \sin(x)##.
My issue isn't that a don't understand the chain rule, it's that I'm investigating one particular substitution method, and you're explaining something other than that.
 
  • #26
Okay, after putting down the computer screen and having a chance to figure out what's going on in this problem, I figured out the crucial step that was left out and causing me confusion.

We start with ##\frac{d}{dx}y(x) = (ax + b)y(x)##. Then we make a substitution ##u = ax + b \implies \frac{u-b}{a} = x.## This now gives us a new equation that I was confused on, but now after going back through it with a new realization:

##\frac{d}{d(\frac{u-b}{a})}y(\frac{u-b}{a}) = uy(\frac{u-b}{a})## which after applying the chain rule becomes

## y'( \frac{u-b}{a} ) = uy( \frac{u-b}{a} ).##

Now at this juncture, PeroK's suggestion now comes into play. We let ##z(u) = y( \frac{u-b}{a} )## and we have, as an intermediate step that I was also asking for, ##y'(\frac{u-b}{a}) = u z(u)##

then we differentiate both sides of our substitution ##z(u) = y( \frac{u-b}{a} )## (applying the chain rule) and to deduce ##y'## in terms of ##z'##:

##\frac{d}{du}z(u) = \frac{d}{du} y(\frac{u-b}{a}) \implies z'(u) = \frac{1}{a} \cdot y'( \frac{u-b}{a}) \implies az'(u) = y'( \frac{u - b}{a} ).##

So now we take that expression to complete our equation in terms of ##z##, leaving us with

##a \cdot z'(u) = u z(u) \implies z'(u) = \frac{u}{a} z(u).##

Now from here it follows that we can solve for ##z(u)## explicitly and then back-substitute to solve for our original equation in terms of ##x##.
 
  • #27
askmathquestions said:
##\frac{d}{d\frac{u-b}{a}}y(\frac{u-b}{a}) = a \frac{d}{du}y(\frac{u-b}{a}) = y'(\frac{u-b}{a})##
Newton's notation of ' for derivative coefficient is ambiguous wrt which parameter we make it when we apply transformation of variables. Do you need to use ' in your method ?
 
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  • #28
askmathquestions said:
##\frac{d}{d(\frac{u-b}{a})}y(\frac{u-b}{a}) = uy(\frac{u-b}{a})## which after applying the chain rule becomes

## y'( \frac{u-b}{a} ) = uy( \frac{u-b}{a} ).##
This isn't an application of the chain rule; it's just a definition of what you mean by ## y'( \frac{u-b}{a} )##. There is no chain rule to apply here because you are differentiating ##y## by its argument. Whether you write its argument as ##x## or ##\frac{u-b}{a}## is immaterial.

The only application of the chain rule I can see in your derivation is here:

askmathquestions said:
then we differentiate both sides of our substitution ##z(u) = y( \frac{u-b}{a} )## (applying the chain rule) and to deduce ##y'## in terms of ##z'##:

##\frac{d}{du}z(u) = \frac{d}{du} y(\frac{u-b}{a}) \implies z'(u) = \frac{1}{a} \cdot y'( \frac{u-b}{a}) \implies az'(u) = y'( \frac{u - b}{a} ).##
Here you have to apply the chain rule because you are differentiating ##y## with respect to ##u##, which is not the argument of ##y##; the argument of ##y## is a function of ##u##, so you have to differentiate that function first.
 

1. What is a substitution technique?

A substitution technique is a method used in scientific experiments to replace one variable with another in order to understand the relationship between the two variables. It is commonly used in fields such as genetics, chemistry, and physics.

2. How does a substitution technique work?

A substitution technique involves replacing one variable with another while keeping all other variables constant. This allows scientists to observe the effects of the substituted variable on the outcome of the experiment.

3. What are the potential sources of error in a substitution technique?

There are several potential sources of error in a substitution technique, including human error, equipment malfunction, and external factors such as temperature or humidity. It is important for scientists to carefully control and monitor these variables to ensure accurate results.

4. Can a substitution technique be used in all scientific experiments?

No, a substitution technique may not be applicable in all scientific experiments. It is most commonly used in experiments where the variables are easily controlled and manipulated, and where the relationship between the variables is well understood.

5. What are the benefits of using a substitution technique?

A substitution technique allows scientists to study the effects of a variable without having to change the entire experimental setup. It also helps to isolate the effects of a single variable, making it easier to draw conclusions and make predictions about the relationship between variables.

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