Differential equations substitution problem

In summary: Try v = ye^x.This was not the substitution I initially went with, but it works. I was able to solve it from here. Thank you for your help.
  • #1
lat77
6
0

Homework Statement


Solve using an appropriate substitution
ydx+(1+ye^x)dy=0


Homework Equations


N/A


The Attempt at a Solution


u=y
du/dx=dy/dx
u+(1+ue^x)du/dx=0

This is where it gets really sticky for me. I can't see it being a separable variable problem because it seems impossible to separate the u and the x. It also seems very unlikely that it is a linear d.e. or a bernoulli d.e. because the integrating factor would be e^∫(1/(1+ue^x)dx ) which seems impossible to do. The answer is e^(-x)=ylny+cy.
 
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  • #2
welcome to pf!

hi lat77! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
lat77 said:
The answer is e^(-x)=ylny+cy.

since they give you the answer,

a "cheat" way is to get the constant on its own

e-x/y = lny + C …

does that help? :smile:
 
  • #3
Not really. But on top of that, I need to know how to do it because I have a test Tuesday, so I need to be able to work the problem forwards since the answers obviously won't be given on the test.
 
  • #4
I tried many things, integrating factor did not seem practical (how to discover integrating factor is e^(-x)/y^2?)). Bernoulii equation does not apply. So, since they say subsititution, I tried what i thought might be natural:

Your substitution u=y is not great, can you see why?

How about u=1+ye^x or u=ye^x.

Unfortunately, these got me close to a Bernoulli differential equation, but no cigar. Please let us know what the method ends up being!

EDIT: Did this come from a textbook or another source, it may help if we have a context of similar problems, a chapter, an ambient discussion et cetera.
 
  • #5
lat77 said:

Homework Statement


Solve using an appropriate substitution
ydx+(1+ye^x)dy=0

Homework Equations


N/A

The Attempt at a Solution


u=y
du/dx=dy/dx
u+(1+ue^x)du/dx=0

This is where it gets really sticky for me. I can't see it being a separable variable problem because it seems impossible to separate the u and the x. It also seems very unlikely that it is a linear d.e. or a bernoulli d.e. because the integrating factor would be e^∫(1/(1+ue^x)dx ) which seems impossible to do. The answer is e^(-x)=ylny+cy.
The substitution u = y does nothing !

While you won't be given a solution on your test, following tim's advice to work the problem "backwards" may give you insight into the process of finding a substitution.
 
  • #6
Try v = ye^x.
lat77 said:
Not really. But on top of that, I need to know how to do it because I have a test Tuesday, so I need to be able to work the problem forwards since the answers obviously won't be given on the test.
The way I came with the substitution was to try separating, but it failed. So I tried v = 1 + ye^x, but it failed. So I tried v = ye^x and it worked. A lot of differential equations don't have immediately obvious methods for solving them and you just have to try methods.

Also, you should really consider why you tried u = y and why there is no hope that could of lead to something.
 
Last edited:
  • #7
Jorriss said:
The way I came with the substitution was to try separating, but it failed. So I tried v = 1 + ye^x, but it failed. So I tried v = ye^x and it worked. A lot of differential equations don't have immediately obvious methods for solving them and you just have to try methods.

Nice, Joriss! :smile:

lat77, that's the way you have to solve these things …

you try something, it doesn't work, you try something else, it doesn't work, you … get the picture? :wink:

just keep bashing your head against that brick wall until something gives! :biggrin:

Just what makes that little old ant
Think he'll move that rubber tree plant?
Anyone knows an ant, can't
Move a rubber tree plant

But he's got high hopes, he's got high hopes
He's got high apple pie, in the sky hopes

So any time you're gettin' low
'stead of lettin' go
Just remember that ant

Oops there goes another rubber tree
Oops there goes another rubber tree
Oops there goes another rubber tree plant!​

Good luck on Tuesday! :smile:
 
  • #8
tiny-tim said:
...

lat77, that's the way you have to solve these things …

you try something, it doesn't work, you try something else, it doesn't work, you … get the picture? :wink:

just keep bashing your head against that brick wall until something gives! :biggrin:

Just what makes that little old ant
Think he'll move that rubber tree plant?
Anyone knows an ant, can't
Move a rubber tree plant

But he's got high hopes, he's got high hopes
He's got high apple pie, in the sky hopes

So any time you're gettin' low
'stead of lettin' go
Just remember that ant

Oops there goes another rubber tree
Oops there goes another rubber tree
Oops there goes another rubber tree plant!​

Good luck on Tuesday! :smile:
Is there somewhere on PF to nominate the post of the day ... or week ... or month ... ?

I think tim deserves a nomination !
 

1. What is a differential equation substitution problem?

A differential equation substitution problem is a type of mathematical problem where a substitution is made in order to solve a differential equation. This substitution can simplify the equation and make it easier to solve.

2. How do you solve a differential equation substitution problem?

To solve a differential equation substitution problem, you must first identify which substitution should be made. This is often done by looking for patterns or similarities in the equation. Once the substitution is made, you can then solve the new equation using standard methods such as separation of variables or integration.

3. What are some common substitutions used in differential equations?

Some common substitutions used in differential equations include trigonometric substitutions, exponential substitutions, and power substitutions. Each of these substitutions is useful for solving different types of differential equations.

4. Can differential equation substitution problems be solved by hand?

Yes, differential equation substitution problems can be solved by hand using various mathematical techniques. However, for more complex equations, it may be necessary to use computer software or numerical methods to obtain an accurate solution.

5. How are differential equation substitution problems used in real-world applications?

Differential equation substitution problems are used in a variety of real-world applications, such as modeling physical systems, predicting population growth, and analyzing financial markets. They are also commonly used in engineering, physics, and economics to solve complex problems and make predictions.

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