Seperable Differential Equation

  • Thread starter mathor345
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  • #1
mathor345
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Homework Statement



[tex]\frac{dy}{dx} = e^{3x-3y} [/tex]

Homework Equations



[tex]\int e^{u}du = e^u + C[/tex]

The Attempt at a Solution



[tex]\frac{dy}{dx} = \frac{e^{3x}}{e^{3y}}[/tex]

[tex]e^{3y} dy = e^{3x} dx[/tex]

[tex] ln (e^{3y}) dy = ln (e^{3x}) dx[/tex]

[tex] 3y dy = 3x dx[/tex]

Integrate...

As you can see I'm doomed to get x = y
 
Last edited:

Answers and Replies

  • #2
36,330
8,289

Homework Statement



[tex]\frac{dy}{dx} = e^{3x-3y} [/tex]

Homework Equations



[tex]\int e^{u}du = e^u + C[/tex]

The Attempt at a Solution



[tex]\frac{dy}{dx} = \frac{e^{3x}}{e^{3y}}[/tex]

[tex]e^{3y} dy = e^{3x} dx[/tex]
OK to here. When you integrate, don't forget the constant of integration.

[tex] ln (e^{3y}) dy = ln (e^{3x}) dx[/tex]
 
  • #3
LeonhardEuler
Gold Member
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[tex]e^{3y} dy = e^{3x} dx[/tex]

[tex] ln (e^{3y}) dy = ln (e^{3x}) dx[/tex]

This last step doesn't work. If you tried to take the log of both sides you would get:
[tex] \ln{(e^{3y} dy)} = \ln{(e^{3x} dx)}[/tex]
which doesn't make sense. Try integrating both sides instead.
 
  • #4
epenguin
Homework Helper
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Look at the last but one line. d/dy of what gives you e3y ?
 
  • #5
mathor345
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This last step doesn't work. If you tried to take the log of both sides you would get:
[tex] \ln{(e^{3y} dy)} = \ln{(e^{3x} dx)}[/tex]
which doesn't make sense. Try integrating both sides instead.

[tex]\int e^{3y} dy = \int e^{3x} dx[/tex]

[tex]\frac{e^{3x}}{3} = \frac{e^{3y}}{3}[/tex]

[tex]e^{3x} + C = e^{3y} + C[/tex]

[tex]e^{3x} + C = e^{3y}[/tex]

[tex]ln (e^{3x} + C) = ln (e^{3y})[/tex]

[tex]ln (e^{3x} + C) = 3y[/tex]

[tex]\frac{1}{3} ln (e^{3x} + C) = y[/tex]


... I think this is the right answer, but it could very well be

[tex]\frac{1}{3} ln (e^{3y} + C) = x[/tex]

Depending on where you put the constant of integration?
 
Last edited:
  • #7
LeonhardEuler
Gold Member
860
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Both answers are correct. Changing where you put the 'C' just changes the value of 'C', not the form of the solution, but the first form is preferable because it gives an explicit function for y.

(If you ever want to know whether your answer is right, just plug it back in to the original differential equation and see if it works)
 
  • #8
mathor345
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Thanks for the help everyone.
 
  • #9
36,330
8,289
[tex]\int e^{3y} dy = \int e^{3x} dx[/tex]

[tex]\frac{e^{3x}}{3} = \frac{e^{3y}}{3}[/tex]
The above should be
[tex]\frac{e^{3x}}{3} = \frac{e^{3y}}{3} + C[/tex]

[tex]e^{3x} + C = e^{3y} + C[/tex]
You're not going to have the same constant on both sides. The best way around having to work with two different constants is to use a single constant on one side, usually the right side.
[tex]e^{3x} = e^{3y} + C[/tex]


[tex]e^{3x} + C = e^{3y}[/tex]
Or you could do it like you did above. That's fine, too.
[tex]ln (e^{3x} + C) = ln (e^{3y})[/tex]

[tex]ln (e^{3x} + C) = 3y[/tex]

[tex]\frac{1}{3} ln (e^{3x} + C) = y[/tex]


... I think this is the right answer, but it could very well be

[tex]\frac{1}{3} ln (e^{3y} + C) = x[/tex]

Depending on where you put the constant of integration?
 

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