Separable differential equation

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Homework Statement



[itex]\frac{du}{dt}[/itex] = [itex]e^{5u + 7t}[/itex]

Solve the separable differential equation for u:

Use the following initial condition: u(0) = 6.

The Attempt at a Solution



I tried to take the natural log of each side but now I'm stuck. How can I separate the equation when both the u and t terms are an exponent? And how can I use u(0) = 6? Sorry if I'm clueless, I missed a bunch of lectures on separable differential equations.
 

Answers and Replies

  • #2
SammyS
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Homework Statement



[itex]\frac{du}{dt}[/itex] = [itex]e^{5u + 7t}[/itex]

Solve the separable differential equation for u:

Use the following initial condition: u(0) = 6.

The Attempt at a Solution



I tried to take the natural log of each side but now I'm stuck. How can I separate the equation when both the u and t terms are an exponent? And how can I use u(0) = 6? Sorry if I'm clueless, I missed a bunch of lectures on separable differential equations.

##a^{x+y}=a^x\cdot a^y ##
 
  • #3
AlephZero
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What do you remember about working with powers? Can you write an expression like ##e^{a+b}## a different way?

Edit: Sammy told you the answer while I was typng :smile:

The general solution to the differential equation will contain an arbitrary constant. You can eliminate that using the initial condition u(0) = 6.
 
  • #4
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Ok so I can rewrite as:

[itex]\frac{du}{dt}[/itex] = [itex]e^{5u}[/itex] * [itex]e^{7t}[/itex]

[itex]e^{-5u}[/itex] du = [itex]e^{7t}[/itex] dt

After finding the antiderivatives I cross-multiplied:

5[itex]e^{7t}[/itex] = -7[itex]e^{-5u}[/itex]

[itex]\frac{-7}{5}[/itex] = [itex]\frac{e^{7t}}{e^{-5u}}[/itex]

[itex]\frac{-7}{5}[/itex] = [itex]e^{7t + 5u}[/itex]

This is now where I'm running into a problem. When I try to natural log both sides to get rid of the e and bring down the 7t + 5u exponent, this means I have ln([itex]\frac{-7}{5}[/itex]) on the other side of the equation, which is unreal. Did I do something wrong or is there some sort of way to work around this?
 
  • #5
Dick
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Ok so I can rewrite as:

[itex]\frac{du}{dt}[/itex] = [itex]e^{5u}[/itex] * [itex]e^{7t}[/itex]

[itex]e^{-5u}[/itex] du = [itex]e^{7t}[/itex] dt

After finding the antiderivatives I cross-multiplied:

5[itex]e^{7t}[/itex] = -7[itex]e^{-5u}[/itex]

[itex]\frac{-7}{5}[/itex] = [itex]\frac{e^{7t}}{e^{-5u}}[/itex]

[itex]\frac{-7}{5}[/itex] = [itex]e^{7t + 5u}[/itex]

This is now where I'm running into a problem. When I try to natural log both sides to get rid of the e and bring down the 7t + 5u exponent, this means I have ln([itex]\frac{-7}{5}[/itex]) on the other side of the equation, which is unreal. Did I do something wrong or is there some sort of way to work around this?

You forgot the +C part when you integrated.
 
  • #6
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Do I add a constant to both sides of the equation?

[itex]e^{-5u}[/itex] + [itex]C_{u}[/itex] = [itex]\frac{-5e^{7t}}{7}[/itex] + [itex]C_{t}[/itex]
 
  • #7
SammyS
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Do I add a constant to both sides of the equation?

[itex]e^{-5u}[/itex] + [itex]C_{u}[/itex] = [itex]\frac{-5e^{7t}}{7}[/itex] + [itex]C_{t}[/itex]
That would be silly.

You could then replace [itex]C_{t}-C_{u}[/itex] with [itex]C[/itex]
 
  • #8
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Got it, thanks guys!
 

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