Separable differential equation

In summary, the equation can be solved for u using the initial condition u(0) = 6. However, when trying to calculate the natural log of each side, you run into a problem due to the e and 7t + 5u exponent in the equation.
  • #1
Loopas
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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.
 
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  • #2
Loopas said:

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
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.
 
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  • #4
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
Loopas said:
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
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
Loopas said:
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]
 
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  • #8
Got it, thanks guys!
 

1. What is a separable differential equation?

A separable differential equation is a type of differential equation where the dependent variable and the independent variable can be separated into two distinct functions, making it easier to solve.

2. How do you solve a separable differential equation?

To solve a separable differential equation, you need to separate the dependent and independent variables on opposite sides of the equation, and then integrate both sides separately. This will result in an equation with only one variable, which can be solved using algebraic methods.

3. What are the applications of separable differential equations?

Separable differential equations are used in various fields of science and engineering, such as physics, chemistry, and economics. They are particularly useful in modeling processes that involve exponential growth or decay, such as population growth or radioactive decay.

4. How do you know if a differential equation is separable?

A differential equation is separable if it can be written in the form dy/dx = f(x)g(y), where f(x) and g(y) are functions of x and y, respectively. In other words, if the dependent and independent variables can be separated into two distinct functions, the equation is separable.

5. Are there any limitations to solving separable differential equations?

While separable differential equations can be solved using relatively simple methods, not all differential equations are separable. In some cases, the equations may be too complex or have additional constraints that prevent them from being separable. In these cases, more advanced techniques, such as numerical methods, may be needed to solve the equations.

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