Separable differential equation

[Loopas]

Homework Statement

\frac{du}{dt} = e^{5u + 7t}

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.

 

[SammyS]

Homework Statement

\frac{du}{dt} = e^{5u + 7t}

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.

$a^{x+y}=a^x\cdot a^y$

 

[AlephZero]

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

The general solution to the differential equation will contain an arbitrary constant. You can eliminate that using the initial condition u(0) = 6.

 

[Loopas]

Ok so I can rewrite as:

\frac{du}{dt} = e^{5u} * e^{7t}

e^{-5u} du = e^{7t} dt

After finding the antiderivatives I cross-multiplied:

5e^{7t} = -7e^{-5u}

\frac{-7}{5} = \frac{e^{7t}}{e^{-5u}}

\frac{-7}{5} = e^{7t + 5u}

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(\frac{-7}{5}) 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?

 

[Dick]

Ok so I can rewrite as:

\frac{du}{dt} = e^{5u} * e^{7t}

e^{-5u} du = e^{7t} dt

After finding the antiderivatives I cross-multiplied:

5e^{7t} = -7e^{-5u}

\frac{-7}{5} = \frac{e^{7t}}{e^{-5u}}

\frac{-7}{5} = e^{7t + 5u}

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(\frac{-7}{5}) 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.

 

[Loopas]

Do I add a constant to both sides of the equation?

e^{-5u} + C_{u} = \frac{-5e^{7t}}{7} + C_{t}

 

[SammyS]

Do I add a constant to both sides of the equation?

e^{-5u} + C_{u} = \frac{-5e^{7t}}{7} + C_{t}

That would be silly.

You could then replace C_{t}-C_{u} with C

 

[Loopas]

Got it, thanks guys!