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:

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

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

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

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:

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

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

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

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!