Can't seem to figure out how to solve for C Diff EQ question

In summary, the conversation discusses solving a separable differential equation for u with an initial condition, using the given equation and initial condition to find the correct value of C, and the mistake of taking the natural log in the process. The solution is found to be u(t)=-\frac{1}{3}\ln \left( -\frac{3}{10}e^{10t}+\frac{3}{10} +e^{-18} \right).
  • #1
mr_coffee
1,629
1
Hello everyone I'm stuck, i got right to the end of this intial value problem and it doens't seem to want to work.

Solve the separable differential equation for u
{du}/{dt} = e^{3 u + 10 t}.
Use the following initial condition: u(0) = 6.
u = ?

I might have screwed up here:
du = e^(3u)*e^(10*t) dt
I took the ln of both sides but can u take the ln of du? that's where i might have screwed up, if it is, what do u suggest?

Here is my work:
http://img75.imageshack.us/img75/613/lastscan0nu.jpg [Broken]

I tried this and it was wrong, i also tried the answers of 1, ln(5), also was wrong, any ideas on where i screwed up? also the problem says find U, not C, but i don't think i have my C right, once i find the correct C, then do i plug it into one of the equations above invovling C? Like this:
u = (e^(10*t^2/2)+e^(ln(5)/3))^3; This is also incorrect i just submitted!
Thjanks!
 
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  • #2
How does

[tex] du =e^{3u} e^{10t} dt [/tex]

goes to

[tex] du = 3u10t dt [/tex]

:confused:
 
  • #3
Here you go, the easy way

Solve the separable differential equation for u:

[tex]\frac{du}{dt} = e^{3u + 10t}[/tex]

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

Rearranging the given differential equation gives

[tex] e^{-3u}du= e^{10t}dt[/tex]

integrate both sides

[tex] \int e^{-3u}du = \int e^{10t}dt [/tex]

to get

[tex]-\frac{1}{3}e^{-3u} = \frac{1}{10}e^{10t} +C[/tex]

solving for u gives

[tex]u(t)=-\frac{1}{3}\ln \left( -\frac{3}{10}e^{10t}-3C\right) [/tex]

to solve for the value of C that satisfies the initial condition u(0)=6 (the easy way,) substitute t=0 and u(0)=6 into

[tex]-\frac{1}{3}e^{-3u} = \frac{1}{10}e^{10t} +C[/tex]

to get

[tex]-\frac{1}{3}e^{-3(6)} = \frac{1}{10}e^{10(0)} +C[/tex]

which simplifies to

[tex]-\frac{1}{3}e^{-18} = \frac{1}{10} +C[/tex]

yielding

[tex]C= -\frac{1}{10} -\frac{1}{3}e^{-18} [/tex]

so the particular solution is

[tex]u(t)=-\frac{1}{3}\ln \left( -\frac{3}{10}e^{10t}-3\left( -\frac{1}{10} -\frac{1}{3}e^{-18} \right) \right) = -\frac{1}{3}\ln \left( -\frac{3}{10}e^{10t}+\frac{3}{10} +e^{-18} \right) [/tex]
 
  • #4
Cyclovenom said:
How does
[tex] du =e^{3u} e^{10t} dt [/tex]
goes to
[tex] du = 3u10t dt [/tex]
:confused:

Yea that's waht I was wondering as well. Taking the natural log when you did is really not advisable, and doesn't help because you get ln(du) which is just going to make things harder for you. You started out pretty well changin the sum in the exponent into the product, but you should notice then that the equation is seperable and fairly simple to integrate once you separate it.
 
  • #5
Thank you everyone for your tips and yes Benion that is exactly what i got too after redoing it! thanks everyone! :)
 

1. What is C Diff EQ?

C Diff EQ stands for "complex differential equation" and refers to a type of mathematical equation that involves both independent and dependent variables, as well as their derivatives.

2. Why is it difficult to solve for C Diff EQ?

C Diff EQs are difficult to solve because they are often nonlinear and have no analytical solution. This means that they cannot be solved using traditional algebraic methods and instead require advanced techniques such as numerical methods or approximations.

3. How can I solve for C Diff EQ?

There are various techniques that can be used to solve for C Diff EQ, such as separation of variables, substitution, and power series methods. It is important to understand the type of equation you are dealing with and choose the appropriate method for solving it.

4. What are some tips for solving C Diff EQ?

Some tips for solving C Diff EQ include identifying the type of equation, understanding the initial conditions, checking for symmetry or other special properties, and using appropriate simplifications or transformations if possible.

5. Where can I find resources for solving C Diff EQ?

There are many online resources and textbooks available for solving C Diff EQ, including step-by-step guides, practice problems, and video tutorials. It may also be helpful to consult with a math tutor or professor for personalized assistance.

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