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I'm desperately trying to find the solution of this simple ODE: $\frac{dx}{dt}= C +\frac{x-a_1}{b_1} + \frac{x-a_2}{b_2} $

Where C is a constant. Someone has a clue?

Thanks for the feedback already: Ok some more info: I think I can solve this by substituting $x$ by $e^{t}$. In that case I get:

$ e^t = C+\frac{e^t-a_1}{b_1} + \frac{e^t-a_2}{b_2}$ But now I'm stuck. Does it mean x is just: $ x (1-1/b_1 -1/b_2)= (C-a_1/b_a -a_2/b_2) $ But then it is no longer depending on t... Iḿ doing something wrong here

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    Welcome to math,SE. What do you find difficult in the question?? Have you tries separating the variables??2012-10-13

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Hints:

  • Write the equation in the form: $x' + c_1 x = c_2$.
  • Use an integrating factor or notice that the equation is separable.

Don't forget to handle any special cases when calculating $c_1$ and $c_2$.

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    @user44558 See [here](http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx), particularly the first example. Which method(s) have you learned so far for solving linear first order differential equations?2012-10-13
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This is a time-invariant linear ODE with constant coefficients. The solution is therefore $x(t) = a+be^{\lambda t}$ for some $a,b,\lambda \in \mathbb{R}$. By substituting that for $x$ in your ODE and comparing the coefficients you can easily find the particular values of $a,b,\lambda$.