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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?? – TheJoker Oct 13 '12 at 13:29


  • 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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Ok, but what would be the solution to $x′+c_1x=c_2$? – user44558 Oct 13 '12 at 14:19
@user44558 See here, particularly the first example. Which method(s) have you learned so far for solving linear first order differential equations? – Ayman Hourieh Oct 13 '12 at 17:32

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 $a,b,\lambda$.

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