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I've been trying to solve this equation $$-\mu u'' + \beta u' = 1$$ where $u(0) = 0$, $u'(1) = 1$. So far the result I have is $$u = \frac{(exp(\frac{\beta x}{\mu})-1)\mu}{\beta exp(\frac{\beta}{\mu})}$$ but when I tried to substitute it on the equation the result is not the same.

What am I missing? I know that to get the solution we have to find two "solutions" the general and the specific one, can anyone explain how to solve this?

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Break the problem up into two parts: getting the general solution of the homogeneous problem as well as finding a particular solution to the nonhomogeneous problem.

Homogeneous ODE: $-\mu u''+\beta u'=0$

Use the standard methods for second order, linear, constant coefficient ODEs to get $u_h(x)=c_1+c_2e^{\beta x/\mu}$, the general solution of the homogeneous problem.

Nonhomogeneous ODE: $-\mu u''+\beta u'=1$

There are multiple methods to find a particular solution $u_p$ of the nonhomogeneous problem. I prefer the method of undetermined coefficients here, but your mileage may vary. Doing so yields $u_p(x)={1\over \beta}x$.

Then, the general solution to the nonhomgeneous problem is $u(x)=u_h(x)+u_p(x)$.

Finally, you would need to apply the two given boundary conditions to find the two constants in $u(x)$, but this is just a little calculus and algebra.

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Yes thanks, I did well for the general solution, the particular solution was the missing part, though. For the boundary conditions is easy also. I will read through the links you posted and understand the method to find the particular solution. Thanks. –  BRabbit27 Dec 15 '12 at 20:44
    
Glad to help. Not saying this to help myself, but rather the site and its users: if you come across and answer (or question!) that you find helpful or interesting, please upvote it by clicking on the up arrow above the number at the top right of an answer/question. This helps others discern the good/helpful postings from the not-so-good/helpful stuff. Thanks. –  JohnD Dec 15 '12 at 20:50
    
Just to clarify it. I did what it says in the link you put and I found that $u_p(x) = Ax^2+Bx+C$ which gives me $u_p(x) = \frac{-x}{\mu}$ ... am I right? –  BRabbit27 Dec 15 '12 at 21:50
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@BRabbit27: That can't be $u_p$ since substituting that back into the left-hand side of the DE would give $0+\beta\cdot(-1/\mu)$ which is not equal to 1 as desired. Instead, think like this: the right-hand side of ODE is a constant, so we will try a constant. That doesn't work (do it!)---and I can explain how you could look at $u_h$ and know $u_p=A$ (constant) won't work---so by the method of undetermined coefficients you "power" up your original guess for $u_p$ to be $u_p=Ax$. Plug that back into the ODE and you will see $A=1/\beta$ (which can also be checked directly). –  JohnD Dec 15 '12 at 22:01
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@BRabbit27: You got it. –  JohnD Dec 15 '12 at 22:58
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Assuming $\mu$ and $\beta$ are constants, you could immediately integrate both sides. However, in this case, it might be easier to proceed like this.

$$u''-\frac\beta\mu u'=-\frac1\mu$$ $$e^{-\frac\beta\mu x}u''-\frac\beta\mu e^{-\frac\beta\mu x}u'=(e^{-\frac\beta\mu x}u')'=-\frac1\mu$$ $$e^{-\frac\beta\mu x}u'=-\frac1\mu x+C$$ $$u'=-\frac1\mu xe^{\frac\beta\mu x}+Ce^{\frac\beta\mu x}$$

We can evaluate our first constant now. Had we just integrated both sides from the beginning, we would have to wait as our initial values are for 2 different values of $x$.

$$1=-\frac1\mu e^{\frac\beta\mu}+Ce^{\frac\beta\mu}$$ $$C=e^{-\frac\beta\mu}+\frac1\mu$$

Are you sure you copied those initial values correctly by the way? Ah well, from here you have $u'=f(x)$. Integrate both sides and use $u(0)=0$ for your final answer.

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