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$$\frac{\partial^2 u}{\partial x^2} + \frac{1}{L} \frac{\partial u}{\partial x} = 0$$

The solution is given as $$u = \exp(-\frac{1}{L}x) + 0.2 ,$$ with far field boundary condition as $u_\infty=0.2$.

Can anyone tell how does one end up with that kind of solution ?

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    $\begingroup$ Try solving the characteristic equation $\endgroup$ – Dylan Dec 19 '18 at 15:39
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First, we can integrate with respect to $x$: $$u' +\frac u L = c_0, \quad c_0\text{ is a constant}$$

This is easy to integrate (hint: look at the derivative $\left(\exp\left(\frac xL\right)u\right)'$), too, it gives $$\exp\left(\frac xL\right)u=Lc_0\exp\left(\frac xL\right) +c_1,$$or $$u=Lc_0 +c_1\exp\left(-\frac xL\right)$$ Plug your boundary conditions for $x\to+\infty$, it would give you $Lc_0=0.2$. You will need some other condition to conclude that $c_1=1$.

edit How to solve a differential equation for constant $a$ and $b$ $$u' + au = b$$ The standard trick is to multiply by a well-chosen exponent in order to obtain "a derivative of a product" on the left-hand side. We know that $(\exp(ct))' = c\exp(ct)$, so what happens if we multiply both sides by $\exp(at)$: $$u'(t)\exp(at) + a\exp(at)u(t) = \left(u(t)\exp(at)\right)' = b\exp(at).$$ Now we have a derivative on the left-hand side and something that does not depend on $u$ on the right-hand side, so we can happily integrate: $$u(t) \exp(at) = \frac{b}{a}\exp(at) + d.$$ Multiply by $\exp(-at)$ to arrive to the final result. This approach has multiple generalizations, but the idea remains the same - multiply the equation to recover a derivative of a product in order to reduce the number of terms containing the derivative of the unknown function.

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  • $\begingroup$ Thanks. Can you briefly explain the second line..how do you get that without looking at the solution? $\endgroup$ – newstudent Dec 19 '18 at 15:26
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    $\begingroup$ @newstudent see edit $\endgroup$ – TZakrevskiy Dec 19 '18 at 16:03

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