# How to solve this differential equation ?

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

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 ?

• Try solving the characteristic equation – Dylan Dec 19 '18 at 15:39

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.