Meteorology Differential Equations Problem Help | Boundaries & Constant Epsilon

[JerzeyDevil]

Hi, I'm a meteorology major and my professor assumes we know how to do differential equations, and I did at one time, but I have seem to forgotten most of what to do in the past few years. I was just wondering if anyone could help me how to solve this problem...she gave us the answer and the starting point but I can't seem to get the answer she gave:

\epsilond^{2}\Psi/dx^{2} + d\Psi/dx = -1

Boundary conditions:
\Psi = 0
\Psi = 0

\epsilon = constant

It may be a bit hard to see in text but its psi(x=0) = psi(x=1) = 0 as the boundary condtions.

Any help would be appreciated!

 

[Mark44]

Hi, I'm a meteorology major and my professor assumes we know how to do differential equations, and I did at one time, but I have seem to forgotten most of what to do in the past few years. I was just wondering if anyone could help me how to solve this problem...she gave us the answer and the starting point but I can't seem to get the answer she gave:

\epsilond^{2}\Psi/dx^{2} + d\Psi/dx = -1

Boundary conditions:
\Psi = 0
\Psi = 0

\epsilon = constant

It may be a bit hard to see in text but its psi(x=0) = psi(x=1) = 0 as the boundary condtions.

Any help would be appreciated!

Let's get rid of the Greek letters first, to make it easier to type.

The differential equation is Cy'' + y' = -1. This is a linear, nonhomogeneous, 2nd-order DE.

The associated homogeneous DE is Cy'' + y' = 0, and its characteristic equation is Cr² + r = 0. Solve the characteristic equation for r to get r₁ and r₂. Your two linearly independent solutions to the homogeneous DE will be e^r₁t and e^r₂t. The complementary solution will be y_c = c₁e^r₁t + c₂e^r₂t

Next, find a particular solution to the DE.

After finding the particular solution, the general solution will be the complementary solution plus the particular solution. Use the initial conditions to find c₁ and c₂.

I haven't filled in many of the details, since you said you knew how to do this sort of thing at one time. Hopefully this will be enough to get you started.