Meteorology Differential Equations Problem Help | Boundaries & Constant Epsilon

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JerzeyDevil
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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:

[tex]\epsilon[/tex][tex]d^{2}[/tex][tex]\Psi[/tex]/dx[tex]^{2}[/tex] + d[tex]\Psi[/tex]/dx = -1

Boundary conditions:
[tex]\Psi[/tex] = 0
[tex]\Psi[/tex] = 0

[tex]\epsilon[/tex] = 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!
 
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JerzeyDevil said:
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:

[tex]\epsilon[/tex][tex]d^{2}[/tex][tex]\Psi[/tex]/dx[tex]^{2}[/tex] + d[tex]\Psi[/tex]/dx = -1

Boundary conditions:
[tex]\Psi[/tex] = 0
[tex]\Psi[/tex] = 0

[tex]\epsilon[/tex] = 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 er1t and er2t. The complementary solution will be yc = c1er1t + c2er2t

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.