Solve One-on-One ODE with Step-by-Step Explanation

[twoflower]

Hi,

I have

$$y'' - y' = \frac{1}{e^{x} + 1}$$

What I did:

$$\lambda^2 - \lambda = 0$$

$$\lambda_1 = 0$$

$$\lambda_2 = 1$$FSS = $[1, e^x]$

$$y = a + be^x$$

$$y' = a' + b'e^x + be^x$$

Putting first two terms zero, I get

$$y'' = b'e^x + be^x$$

$$y'' - y' = b'e^x + be^x - be^x = b'e^x = \frac{1}{e^x + 1}$$

$$b' = \frac{1}{e^{2x} + e^x}$$

From the second condition

$$a' + b'e^x = 0$$

I get

$$a' = -\frac{e^x}{e^{2x} + e^x} = 1 - \frac{e^x}{1+e^x}$$

Anyway,

$$a' + b'e^x = 0$$

now isn't true. Where do I do the mistake?

Thank you very much.

 

[hotvette]

$$a' = -\frac{e^x}{e^{2x} + e^x} = 1 - \frac{e^x}{1+e^x}$$

Your mistake is in simplifying a'. Check it again.

 

[twoflower]

Your mistake is in simplifying a'. Check it again.

Thank you hotvette! I got it.