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

[twoflower]

Hi,

I have

<br /> y&#039;&#039; - y&#039; = \frac{1}{e^{x} + 1}<br />

What I did:

<br /> \lambda^2 - \lambda = 0<br />

<br /> \lambda_1 = 0<br />

<br /> \lambda_2 = 1<br />FSS = [1, e^x]

<br /> y = a + be^x<br />

<br /> y&#039; = a&#039; + b&#039;e^x + be^x<br />

Putting first two terms zero, I get

<br /> y&#039;&#039; = b&#039;e^x + be^x<br />

<br /> y&#039;&#039; - y&#039; = b&#039;e^x + be^x - be^x = b&#039;e^x = \frac{1}{e^x + 1}<br />

<br /> b&#039; = \frac{1}{e^{2x} + e^x}<br />

From the second condition

<br /> a&#039; + b&#039;e^x = 0<br />

I get

<br /> a&#039; = -\frac{e^x}{e^{2x} + e^x} = 1 - \frac{e^x}{1+e^x}<br />

Anyway,

<br /> a&#039; + b&#039;e^x = 0<br />

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

Thank you very much.

 

[hotvette]

<br /> a&#039; = -\frac{e^x}{e^{2x} + e^x} = 1 - \frac{e^x}{1+e^x}<br />

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.