MAT244--2018F > Term Test 1

TT1 Problem 2 (morning)

**Weina Zhu**:

I am trying to go the cosh~sinh way, but cannot continue when finding the second solution, can someone help me out of this?

Thanks in advance. I tried to see classmates' post, but they gave the answer directly;D

**Samarth Agarwal**:

Integration of (cosh(x) + 1)/(e^x + 1)^2 wrt dx = -e^(-x)/2 + c

Therefore g = (e^x + 1)(-e^(-x)/2 + c)

**Victor Ivrii**:

Monika did everything right, I also commend her comment to Doris' post.

Weina

You are right that equation is

$$-2 (\cosh(x)+1)y''+2\sinh(x)y'+2y=0$$

and we can rewrite it as

$$ (\cosh(x)+1)y''-\sinh(x)y'-y=0$$

and then

$$W=C\exp\Bigl(\int \frac{\sinh(x)}{\cosh(x)+1}\,dx\Bigr) = C\exp\Bigl(\ln (\cosh(x)+1)\Bigr)=C_1(\cosh(x)+1).$$

But it is not another way, just another form.

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