Question

Find another solution if y1=cosh(x) is a solution to y''-y=0 using reduction of order method

Answer 1

let y=u(x)y1

y=ucosh(x)
y'=usinh(x)+u'cosh(x)
y''=ucosh(x)+u'sinh(x)+u'sinh(x)+u''cosh(x)

y''-y=0=ucosh(x)+2u'sinh(x)+u''cosh(x)-ucosh(x)=2u'sinh(x)+u''cosh(x)

Answer 2

letting w=u' , you get the linear equation

w'cosh(x)-2sinh(x)w=0

Answer 3

Did you manage to find u(x) or w(x)?

Answer 4

i feel stupid -.- i solved to get u but never got y

Answer 5

i ended up with

\[cosh^2(x)w=c\]

=\[c\int{}{}sec^2h(x)\]=ctanh(x)

Answer 6

don't be, I'm more lost than you are most of the time lol

Answer 7

so w(x) = tanh(x)?

Answer 8

y=ucosh(x)=tanh(x)cosh(x)=sinh(x)

Answer 9

yes

Answer 10

the constant isn't really need since you're just trying to find the general

Answer 11

looks good and it looks like you definitely know what you're doing

Answer 12

i gotcha

Answer 13

no i didn't i totally was sitting here for hours trying to figure out what i did wrong but i hadn't done anything wrong. I just forgot to solve for y. I had u

Answer 14

lol glad you found the culprit

Answer 15

onto chapter 4.3 lol

Answer 16

I don't understand how, in your second post, one of the terms became negative. How did 2sinh(x)mu'(x) become negative after w(x) was substituted for mu'(x)?