- #1

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solve using method of variation of parameters

y''-y = 2/(1+e^x)

y'' ==> second order

y''-y = 2/(1+e^x)

y'' ==> second order

- Thread starter plucker_08
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- #1

- 54

- 0

solve using method of variation of parameters

y''-y = 2/(1+e^x)

y'' ==> second order

y''-y = 2/(1+e^x)

y'' ==> second order

- #2

HallsofIvy

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It should be easy to see that e

The "variation of parameters" is to look for a solution of the form

y(x)= u(x)e

In fact, there are an infinite number of u(x), v(x) that would work.

Differentiating, y'= u'(x)e

ASSUME that u'(x)e

(Since there are an infinite number of u(x), v(x) that would work above, this is just "narrowing the search".)

With that assumption y'= u(x)e

Differentiating again, y"= u'(x)e

Plug that into the original equation and, because e

u'e

Integrate those solutions to find u and v and plug into y(x)= u(x)e

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