Differential Equation HelpWatch

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#1
Can anybody help me solve this:

Code:
`dy/dx = 1/( e^y - x )`
0
14 years ago
#2
i can't see how to do this by one of the standard a level methods -- however, i might be wrong -- but you can get somewhere with it "by inspection".

look at the right hand side

for
1/(e^y - x)
to take a "nice" form you need e^y to be some multiple of x.

So assume y=ln(kx) where k is just some number to be worked out and see what happens

left hand side
==========

y=ln(kx) => dy/dx = k/kx = 1/x

right hand side
===========

y=ln(kx) => 1/(e^y - x) = 1/(e^(ln(kx)) - x) = 1/(kx - x) = 1/((k-1)x)

so, if the differential equation is satisfied
==============================

1/x = 1/((k-1)x) => 1 = k -1 => k = 2

Hence solution
===========

y = ln(2x)

however
======

the solution should have a constant of integration in it somewhere...i have missed this out in the above...try putting it a few places and see which one cancels on both sides might work...there again maybe not...hopefully the above is some help, tho. the following might help also
http://www.efunda.com/math/ode/ode1_implicit.cfm
0
#3
Nice solution. Thanks for the help.

I think you are right about the constant, it doesn't seem to work for y = ln(2Cx) or y = ln(2x) + C because the left hand side will always be 1/x and the right will only be 1/x for k = 2
0
14 years ago
#4
(Original post by mikesgt2)
Can anybody help me solve this:

Code:
`dy/dx = 1/( e^y - x )`
This is easy to solve if we note that dy/dx = 1/(dx/dy) and use this to change the independent variable. Then we use the standard integrating factor technique.

dy/dx = 1/(e^y - x) <=> dx/dy = e^y - x

dx/dy = e^y - x
dx/dy + x = e^y
(e^y)(dx/dy) + (e^y)x = e^(2y)
x(e^y) = Integral(e^(2y) dy)
x(e^y) = (1/2)e^(2y) + C
x = (1/2)(e^y) + C(e^(-y))

I hope this helps,
0
#5
Thanks... that is a lot nicer.
0
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