# Differential Equations Help!Watch

#1
By substuting v=1/y solve the differntial equation:

y' +2y= xy^2

Any help very much appreciated as well as reputation points earned. LOL
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#2
Any1?
0
12 years ago
#3
v=1/y

so y=1/v

this means y' = -1/(v^2)

and y^2 = 1/(v^2)

y' +2y= xy^2

-1/(v^2)+2/v=x/(v^2)

cross multiply the v^2 and cancel to give

-1+2/v=x
x=(2/v)-1

now integrate both sides to get

.5x^2=2lnv - v +c

but v=1/y so sub this in

.5x^2=2ln(1/y) - (1/y) +c

x^2=4 ln (1/y) - (1/y) +c

x = sqrt (ln (1/y)^4 - (1/y) +c)

if u have boundary conditions u can now find a particular solution and eleiminate c

hope this helps - please rep me if it does - i'm almost positive again!!!
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