Solve dy/dx−1/2(1+1/x)y+3/xy^3=0 ?

Ciolan3u 2022-09-11 Answered
Solve d y d x - 1 2 ( 1 + 1 x ) y + 3 x y 3 = 0 ?
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Answers (1)

Kimberly Evans
Answered 2022-09-12 Author has 13 answers
Making the substitution u ( x ) = 1 y ( x ) and u ( x ) = - 1 y 2 ( x ) y ( x )
we have
( 1 + x ) u 2 + 2 x u u - 6 2 x u 3 = 0 and now supposing that 2 x u 3 0 we have
( 1 + x ) u 2 + 2 x u u - 6 = 0 Now making z = u 2 we have the new linear differential equation
x z + ( 1 + x ) z - 6 = 0
This differential equation can be easily solved giving
z = 1 x ( 6 - C e - x ) where
and then
u = ± C e - x + 6 x and finally
y = 1 u = ± 1 C e - x + 6 x
NOTE:
y(x)=0 is also solution.

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