solve x dy/dx + y = xy^3
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Making the change of variable y=1z we have the new version
dydx+y−xy3=0→x−z2+zz'z3=0 or
x−z2+zz'=0
Now calling ξ=z2 we have
x−ξ+12ξ'=0
Solving for ξ we obtain easily
ξ=12+x+Ce2x=z2 then
z=±√12+x+Ce2x=1y then finally
y=±1√12+x+Ce2x
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