What is the general solution of the differential equation # dy/dx + y = xy^3 #?

Question

Isabella Ackerman · Accepted Answer

See below.

Making the change of variable #y = 1/z# we have the new version #dy/dx + y - xy^3=0 -> (x - z^2 + z z')/z^3=0# or #x - z^2 + z z' = 0# Now calling #xi = z^2# we have #x-xi+1/2 xi'=0# Solving for #xi# we obtain easily #xi =1/2+x+C e^(2x) = z^2# then #z = pm sqrt(1/2+x+C e^(2x) ) = 1/y# then finally #y = pm 1/sqrt(1/2+x+C e^(2x) )#

Paisley Johnson · Answer

#y = +-1/sqrt(x+1/2+ce^(2x)#

Given: #dy/dx + y = xy^3#

Multiply both sides by #y^-3#:

#dy/dxy^-3 +y^-2 = x" [1]"#

Let #u = y^-2#, then #(du)/dx = -2y^-3dy/dx#

Writing the differential in a form that is suitable for substitution into equation [1]:

#-1/2(du)/dx = dy/dxy^-3#:

Perform the substitutions:

#-1/2(du)/dx + u = x#

Multiply the equation by -2:

#(du)/dx - 2u = -2x" [2]"#

This is the well known form:

#(dy)/dx +P(x)y = Q(x)#

Where #P(x) = -2#

The integrating factor is:

#mu(x) = e^(int-2dx)#

#mu(x) = e^(-2x)#

Multiply both sides of equation 2 by #mu(x)#:

#(du)/dxe^(-2x) - 2e^(-2x)u = -2xe^(-2x)#

We know that the left side integrates to #mu(x)u#, therefore, we need only integrate the right side:

#e^(-2x)u = -2intxe^(-2x)dx#

#e^(-2x)u = -2(xe^(-2x)/-2+1/2inte^(-2x)dx)#

#e^(-2x)u = (x+1/2)e^(-2x)+C#

#u = x+ 1/2+ ce^(2x)#

Reverse the substitution:

#y^-2 = x+ 1/2+ ce^(2x)#

#y = +-1/sqrt(x+1/2+ce^(2x)#

Cameron Alexander · Answer

# y^2 = 2/(2x+1+Ce^(-2x)) #

Alternatively:

# y = +-sqrt(2)/sqrt(2x+1+Ce^(-2x)) #

We have: # dy/dx + y = xy^3 # This is a Bernoulli equitation which has a standard method to solve. Let: # u = y^(-2) => (du)/dy = -2y^(-3) # and #dy/(du) = -y^3/2 # By the chain rule we have; # dy/dx = dy/(du) * (du)/dx # Substituting into the last DE we get; # dy/(du) (du)/dx + y = xy^3 # # :. -y^3/2 (du)/dx + y = xy^3 # # :. - (du)/dx + 2/y^2 = 2x # # :. - (du)/dx + 2u = 2x # # :. (du)/dx - 2u = -2x # So the substitution has reduced the DE into a first order linear differential equation of the form: # (d zeta)/dx + P(x) zeta = Q(x) # We solve this using an Integrating Factor # I = exp( \ int \ P(x) \ dx ) # # \ \ = exp( int \ (-2) \ dx ) # # \ \ = exp( -2x ) # # \ \ = e^(-2x) # And if we multiply the last by this Integrating Factor, #I#, we will have a perfect product differential; # e^(-2x)(du)/dx - 2ue^(-2x) = -2xe^(-2x) # # d/dx(e^(-2x)u) = -2xe^(-2x) # Which is now a trivial separable DE, so we can "separate the variables" to get: # e^(-2x)u = int \ -2xe^(-2x) \ dx# And integrating by parts (skipped step) gives us: # e^(-2x)u = 1/2(2x+1)e^(-2x) + c# Restoring the substitution we get: # e^(-2x)y^(-2) = 1/2(2x+1)e^(-2x) + c# # :. y^(-2) = 1/2(2x+1) + ce^(-2x)# # :. 1/y^2 = 1/2(2x+1+Ce^(-2x))# # :. y^2 = 2/(2x+1+Ce^(-2x))#

Elijah Abram · Answer

undefined The general solution of the differential equation dy/dx + y = xy^3 is:

y(x) = [1 / (1 - x^2)]^(1/2)

This solution is obtained by separating variables and integrating both sides of the equation.