What is the general solution of the differential equation ? # 2xdy/dx = 10x^3y^5+y #

Question

Emma Aldridge · Accepted Answer

# y = (sqrt(x))/root(4)((C - 4x^5)) #

We have: # 2xdy/dx = 10x^3y^5+y # We can rewrite as: # dy/dx = 5x^2y^5+y/(2x) # # :. dy/dx - y/(2x) = 5x^2y^5 # This is a Bernoulli equitation which has a standard method to solve. Let: # u = y^(-4) => (du)/dy = -4y^(-5) # and #dy/(du) = -y^5/4 # By the chain rule we have; # dy/dx = dy/(du)*(du)/dx # Substituting into the last DE we get; # (-y^5/4)((du)/dx) -y/(2x) = 5x^2y^5 # # :. (du)/dx + 2y^(-4)/(x) = -20x^2 # # :. (du)/dx + 2/xu = -20x^2 # 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/x \ dx ) # # \ \ = exp( 2 lnx ) # # \ \ = exp( lnx^2 ) # # \ \ = x^2 # And if we multiply the last by this Integrating Factor, #I#, we will have a perfect product differential; # :. (du)/dx + 2/xu = -20x^2 # # :. x^2(du)/dx + 2xu = -20x^4 # # :. d/dx (x^2u) = -20x^4 # Which is now a trivial separable DE, so we can "separate the variables" to get: # x^2u = int \ -20x^4 \ dx# And integrating gives us: # \ \ \ \ \ x^2u = -20x^5/5 + C # # :. x^2u = -4x^5 + C # Restoring the substitution we get: # \ x^2y^(-4) = -4x^5 + C # # :. y^(-4) = (C - 4x^5)/x^2 # # :. \ \ \ y^4 = x^2/(C - 4x^5) # # :. \ \ \ \ \ y = (x^2/(C - 4x^5))^(1/4) # # :. \ \ \ \ \ y = (sqrt(x))/root(4)((C - 4x^5)) #

Evelyn Agard · Answer

undefined The general solution of the given differential equation is y(x) = [(x^5 - 1) / (5x^4)]^(1/4)