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

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)) #