# What is the general solution of the differential equation #xy'' = y'+x(y')^2#?

##### 1 Answer

# y = C-ln |x^2+B| #

#### Explanation:

We have:

# xy'' = y'+x(y')^2 # ..... [A]

Which is a Second Order Non-Linear ODE. As there is no term in

# v = y' => v'=y'' #

Substituting into the DE [A] we get:

# xv' = v+xv^2 # ..... [B]

Which has reduced the initial ODE to a First Order Non-Linear ODE (in this case a Bernoulli Equation), so now we attempt a second substitution of the form:

Substituting into the last DE [B], in conjunction with the chain rule we get;

# x(dv)/(dx) = v+xv^2 #

# :. x(dv)/(du) * (du)/(dx) = v+xv^2 #

# :. x(-v^2) * (du)/(dx) = v+xv^2 #

# :. - (du)/(dx) = 1/(xv)+1 #

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

# :. (du)/(dx) + u/x = -1 # ..... [C]

So the second 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 \ (1/x) \ dx ) #

# \ \ = exp( lnx ) #

# \ \ = x #

And if we multiply the last equation [C] by this Integrating Factor,

# \ \ \ x(du)/(dx) + u = -x #

# :. -d/dx ( xu ) = x #

Which is now a trivial separable DE, so we can integrate to get:

# -xu = int \ x \ dx #

# \ \ \ \ \ \ \ = x^2/2 + K #

And restoring the

# -x(1/v) = x^2/2 + K #

# :. -x/v = (x^2+2K)/2 #

# :. -v/x = 2/(x^2+A) #

# :. v = -(2x)/(x^2+A) #

And restoring the

# y' = -(2x)/(x^2+B) #

Which again is a First Order separable ODE, so again we *"separate the variable"* to get:

# int \ dy = -int \ (2x)/(x^2+B) \ dx#

Integrating we get:

# y = -ln |x^2+B| + C#

Which is the GS of [A]