Solve dy/dx+y/x=y² by bernollies equation ?

2 Answers
Jul 1, 2018

Answer:

The general solution is #y=1/(-xlnx+Cx)#

Explanation:

The Bernouilli ODE is of the form

#y'+p(x)y=q(x)y^n#

The general solution is obtained by substituting

#v=y^(1-n)#

and solving

#1/(1-n)v'+p(x)v=q(x)#

Here,

The equation is

#dy/dx+y/x=y^2#

#p(x)=1/x#

#q(x)=1#

#n=2#

Divide both sides by #y^2#

#1/y^2dy/dx+1/(yx)=1#.....................#(1)#

Let #v=1/y#, #=>#, #yv=1#

Differentiating both side wrt #x#

#vdy/dx+y(dv)/dx=0#

#dy/dx=-y/v(dv)/dx#

Substituting in #(1)#

#-v^2y/v(dv)/dx+v/x=1#

#(dv)/dx-v/x=-1#...................#(2)#

The integrating factor is

#IF=e^(int(-1/x)dx) =e^(-lnx)=1/x#

Multiply #(2)# by #IF#

#1/x(dv)/dx-v/x^2=-1/x#

#d/dx(v/x)=-1/x#

Integrating both sides

#v/x=-lnx+C#

#v=-xlnx+Cx#

Substituting back #v=1/y#

#=>#, #1/y=-xlnx+Cx#

#y=1/(-xlnx+Cx)#

Jul 1, 2018

Answer:

#y = 1/(x(C - ln absx ))#

Explanation:

Bernoulli General Form:

  • #y'+P(x)y=Q(x)y^n qquad "with "{(P = 1/x),(Q = 1),(n = 2):}#

Standard substitution:

  • #z(x) = y^(1-n) = 1/y qquad :. z' = -1/y^2 y' #

So the DE #y'+y/x=y^2# becomes:

# - y^2 \ z' + y/x = y^2#

# z' - 1/(xy) = - 1 implies z' - z/x = - 1#

This is now linear, and can be solved by Integration factor:

  • #exp (int -1/x dx) = 1/x#

#1/x(z' - z/x) = - 1/x#

#(1/xz)^' = - 1/x#

#1/xz = - ln absx + C#

#z = x(C - ln absx) = 1/y#

#y = 1/(x(C - ln absx ))#