Question

Solve `dy/dx - 1/2(1+1/x)y+3/xy^3 = 0` ?

Answer 1

See below.

Explanation:

Making the substitution `u(x) = 1/(y(x))` and `u'(x) = -1/(y^2(x)) y'(x)`

we have

`( (1 + x) u^2 + 2 x u u'-6)/(2 x u^3)=0` and now supposing that `2 x u^3 ne 0` we have

`(1 + x) u^2 + 2 x u u'-6 = 0` Now making `z = u^2` we have the new linear differential equation

`x z'+(1+x)z-6=0`

This differential equation can be easily solved giving

`z =1/x(6-C e^-x)` where

and then

`u = pm sqrt( (C e^-x + 6)/x)` and finally

`y = 1/u = pm 1/sqrt( (C e^-x + 6)/x)`

NOTE:

`y(x)=0` is also solution.

Answer 2

This is a Bernoulli differential equation ; it can be recognized by the form:

`y'+ P(x)y= Q(x)y^n`

Please see the explanation.

Explanation:

Given: `dy/dx- 1/2 (1+1/x)y + 3/x y^3 =0`

Subtract `3/xy^3` from both sides:

`dy/dx- 1/2 (1+1/x)y = -3/x y^3`

Now we have the form:

`y'+ P(x)y= Q(x)y^n`

where `P(x) = - 1/2 (1+1/x)` and `Q(x) = -3/x`

The solution process of the Bernoulli differential equation is well known:

Multiply by `y^-n`:

`y'y^-3- 1/2 (1+1/x)y^-2 = -3/x`

Let `u = y^-2` then `u' = -2y^-3y'` which is better written as `-1/2u'=y'y^-3`:

`-1/2u'- 1/2 (1+1/x)u = -3/x`

Multiply both sides of the equation by -2:

`u'+(1+1/x)u = 6/x" [1]"`

We know that this form is solved by an integrating factor:

`I(x) = e^(int(1+1/x)dx)`

`I(x) = e^(x+ln|x|)`

`I(x) = xe^x`

Multiply equation [1] by `I(x)`:

`(u'+(1+1/x)u)xe^x = (6/x)xe^x`

`(u'+(1+1/x)u)xe^x = 6e^x`

Integrate both sides:

`int(u'+(1+1/x)u)xe^xdx = int6e^xdx`

We know that the left side integrates to `uI(x)` and the right side is trivial:

`uxe^x= 6e^x+ C`

Divide by `I(x)`:

`u = (6e^x)/(xe^x) + C/(xe^x)`

`u = (6+Ce^-x)/x`

Reverse the substitution:

`y^-2 = (6+Ce^-x)/x`

`y^2=x/(6+Ce^-x)`

`y = +-sqrt(x/(6+Ce^-x))`

Answer 3

`y^2 = (xe^x)/(6e^x + C)`

Alternatively:

`y =+-sqrt( (xe^x)/(6e^x + C))`

Explanation:

We have:

`dy/dx - 1/2(1+1/x)y+3/xy^3 = 0`

This is a Bernoulli equation 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 - 1/2(1+1/x)y+3/xy^3 = 0`

`:. -y^3/2 (du)/dx - 1/2(1+1/x)y+3/xy^3 = 0`

`:. (du)/dx + (1+1/x)/y^2-6/x = 0`

`:. (du)/dx + (1+1/x)u-6/x = 0`

`:. (du)/dx + (1+1/x)u = 6/x`

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 \ (1+1/x) \ dx )`
`\ \ = exp( x+lnx )`
`\ \ = e^(x)e^lnx`
`\ \ = xe^x`

And if we multiply the last by this Integrating Factor, `I`, we will (by virtue of the IF) have a perfect product differential;

`xe^x(du)/dx +xe^x (1+1/x)u = (6xe^x)/x`

`:. xe^x(du)/dx + (xe^x + e^x)u = 6e^x`

`:. d/dx(xe^xu) = 6e^x`

Which is now a trivial separable DE, so we can "separate the variables" to get:

`xe^xu = int \ 6e^x \ dx`

And integrating gives us:

`xe^xu = 6e^x + C`

Restoring the substitution we get:

`(xe^x)/y^2 = 6e^x + C`

`:. y^2/(xe^x) = 1/(6e^x + C)`

`:. y^2 = (xe^x)/(6e^x + C)`