Question 1196935
<pre>

{{{"y'"-2y=y^2}}}

This is a Bernoulli differential equation with n = 2, the exponent of y

let u = y<sup>1-n</sup> = y<sup>1-2</sup> = y<sup>-1</sup>, then

{{{"u'"=-y^(-2)*"y'"}}}

We can make the first term of the original equation become u' by 
multiplying through by {{{-y^(-2)}}}

{{{-y^(-2)*"y'"-(-y^(-2))2y=(-y^(-2))y^2}}}

{{{-y^(-2)*"y'"+2y^(-1)=-y^0}}}

{{{-y^(-2)*"y'"+2y^(-1)=-1}}}

Now we also see that since y<sup>-1</sup> = u, the above becomes

{{{"u'"+2u=-1}}}

{{{du/dx+2u=-1}}}

{{{du+2udx=-dx}}}

That is separable as well as a linear differential equation. I will
solve it as a linear differential equation, with P = 2 and Q = -1.
so we find the integrating factor 

{{{matrix(2,1,"",e^int(P*dx))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",e^int(2*dx))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",e^(2x))}}} 
  
Multiplying through by e<sup>2x</sup>, 

{{{e^(2x)du+2u*e^(2x)dx=-e^(2x)*dx}}}

The left side is the differential of the product (e<sup>2x</sup>)(u),
so we can integrate both terms on the left together.  

{{{int((e^(2x)du+2u*e^(2x)dx))}}}{{{""=""}}}{{{int(-e^(2x)*dx)}}}

On the right we take out the negative sign, insert 2 and take out 1/2:

{{{e^(2x)*(u)}}}{{{""=""}}}{{{expr(-1/2)int(e^(2x)*2*dx)}}}

{{{e^(2x)*(u)}}}{{{""=""}}}{{{expr(-1/2)e^(2x)}}}{{{""+""}}}{{{C}}}

We are told y=3 when x=0

u = y<sup>-1</sup>, substituting y=3

u = 3<sup>-1</sup> = {{{1/3}}}

{{{e^(2*0)*(1/3)}}}{{{""=""}}}{{{expr(-1/2)e^(2*0)}}}{{{""+""}}}{{{C}}}

{{{1*(1/3)}}}{{{""=""}}}{{{expr(-1/2)*1}}}{{{""+""}}}{{{C}}}

{{{1/3}}}{{{""=""}}}{{{-1/2}}}{{{""+""}}}{{{C}}}

{{{1/3+1/2}}}{{{""=""}}}{{{C}}}

{{{5/6}}}{{{""=""}}}{{{C}}}

{{{e^(2x)*(u)}}}{{{""=""}}}{{{expr(-1/2)e^(2x)}}}{{{""+""}}}{{{5/6}}}

Since  u = y<sup>-1</sup>,  

{{{e^(2x)*(y^(-1))}}}{{{""=""}}}{{{expr(-1/2)e^(2x)}}}{{{""+""}}}{{{5/6}}}

Multiply through by 6

{{{6e^(2x)*(y^(-1))}}}{{{""=""}}}{{{-3e^(2x)+5}}}

{{{6e^(2x)*(1/y)}}}{{{""=""}}}{{{-3e^(2x)+5}}}

Multiply through by y

{{{6e^(2x)}}}{{{""=""}}}{{{-3ye^(2x)+5y}}}


{{{3ye^(2x)+6e^(2x)}}}{{{""=""}}}{{{5y}}}

Factor out 3e<sup>2x</sup> on the left

{{{3e^(2x)(y+2)}}}{{{""=""}}}{{{5y}}}

Swap sides:

{{{5y}}}{{{""=""}}}{{{3e^(2x)(y+2)}}}

{{{5y}}}{{{""=""}}}{{{3(y+2)e^(2x)}}}

Edwin</pre>