Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 66

Given equation is yy'-2y^2=e^x
=> y' -2y=e^x y^(-1)
An equation of the form y'+Py=Qy^n
is called the Bernoulli equation .
so, to proceed to solve this equation we have to transform the equation into a linear equation form of first order as follows
=> y' (y^-n) +P y^(1-n)=Q
let u= y^(1-n)
=> (1-n)y^(-n)y'=u'
=> y^(-n)y' = (u')/(1-n)
so ,
y' (y^-n) +P y^(1-n)=Q
=> (u')/(1-n) +P u =Q
so this equation is now of the linear form of first order
Now,
From this equation ,
y' -2y=e^x y^(-1)
and
y'+Py=Qy^n
on comparing we get
P=-2 Q=e^x , n=-1
so the linear form of first order of the equation y' -2y=e^x y^(-1) is given as

=> (u')/(1-n) +P u =Q where u= y^(1-n) =y^2
=> (u')/(1-(-1)) +(-2)u =e^x
=> (u')/2 -2u=e^x
=> (u')-4u = 2e^x

so this linear equation is of the form
u' + pu=q
p=-4 , q=2e^x
so I.F (integrating factor ) = e^(int p dx) = e^(int -4dx) = e^(-4x)

and the general solution is given as
u (I.F)=int q * (I.F) dx +c
=> u(e^(-4x))= int (2e^x) *(e^(-4x)) dx+c
=> u(e^(-4x))= int (2e^x) *(e^(-4x)) dx+c
=> u(e^(-4x))= 2 int (e^(-3x)) dx+c
=>u(e^(-4x))= 2 int (e^(-3x)) dx+c
=>u(e^(-4x))= 2 (1/(-3)*e^(-3x))+c as int e^(ax) dx = 1/a e^(ax).
=>u(e^(-4x))= (-2/3)*e^(-3x)+c
=> u = ((-2/3)*e^(-3x)+c)/(e^(-4x))
but u= y^2 so ,
y^2 = ((-2/3)*e^(-3x)+c)/(e^(-4x))
y= sqrt((-2/3e^(-3x)+c)/(e^(-4x)))
=sqrt((-2/3e^(-3x)+c)*(e^(4x)))
= sqrt((-2/3e^(x)+ce^(4x)))
=e^(x/2)sqrt((-2+3ce^(3x))/3)
is the general solution.