# yy' - 2y^2 = e^x Solve the Bernoulli differential equation.

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.