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=...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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.