# y' - y = y^3 Solve the Bernoulli differential equation. Given equation is y' -y =y^3

An equation of the form y'+Py=Qy^n

is called as the Bernoullis 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)

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Given equation is y' -y =y^3

An equation of the form y'+Py=Qy^n

is called as the Bernoullis 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' -y =y^3

and

y'+Py=Qy^n

on comparing we get

P=-1 , Q=1 , n=3

so the linear form of first order of the equation y' -y =y^3  is given as

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

=> (u')/(1-3) +(-1) u =1

=> (-u')/2 -u=1

=> u'+2u = -2

so this linear equation is of the form

y' + py=q

p=2 , q=-2

so I.F (integrating factor ) = e^(int p dx) = e^(int 2dx) = e^(2x)

and the general solution is given as

u (I.F)=int q * (I.F) dx +c

=> u(e^(2x))= int (-2) *(e^(2x)) dx+c

=> u (e^(2x))= (-2) int (e^(2x)) dx+c

let us solve int (e^(2x)) dx

=>let t= 2x

dt = 2dx

=> int (e^(t)) dt/2

=>1/2 (int (e^(t)) dt) = 1/2 e^t = (e^(2x))/2

so, int (e^(2x)) dx =(e^(2x))/2

so  ,now

u (e^(2x))= (-2) ((e^(2x))/2)+c

=>u (e^(2x))= -(e^(2x))+c

=> u = ((-(e^(2x)))+c)/(e^(2x))

but u=y^-2

so,

y^-2=((-(e^(2x)))+c)/(e^(2x))

=> y^2 = (e^(2x))/((-(e^(2x)))+c)

=> y = sqrt((e^(2x))/((-(e^(2x)))+c))

=> y = e^x/(sqrt(c-e^(2x)))

the general solution.

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