y' + xy = xy^-1 Solve the Bernoulli differential equation.

Given equation is y'+xy=xy^(-1)
 
An equation of the form y'+Py=Qy^n
is called as 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'+xy=xy^(-1)
and
y'+Py=Qy^n
on comparing we get
P=x , Q=x , n=-1
so the linear form of first order of the equation y'+xy=xy^(-1) is given as
 
=> (u')/(1-n) +P u =Q where u= y^(1-n) =y^2
=> (u')/(1-(-1)) +(x)u =x
=> (u')/2 +xu=x
=> u'+2xu = 2x
 
so this linear equation is of the form
u' + pu=q
p=2x , q=2x
so I.F (integrating factor ) = e^(int p dx) = e^(int 2x dx) = e^2(x^2)/2 = e^(x^2)
 
and the general solution is given as
u (I.F)=int q * (I.F) dx +c
=> u(e^(x^2))= int (2x) *(e^(x^2)) dx+c
=> u(e^(x^2))=  int (e^(x^2)) 2xdx+c
 
let us first solve
int e^(x^2) 2xdx
so , let t =x^2
dt = 2xdx
int e^(x^2) 2xdx = int e^(t) dt = e^t = e^(x^2)
 
so now => ue^(x^2)=  e^(x^2)+c
=>u=((e^(x^2))+c)/(e^(x^2))
 = 1 +ce^(-x^2)
but
u=y^2 ,so
y^2=(1 +ce^(-x^2))
y= sqrt (1 +ce^(-x^2))
is the general solution.

Comments

Popular posts from this blog

In “Fahrenheit 451,” what does Faber mean by “Those who don’t build must burn. It’s as old as history and juvenile delinquents”?

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 34

What was the effect of World War II on African Americans?