Find the general solution of following Riccati equation y y

Find the general solution of following Riccati equation: y\' = y/x + x^3 y^2 - x^5

Solution

Given that

y\'=y/x+x³y²-x⁵--------------------------(1)   

(1) is in the form of y\'=q₀(x)+q₁(x).y+q₂(x).y²

comparing (1) with above eqn, we get

q₀(x)=-x^{5 ;}q₁(x)= 1/x ;q₂(x)=x³

also we know that y=x is a solution

now let z=1/(y-y₁)=1/(y-x)

and also

dz/dx = -(q₁(x)+2y₁q₂(x))z-q₂(x)

=-[1/x+2x.x³]z-x³

dz/dx =-[1/x+2x⁴]z-x³

dz/dx +[1/x+2x⁴]z = -x^3,which is a linear diff.eqn.

which is in the form of

dz/dx+Pz=Q;

where P=1/x+2x⁴ ;Q=-x³

now I.F =exp[integral(Pdx)]

=exp[int(1/x+2x⁴)]

=x.exp(2x⁵/5)

so that the General solution is

z.(I.F) =int[Q(x).(I.F)dx]+C

ie.,

z.(x.exp[2x⁵/5])=int[-x³.x.exp(2x⁵/5)]

z.(x.exp[2x⁵/5])=int[-x⁴.exp(2x⁵/5)]

   =-1/2[exp[2x⁵/5] +C [Reason: let 2x⁵/5=t

5(2x⁴/5)dx=dt

   2x⁴dx =dt, so int[-x⁴.exp(2x⁵/5)]=int[-exp(t).dt/2]=-(1/2).expt]

substituting z value,we get

x.exp(2x⁵/5) =-(1/2).exp(2x⁵/5)+C is the required solution of given eqn.