I need a solution for this equation : y'-y^2 +2e^x y=e^2x +e^x

Question

thanks for support

Answer 1

The given differential equation is a Riccati one.

 If one solution of a 2nd order ODE is known, then it is known that another solution can be obtained by

quadrature, i.e., a simple integration. The same holds for the Riccati equation. If one particular

solution y₁ can

be found, the general solution is obtained as y = y₁ + u.

It can be readily verified that y₁= e^x is a solution. Then by substituting y = e^x + u we have

e^x + u^' = ( e^x + u )² − 2 e^x ( e^x + u ) + e^x + e^2x

u' = e^2x + 2 e^x u + u² −2 e^2x + 2 e^x u + e^2x

u' = u²

du / u² = dx

u = -1 / (x +c )

Hence the general solution is

y = e^x − 1 / (x +c ) , c ∈ ℜ