Real General solution of differential

Question

For next Riccati d.e $ y' = (y^2) - 2 x^2 y + (x^4) + 2 x + 4 $ with DSolve i get Complex general solution

Opres = DSolve[y'[x] == y[x]^2-2x^2*y[x]+x^4+2x+4, y[x], x]

$\left\{\left\{y(x)\to \frac{1}{c_1 e^{4 i x}-\frac{i}{4}}+x^2-2 i\right\}\right\}$

Like you told me here in some my posts that DSolve working with Complex numbers and etc, i want you ask next:

On my universitet in Serbia in our definition for General Solution we are taking only REAL"S General Solution. My teacher told that we took that definition from Rusian's book's, and we do not take complex general solutions.

My question: Is it posible command DSolve to do only with real's numbers and give me only real's general solution ?

Tnx.

Answer 1

Will this be enough?

ComplexExpand performs operations assuming variables will be real.

sol=ComplexExpand[y[x]/.DSolve[y'[x] == y[x]^2-2x^2*y[x]+x^4+2x+4, y[x], x][[1]]]

which gives you

x^2 + (C[1]*Cos[4*x])/(C[1]^2*Cos[4*x]^2 + (-1/4 + C[1]*Sin[4*x])^2) + 
I*(-2 + 1/(4*(C[1]^2*Cos[4*x]^2 + (-1/4 + C[1]*Sin[4*x])^2)) - 
(C[1]*Sin[4*x])/(C[1]^2*Cos[4*x]^2 + (-1/4 + C[1]*Sin[4*x])^2))

The second term appears to be the complex component.

Try solving to make that second term==0.

Simplify[Solve[(-2 + 1/(4*(C[1]^2*Cos[4*x]^2 + (-1/4 + C[1]*Sin[4*x])^2)) -
  (C[1]*Sin[4*x])/(C[1]^2*Cos[4*x]^2 + (-1/4 + C[1]*Sin[4*x])^2))==0,C[1]]]

which gives you

{{C[1] -> -1/4}, {C[1] -> 1/4}}

If you substitute those into your original solution then

sol/.C[1]->-1/4//FullSimplify

gives you

2 + x^2 - 4/(1 + Tan[2*x])

and

sol/.C[1]->1/4//FullSimplify

gives you

2 + x^2 + 4/(-1 + Cot[2*x])

both of which appear to be real solutions.

Please check all this very carefully to make certain there are no mistakes.