# Complex shape of general solution of differential equation [closed]

I am abit confused, in my definition for General Solution (definition which we used on Universitet) it say's for " $$C \in Reals$$ " and ever when i solving ordinary differential equation of first order i never get general solution or general integral in Complex shape ( with i ... ).

For example Riccati differential equation $$y' = (y^2) - 2 x^2 y + (x^4) + 2 x + 4$$

I was solving on my study like this

https://imgur.com/a/ngnMdDv


$$arctg(\frac{y-x^2}{2})-2x=C$$ and there i stop on my class .

When i try with DSolve , DSolve get me DSolve[y'[x] == y[x]^2-2x^2*y[x]+x^4+2x+4, y[x], x]

$$\left\{\left\{y(x)\to \frac{1}{c_1 e^{4 i x}-\frac{i}{4}}+x^2-2 i\right\}\right\}$$ Complex General solution in explicit shape !

When i try to Derivate that and etc i see that what DSolve gave me is rly Solution , but in my definition of General Solution in d.e its not!

My question is :

1. Is that what DSolve gived me General Solution, what is your definition of General Solution

On my studing , we use next definition : Function $$y=\varphi(x,c), x \in (a,b) , C \in H \subset R$$, where is C parametar is GENERAL SOLVE d.e y'=f(x,y) if :

1) equation $$y=\varphi(x,c)$$ can be solved for C, like $$C=\Psi(x,y), \forall (x,y) \in G$$ (G is region of unique)

2)Function $$y=\varphi(x,c)$$ is solve d.e $$y'=f(x,y)$$ in G for every $$C \in H$$ where is $$C=\Psi(x_y,y_0)$$ for any $$(x_0,y_0) \in G$$

1. Why DSolve gave me General Solution in that Complex shape , how DSolve works? If DSolve worked with subtitute, it would get General integral like me $$arctg(\frac{y-x^2}{2})-2x=C$$ , then because DSolve want this equation in EXPLICIT shape , next step would be $$arctg(\frac{y-x^2}{2})=C + 2x$$ then $$\frac{y-x^2}{2}=Tangens(C + 2x)$$ and explicit form $$y=2* Tangens(C + 2x) + x^2$$ ..
• Nasser, i know that in second-order differential equations we get Complex, but when i working manual first-order differential equations , i never saw general solution in Complex shape. I will edite now my question (explain how get General Integral and write definition of general solution ) . – Милош Вучковић Dec 2 '18 at 2:17
• Well if i didnt make any mistake in my step's it's for sure Solution ... – Милош Вучковић Dec 2 '18 at 3:07
• I find the question hard to read. Is the main question why M gives a solution that is a complex-valued function of a complex-variable instead of the form you desire as a real-valued function of a real variable? (The exact solvers of M work over the complex numbers by default.) Or are you claiming the DSolve solution is not a general solution? – Michael E2 Dec 2 '18 at 15:04

I'm not sure what's being asked, but I can think of two things to answer:

First, Mathematica's exact solvers solve systems over the complex numbers.

Second, if a way to transform the DSolve solution to the OP's desired form is being sought, if only to show the relationship between them, here is a sequence of steps that accomplishes that:

dsol = First@ DSolve[ode = y'[x] == y[x]^2 - 2 x^2*y[x] + x^4 + 2 x + 4, y, x]
cvt = First@ Solve[ 2 Tan[C[2] + 2 x] + x^2 == (y[x] /. dsol) /. {x -> 0}, C[1]]
intermediate = dsol /. cvt
y == (y[x] /. intermediate) // ExpToTrig // Simplify
(*
{y -> Function[{x}, -2 I + x^2 + 1/(-(I/4) + E^(4 I x) C[1])]}
{C[1] -> -(1/4) I Cos[2 C[2]] + 1/4 Sin[2 C[2]]}
{y -> Function[{x}, -2 I + x^2 +
1/(-(I/4) + E^(4 I x) (-(1/4) I Cos[2 C[2]] + 1/4 Sin[2 C[2]]))]}
x^2 + 2 Tan[2 x + C[2]] == y
*)


One of the differences in the two forms is that the constant C[1] of DSolve ranges over $${\Bbb{CP}}^1$$ and the OP's constant C[2] ranges over a generic subset of $$S^1 \equiv {\Bbb R}/2\pi{\Bbb Z}$$ or $$S^1 \times \big(i\,{\Bbb{R}\cup\{\pm i\infty\}\big)}$$. Topologically, the DSolve solution is perhaps simpler.