Solving Differential Equation (x^2 + x + 1)y'' + (4x + 1)y' + 2y = 1

[MHD93]

Hello
Don't think this is homework, because I just want to know whether or not it has a solution

(x^2 + x + 1)y'' + (4x + 1)y' + 2y = 1

Thanks

 

[snipez90]

If you specify initial data y(x_0) = y_0 and y'(x_0) = y_1 for y over an interval [a,b], then y has a unique solution.

 

[JJacquelin]

I just want to know whether or not it has a solution

Stictly referring to the wording of the question, the answer is :
It has not only a solution, but a lot of solutions.
Some can be expressed with elementary function, others with special functions.

 

[epenguin]

If you know a little trick called Liebniz' theorem you can recognise the left-hand side as being nearly
[(x² + x + 1)y]'', in fact the equation is

[(x² + x + 1)y]'' - y' = 1

The first integration is then easy and I think the second can be done too.

 

[HallsofIvy]

The fact that the leading coefficient,x^2+ x+ 1 has no real zeros tells you this equation has a unique solution for any initial values.

 

[JJacquelin]

Hi !
I suppose that Mohammad_93 was already aware that his ODE has a solution which depends on the initial values.
I suppose that his question was to know if all those solutions are easy, or not, to be expressed in terms of usual functions so that he would be able, or not, to solve the equation.
From this perspective, the answer is :
Some solutions of the ODE can be expressed in terms of the combination of a finite number of elementary functions. So, it is easy to find the particular solutions of this kind.
Some solutions of the ODE can be expressed only in terms of special functions ( Hypergeometric or Incomplete Beta functions). So, these cases require more advanced mathematical knowledge.
As a consequence, the ability to solve the given ODE depends both on the initial values and on the mathematical level.