How to solve ODE y'' - 4y' + y - x = 0 using rk4?

[jmarcell...@ufpi.edu.br]

How to solve ODE y'' - 4y' + y - x = 0 using rk4? :(

[David Joyner]

On Mon, Sep 5, 2016 at 9:33 PM, <jmarcell...@ufpi.edu.br> wrote:
> How to solve ODE y'' - 4y' + y - x = 0 using rk4? :(
>

http://doc.sagemath.org/html/en/reference/calculus/sage/calculus/desolvers.html
lists dsolve_rk4.
Was there a specific problem you encountered?

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[jmarcell...@ufpi.edu.br]

I could not define a way to insert the arguments in RK4 command :(

[jmarcell...@ufpi.edu.br]

I tryed

# w = y''
# z = y'

var('x, w, z')

desolve_system_rk4( [w , 4*z + x] , [w,z] , ics = [0,1,0], ivar = x, end_points = 10, step = 0.1)

But does no work correty :(

[jmarcell...@ufpi.edu.br]

I got a way to solve the differential equation. If wrong please correct;)

To solve this differential equation we must first make a variable substitution to reduce the differential equation for a first order and thus create a system of ODEs. variable change:

w1 = y
w2 = y
w3 = y '

Thus, deriving the above variables, we have a system of equations of the form:

w1 '= w2
w2 '= f (x, w1, w2) = 4 * w2 + x

desolve_system_rk4([w2, 4*w2 + x ],[w1 , w2], ics = [0,1,0],ivar = x, step = 0.1, end_points = 2)