Two ODE's

ferry2

Hi, friends .

These are two equations, which were unable to resolve. Hope to help me. Note: this is not home, I just want to see how to resolve the equations. Thank answered.

[tex](2y-x+1)dx-(x-3y^2)dy=0[/tex]

Find the common solution of the Euler's eqution:

[tex](2x+1)^2y''-2(2x+1)y'+4y=0,[/tex] [tex]x>-\frac{1}{2}[/tex]

ross_tang

Hello ferry2, I have solved your 2nd equation here:

How to solve (2x+1)^2 y'' -2 (2x+1) y' +4 y = 0?

And the solution is given by:

[tex]y(x) = C_1(2x+1) +C_2 (2x+1) \ln (2x+1)[/tex]

ferry2

Thanks a lot Ross Tang! Can you tell something about first equation?

ross_tang

I tried various method in solving the 1st equation, but without any success. Sorry.

JJacquelin

Hello !

May be a typo in the 1st equation ? No difficulty if (2x-y+1) instead of (2y-x+1).
2nd equation : Let t=ln(2x+1) leads to
d²y/dt² -dy/dt +y =0
y(t) = exp(t)*(a*t+b)
and y(x) according to ross_tang formula.

ferry2

It is possible there have been a typo. Thank you both.