## Two ODE's

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.

$$(2y-x+1)dx-(x-3y^2)dy=0$$

Find the common solution of the Euler's eqution:

$$(2x+1)^2y''-2(2x+1)y'+4y=0,$$ $$x>-\frac{1}{2}$$
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 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: $$y(x) = C_1(2x+1) +C_2 (2x+1) \ln (2x+1)$$
 Thanks a lot Ross Tang! Can you tell something about first equation?

## Two ODE's

I tried various method in solving the 1st equation, but without any success. Sorry.
 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.
 It is possible there have been a typo. Thank you both.