Eliminate the Arbitrary Constant

Question

y=Ax^2+Be^(4x)

there are only two AC (A&B) so I can only get up to the second derivative

Answer 1

I'm not sure what your question is.

Since y = Ax² + Be^4x,

so y' = 2Ax + 4Be^4x.

Therefore

0 = dA/dx = d((y' – 4y)/(2x(1 – 2x)))/dx.

From here you can easily get a 2nd-order homogeneous ordinary differential equation for y(x) whose coefficients are all quadratic polynomials in x.

Likewise 0 = dB/dx = d(e^–4x(y' – 2y/x)/(4 – 2/x))/dx leads to the same ODE for y(x).