Solving $(y'' + 2iy' + 1)y = 0$

Question

$$(y'' + 2iy' + 1)y = 0$$

I tried to solve this problem by substituting $y=e^{mx}$ to find the auxiliary equation and solve further. But I couldn't solve it.

Answer 1

The first solution is given by $$y(x)=0$$ then we Substitute $$y(x)=e^{\lambda x}$$ for the homogeneous part. For a particular solution make the ansatz $$y_p=a_1x$$ For your Control: the solution is given by $$y(x)=C_1+\frac{ix}{2}$$

Answer 2

This is a linear differential equation and can be solved as

$$ y = y_h+y_p\\ y''_h+2iy'_h=0\\ y''_p+2iy'_p=1 $$

The homogeneous solution is obtained by substituting $y_h = e^{\lambda x}$ giving

$$ \lambda^2+2i\lambda = 0 \Rightarrow \lambda\in \{0,-2i\} $$

and then

$$ y_h = \frac{C_0}{2}e^{-2ix}+C_1 $$

and for $y_p$ we have

$$ y_p = \frac{i x}{2} $$

and finally

$$ y = \frac{i x}{2}+ \frac{C_0}{2}e^{-2ix}+C_1 $$

joined with $y = 0$

Answer 3

$$(y'' + 2iy' + 1)y = 0$$

you have $y=0$ as a solution and also $$y'' + 2iy' + 1 = 0$$ Substitute $z=y'$ $$z+2iz=-1$$ Use $e^{2ix}$ as integrating factor $$(ze^{2ix})'=-e^{2ix} \implies ze^{2ix}=-\int e^{2ix} dx=-e^{2ix}/2i+K$$ $$\implies y'=Ke^{-2ix}-\frac 1 {2i}$$ Integrate to get $y(x)$ $$y=C_1e^{-2ix}+C_2+\frac {ix}2$$