I'm currently trying to solve an ordinary differential equation using complex arithmetic. The equation is

$$y''+y=e^{-x}\left [ \cos(2x)-3\sin(2x) \right ]=g(x)$$

Summarizing, I see that $g(x)$ is of the special form: $$g(x)=e^{\alpha x}(a_0\cos\beta x+b_0\sin\beta x)$$ where $\alpha=-1$, $\beta=2$,$a_0=1$, and $b_0=-3$. We have that

$$g(x)=\Re \left \{ G(x) \right \}$$ where, summarizing, $G(x)$ is of the form $$G(x)=e^{(\alpha -i\beta )x}(a_0+ib_0)\rightarrow e^{(-1-2i)x}(1-3i)$$ So now I'm supposing that we can find a complex-valued solution $Y$ to the equation

$$L\left [ Y \right ]=Y''+Y=G(x)=e^{(-1-2i)x}(1-3i)$$

According to my textbook, the method of undetermined coefficients asserts that any differential equation of the form

$$L\left [ Y \right ]=e^{(\alpha \pm i\beta )x}\left [ (a_n+ib_n)x^n+...+(a_1+ib_1)x+(a_0+ib_0) \right ]$$

has a solution of the form

$$Y_p(x)=x^se^{(\alpha \pm i\beta )x}\left [ A_nx^n+...+A_1x+A_0\right ]$$,

where $A_n,...,A_0$ are complex constants and $s$ is the smallest nonnegative integer such that no term in this equation is a complex solution to the corresponding homogeneous equation $L\left [ Y \right ]=0$.

I'm confused on how to apply these statements to this particular problem. Seeing this done as an example would be very appreciated.

  • 1
    $\begingroup$ I ended up figuring it out, simply a matter of understanding what was written. $\endgroup$ Commented Nov 11, 2021 at 5:29

1 Answer 1


I suppose that your problem is with the particular solution. If we have an ODE such as

$$ y''+y = c_0 e^{(\alpha+i\beta)x} $$ a particular solution has the structure $y_p = c_1 e^{(\alpha+i\beta)x}$ and after substituting we have

$$ c_1(\alpha+i\beta)^2+c_1 = c_0 $$

and thus

$$ y_p = \frac{c_0}{(\alpha+i\beta)^2+1}e^{(\alpha+i\beta)x} $$


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