# Where is the solution $y=0$ when solving a linear homogeneous constant-coefficient $n^{th}$-order ODE?

In my introductory differential equations class, we have been learning about solutions to the constant-coefficient $$n^{th}$$-order homogeneous differential equation $$\sum_{k=0}^n a_k\frac{d^ky}{dx^k}=0$$; the method to solve these has been to find the roots $$r=r_1,r_2,\dots,r_n$$ of the characteristic polynomial $$\sum_{k=0}^n a_kr^k=0$$ and use them to construct solutions of the form $$y_k=e^{r_kx}$$. In particular, we have been studying complex roots to the characteristic polynomial. We have shown that for a root $$r_k=a+bi$$, $$y=e^{ax}\cos(bx)$$ and $$y=e^{ax}\sin(bx)$$ are both solutions to the differential equation.

I noticed that if the root $$r_k$$ is real, then plugging in $$b=0$$ to the two equations for the complex solutions does in fact yield the solution $$y=e^{ax}=e^{r_kx}$$, as expected. However, it also yields an extra solution of $$y=0$$, as $$e^{ax}\sin(0x)=0$$. This solution obviously works for any linear homogeneous ordinary differential equation, as plugging in $$y=0$$ will cause all terms in the equation to go to zero. However, this information is lost in the real solution to the differential equation, which only gives the exponential solution. Why is this extra solution lost?

• Exponential solution… with a coefficient.
– user65203
Feb 25, 2019 at 21:31

When solving such equations, we obtain $$n$$ particular solutions $$e^{r_kx}$$. But a general solution is a linear combination of them, i.e. $$y=\sum_{k=1}^n C_ke^{r_kx}.$$ If all coefficients are zero, then we get $$y=0$$.
By linearity, if $$y$$ is a solution, so is $$\lambda y$$, where $$\lambda=0$$ is possible.