Solving heat equation in the given form which satisfy certain conditions

Question

I'm trying to solve the following heat equation: \begin{align} \frac{\partial u}{\partial t} = m\frac{\partial^2u}{\partial x^2} \\ u(0,x) = f(x) \\ u(t,0) = u(t, L) = 0 \end{align} where $m > 0$ The question asks us to find all solutions of the form $u(t,x) = \exp(-\lambda t)f(x)$ where $\lambda \in R$. I get that $ u(0,x) = exp(0)f(x) = f(x)$ is always met so we can ignore this condition. And $u(t,0) = u(t,L) = f(x) \Rightarrow f(0) = f(L) = 0$. Substituting $\frac{du}{dt} = m\frac{d^2u}{dx^2}$, we get that $f''(x) = -\frac{ \lambda}{m}f(x)$. I then tried to take different cases of $\lambda$. So we get that if $\lambda = 0$ then $f(x)$ is constant which is not possible. However, if $\lambda != 0 $, we either get real or complex roots. Suppose $\lambda < 0$, then we would get real roots for the second order differential equation $f''(x)$ so it will be of the form $f(x) = c_{1}e^{\sqrt{-\frac{\lambda}{m}}x} + c_{2}e^{-\sqrt{-\frac{\lambda}{m}}x}$. Now since $f(0) = 0$, we get that $f(0) = c_{1} + c_{2} = 0 \Rightarrow c_{1} = -c_{2}.$ Also since $f(L) = 0$, we get that $f(L) = c_{1}e^{\sqrt{-\frac{\lambda}{m}}L} + c_{2}e^{-\sqrt{-\frac{\lambda}{m}}L} = c_{2}\left(e^{-\sqrt{-\frac{\lambda}{m}}L} - e^{\sqrt{-\frac{\lambda}{m}}L}\right) = 0 \Rightarrow c_{2} = 0$ so $c_{1} = 0$ and $f(x) = 0$ and $U(t,x) = 0$. Could somebody please tell me where I'm going wrong?