--- Already managed to solve it.
Just to give the whole context:
Since the fin has constant cross-section along its length and the area and perimeter are also constant, I can use:
$$\frac{\partial² T(x) }{\partial x²}=\frac{hP(T(x)-T_e)}{kA} (1) $$
$$m²=\frac{hP}{kA} $$
$$ϴ(x) = T(x)-T_e$$
$$\frac{\partial² ϴ(x) }{\partial x²}=\frac{\partial² T(x) }{\partial x²}$$
Substitute all of that in (1):
$$\frac{\partial² ϴ(x) }{\partial x²}-m²ϴ=0$$
$$ϴ(x)=c_1e^{mx}+c_2e^{-mx}$$
I know that if: $$x = 0, T = T_w, ϴ = T_w - T_e$$
Now I have to find the constants. I'm assuming it involves trying to substitute cosh(mx) and sinh(mx) into this but I'm not having much luck. What came into my mind was:
$$cosh(mx)=\frac{e^{mx}+e^{-mx}}{2}$$
$$e^{mx}=2cosh(mx)-e^{-mx}$$
$$sinh(mx)=\frac{e^{mx}-e^{-mx}}{2}$$
$$e^{-mx}=e^{mx}-2sinh(mx)$$
Substituting that on (2)
$$ϴ(x)=c_12cosh(mx)-c_1e^{-mx}+c_2e^{mx}-c_22sinh(mx)$$
If x = 0
$$ϴ_w=c_12cosh(m0)-c_1e^{-m0}+c_2e^{m0}-c_22sinh(m0)$$
$$ϴ_w=2c_1-c_1+c_2$$
$$ϴ_w=c_1+c_2$$
Sadly I can't see how to progress from here.
$$cosh(mx)=\frac{e^{mx}+e^{-mx}}{2}$$
$$e^{mx}=2cosh(mx)-e^{-mx}$$
$$sinh(mx)=\frac{e^{mx}-e^{-mx}}{2}$$
$$e^{-mx}=e^{mx}-2sinh(mx)$$
$$e^{mx}=2cosh(mx)-e^{mx}+2sinh(mx)$$
$$e^{mx}=cosh(mx)+sinh(mx)$$
$$e^{-mx}=cosh(mx)+sinh(mx)-2sinh(mx)$$
$$e^{-mx}=cosh(mx)-sinh(mx)$$
$$ϴ(x)=c_1(cosh(mx)+sinh(mx))+c_2(cosh(mx)-sinh(mx))$$
$$ϴ(x)=(c_1+c_2)cosh(mx) + (c_1-c_2)sinh(mx)$$
$$ϴ(x)=c_1cosh(mx) + c_2sinh(mx)$$
$$x = 0, T = T_w, ϴ_w = T_w - T_e$$
$$ϴ_w = c_1cosh(m0) + c_2sinh(m0)$$
$$c_1=ϴ_w$$
$$ϴ(x) = ϴ_wcosh(m0) + c_2sinh(m0)$$
$$\frac{dϴ(x)}{dx}=ϴ_wmsinh(m0) + c_2mcosh(m0)$$
Fourier law: $$q(x) = -kA_c\frac{dT(x)}{dx}=-kA_c\frac{dϴ(x)}{dx}$$
Newton law of cooling: $$q(x) = hA_c(T-T_e) = hA_cϴ$$
$$-kA_c\frac{dϴ(x)}{dx} = hA_cϴ$$
$$-k\frac{dϴ(x)}{dx} = hϴ$$
$$-k(ϴ_wmsinh(mx) + c_2mcosh(mx)) = h(ϴ_wcosh(mx) + c_2sinh(mx))$$
$$c_2 = \frac{-ϴ_w(\frac{km}{h}sinh(mx)+cosh(mx)}{\frac{km}{h}cosh(mx)+sinh(mx)}$$
$$ϴ(x) = ϴ_wcosh(m0) + (\frac{-ϴ_w(\frac{km}{h}sinh(mx)+cosh(mx)}{\frac{km}{h}cosh(mx)+sinh(mx)})sinh(m0)$$
$$ϴ_w = T_w-T_e$$
$$ϴ = T-T_e$$
$$T(x) = (T_w-T_e)[cosh(mx) + (\frac{-(\frac{km}{h}sinh(mx)+cosh(mx)}{\frac{km}{h}cosh(mx)+sinh(mx)})sinh(mx)]+T_e$$
I can't believe I managed to solve this, lol