# Problem with determining the constants of the general solution of a differential equation

This should be easy, but for some reason I don't succeed in determining the constants of the solution of a differential equation. This general solution is $$\theta(x) = C_1e^{mx}+C_2e^{-mx}$$ or $$\theta(x) = A\cosh(mx)+B\sinh(mx)$$ And the following conditions apply: $$\theta(0) = \theta_0\\ \left .-k\frac{d\theta}{dx}\right |_{x=L}=h\theta_L = h\theta(L)$$ Where $m^2 = \frac{hP}{kA}$ and the solution should be: $$\theta = \theta_0\frac{\cosh(m[L-x])+(h/mk)\sinh[m(L-x)]}{\cosh(mL)+(h/mk)\sinh(mL)}$$ But I don't succeed in obtaining this solution. The question comes from a physical situation, but I thought that, although all the symbols have a physical meaning, it was better to ask it here than in the physics part of the site. It comes from this paper, at page 236-237: $\theta$ stands for $T-T_\infty$. The solution is equation (17-40). But I don't think that these physical details are needed to solve the question.

In my attempts I solved it already until I get: $$\theta = \theta_0 \left [\cosh(mx)-\frac{h\theta_L\sinh(mx)}{\theta_0km\cosh(mL)}-\frac{\sinh(mL)}{\cosh(mL)}\right ]$$

but now it feels like it is impossible to proceed, because I see no way to get rid of the $\theta_L$. I don't see any mistakes though in the calculations and I hope that someone can clarify this for me.

You need $\theta'(L)=-h\theta(L)/k$, so calculate $$\theta'(x)=Am\sinh (mx) + Bm\cosh (mx)$$ and $\theta'(L)=Am\sinh(mL)+Bm\cosh(mL)$, whereas $$\theta(L)=A\cosh(mL)+B\sinh(mL)$$ So you need $$A(m\sinh(mL)+\frac{h}{k}\cosh(mL))=-B(\frac{h}{k}\sinh(mL)+m\cosh(mL))$$ On the other side, $$\theta(0)=A$$ so $A=\theta_0$ and $$B=-\theta_0\frac{m\sinh(mL)+\frac{h}{k}\cosh(mL)}{\frac{h}{k}\sinh(mL)+m\cosh(mL)}$$ So $$\theta=\frac{\theta_0}{\frac{h}{k}\sinh(mL)+m\cosh(mL)}(\frac{h}{k}\cosh(mx)\sinh(mL)+m\cosh(mx)\cosh(mL) -m\sinh(mL)\sinh(mx)-\frac{h}{k}\cosh(mL)\sinh(mx))=$$ $$=\frac{\theta_0}{\frac{h}{k}\sinh(mL)+m\cosh(mL)}(\frac{h}{k}\sinh(m(L-x))+m\cosh(m(L-x)))$$ which is yours up to dividing denominator and numerator by $m$. I used just the standard formulae of addiction and subtraction for hyperbolic functions.