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Consider heat equation $$\frac{\partial h}{\partial t}= \text{D}\frac{\partial^{2} h}{\partial x^{2}}$$ I want to solve heat\diffusion equation with this Neumann boundary condition $$\frac{dh(0,t)}{dx}=\frac{1}{a}{g(0,t)-h(x,t)}$$ Fourier cosine transfer for second derivative of x is given as $$F_{c}(\frac{d^{2}h(x,t)}{dx^{2}})= U_{c}(w,t)= -w^{2}U_{c}(w,t)-\sqrt{\frac{2}{\pi}}\frac{dh(0,t)}{dx} $$ Consider Fc to be fourier cosine operator which converts h(x,t) into Uc(w,t). Fourier cosine transfer of time derivative is given as $$F_{c}(\frac{dh}{dt})=\frac{d}{dt}U_{c}{(w,t)}$$ The problem is, by solving heat equation using Fourier cosine transform with above mentioned neumann boundary condition we will end up having h(x,t) at both sides of resulting equation. $$\frac{\partial}{\partial t}U_{c}{(w,t)}=\text{D}(-w^{2}U_{c}(w,t)-\sqrt{\frac{2}{\pi}}\frac{\partial h(0,t)}{\partial x})$$ by substituting values of above mentioned boundary condition we get following ODE $$\frac{d}{dt}U_{c}{(w,t)}=\text{D}(-w^{2}U_{c}(w,t)-\sqrt{\frac{2}{\pi}}\frac{1}{a}{g(0,t)-h(x,t)})$$ Because Uc is a transformed form of h(x,t) which we just assumed and it will be transformed back when we apply inverse fourier transform How can I solve this ODE so that h(x,t) can be later taken to LHS. We have initial condition $$h(0,t)=0 $$

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    $\begingroup$ This link describes problem arising by solving this last ODE. $\endgroup$ – Ather Cheema Mar 2 '16 at 18:57

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