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 Problem Statement

$$u_t=ku_{xx}+h \; \;\; \; \; 0<x<1 \; \; \,\; t>0$$
$$u(x,0)=u_0(1\cos{\pi x}) \; \;\; \; \; 0\leq x \leq 1$$
$$u(0,t)=0 \; \;\; \; \; u(1,t)=2u_0 \; \;\; \; \; t\geq0$$
 Relevant Equations

$$u(x,t)=\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k}\right)x\frac{4h}{k\pi}e^{k\pi^2 t}\sin{\pi x}+ \sum_{n=2}^\infty{a_n e^{kn^2\pi^2 t}\sin{n\pi x}} $$
$$a_n=\frac{2u_0}{n\pi}[1+(1)^n]+\frac{2u_0n}{(n^21)\pi}[1(1)^n]+\frac{2h}{k\pi^3 n^3}[(1)^n1]$$
Hello, I posted the same in the partial differential equations section but I'm not getting responses and maybe this section is better for help with homework. I have to solve this problem:
$$u_t=ku_{xx}+h \; \;\; \; \; 0<x<1 \; \; \,\; t>0$$
$$u(x,0)=u_0(1\cos{\pi x}) \; \;\; \; \; 0\leq x \leq 1$$
$$u(0,t)=0 \; \;\; \; \; u(1,t)=2u_0 \; \;\; \; \; t\geq0$$
So I know that I can split the solution in two (I don't know the reason. I would appreciate a short explanation because we introduce a new fucntion depending just on x and because the sum of the both solutions are also a solution):
$$u(x,t)=v(x,t)+U(x)$$
This give us two different problems.
$$kU''(x)+h=0$$
and
$$v_t=kv_{xx}$$
$$\text{I.C}\begin{cases} v(x,0)=u_0(1\cos{\pi x})U(x)\end{cases}$$
$$\text{B.C}\begin{cases} v(0,t)=0 \\ v(1,t)=0 \end{cases}$$
First I can sol the U(x) which is giving me:
$$kU''(x)+h=0$$
$$U(x)=\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k} \right)x$$
Now I solve the PDE with the new boundary conditions using the separation of variables and I get:
$$X_n(x)=B_n\sin{n\pi x}$$
$$T_n(t)=C_ne^{kn^2\pi^2 t}$$
$$v(x,t)=\sum_{n=1}^\infty{=a_ne^{kn^2\pi^2 t}\sin{n\pi x}}$$
If I apply the initial condition:
$$v(x,0)=\sum_{n=1}^\infty{=a_n\sin{n\pi x}}=u_0(1cos(\pi x))U(x)$$
So finally the coefficients should be given by the expression:
$$a_n=2\int_0^1 (u_0(1cos(\pi x))U(x))\sin(n \pi x)dx$$
But i'm not gettin the supposed solution which is:
$$a_n=\frac{2u_0}{n\pi}[1+(1)^n]+\frac{2u_0n}{(n^21)\pi}[1(1)^n]+\frac{2h}{k\pi^3 n^3}[(1)^n1]$$
Actually the supposed solution to the problem is:
$$u(x,t)=\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k}\right)x\frac{4h}{k\pi}e^{k\pi^2 t}\sin{\pi x}+ \sum_{n=2}^\infty{a_n e^{kn^2\pi^2 t}\sin{n\pi x}} $$
As you can see the expansion series starts at 2 instead of one so how am I supposed to get the term for n=1 because you can not get it from a_n
$$u_t=ku_{xx}+h \; \;\; \; \; 0<x<1 \; \; \,\; t>0$$
$$u(x,0)=u_0(1\cos{\pi x}) \; \;\; \; \; 0\leq x \leq 1$$
$$u(0,t)=0 \; \;\; \; \; u(1,t)=2u_0 \; \;\; \; \; t\geq0$$
So I know that I can split the solution in two (I don't know the reason. I would appreciate a short explanation because we introduce a new fucntion depending just on x and because the sum of the both solutions are also a solution):
$$u(x,t)=v(x,t)+U(x)$$
This give us two different problems.
$$kU''(x)+h=0$$
and
$$v_t=kv_{xx}$$
$$\text{I.C}\begin{cases} v(x,0)=u_0(1\cos{\pi x})U(x)\end{cases}$$
$$\text{B.C}\begin{cases} v(0,t)=0 \\ v(1,t)=0 \end{cases}$$
First I can sol the U(x) which is giving me:
$$kU''(x)+h=0$$
$$U(x)=\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k} \right)x$$
Now I solve the PDE with the new boundary conditions using the separation of variables and I get:
$$X_n(x)=B_n\sin{n\pi x}$$
$$T_n(t)=C_ne^{kn^2\pi^2 t}$$
$$v(x,t)=\sum_{n=1}^\infty{=a_ne^{kn^2\pi^2 t}\sin{n\pi x}}$$
If I apply the initial condition:
$$v(x,0)=\sum_{n=1}^\infty{=a_n\sin{n\pi x}}=u_0(1cos(\pi x))U(x)$$
So finally the coefficients should be given by the expression:
$$a_n=2\int_0^1 (u_0(1cos(\pi x))U(x))\sin(n \pi x)dx$$
But i'm not gettin the supposed solution which is:
$$a_n=\frac{2u_0}{n\pi}[1+(1)^n]+\frac{2u_0n}{(n^21)\pi}[1(1)^n]+\frac{2h}{k\pi^3 n^3}[(1)^n1]$$
Actually the supposed solution to the problem is:
$$u(x,t)=\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k}\right)x\frac{4h}{k\pi}e^{k\pi^2 t}\sin{\pi x}+ \sum_{n=2}^\infty{a_n e^{kn^2\pi^2 t}\sin{n\pi x}} $$
As you can see the expansion series starts at 2 instead of one so how am I supposed to get the term for n=1 because you can not get it from a_n