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$$e^{x\sqrt{1+t}}=\sum \limits_{k=0}^\infty \frac{U_k(x)t^k}{k!}$$

$$\frac{\partial}{\partial t }(e^{x\sqrt{1+t}})=\frac{\partial}{\partial t }(\sum \limits_{k=0}^\infty \frac{U_k(x)t^k}{k!})$$

$$\frac{1}{2\sqrt{1+t} }e^{x\sqrt{1+t}}=\sum \limits_{k=0}^\infty \frac{U_{k+1}(x)}{z}\frac{t^{k}}{{k}!}$$

$$\frac{\partial}{\partial x }(\frac{1}{2\sqrt{1+t} }e^{x\sqrt{1+t}})=\frac{\partial}{\partial x }(\sum \limits_{k=0}^\infty \frac{U_{k+1}(x)}{x}\frac{t^{k}}{{k}!})$$

$$\frac{e^{x\sqrt{1+t}}}{2}=\sum \limits_{k=0}^\infty \frac{\partial}{\partial x }(\frac{U_{k+1}(x)}{x})\frac{t^{k}}{{k}!}$$

$$\sum \limits_{k=0}^\infty \frac{U_k(x)t^k}{k!}=2\sum \limits_{k=0}^\infty \frac{\partial}{\partial x }(\frac{U_{k+1}(x)}{x})\frac{t^{k}}{{k}!}$$

$$U_n(x)=2\frac{\partial}{\partial x }(\frac{U_{n+1}(x)}{x}) \tag 1$$


$$e^{x\sqrt{1+t}}=\sum \limits_{k=0}^\infty \frac{U_k(x)t^k}{k!}$$

$$\frac{\partial}{\partial x }(e^{x\sqrt{1+t}})=\frac{\partial}{\partial x }(\sum \limits_{k=0}^\infty \frac{U_k(x)t^k}{k!})$$

$$\frac{\partial}{\partial x }(\sqrt{1+t}e^{x\sqrt{1+t}})=\frac{\partial}{\partial x }(\sum \limits_{k=0}^\infty \frac{\partial (U_k(x))}{\partial x }\frac{t^k}{k!})$$

$$(1+t)e^{x\sqrt{1+t}}=\sum \limits_{k=0}^\infty \frac{\partial^2 (U_k(x))}{\partial x^2 }\frac{t^k}{k!}$$

$$\sum \limits_{k=0}^\infty \frac{U_k(x)t^k}{k!}+\sum \limits_{k=0}^\infty \frac{U_k(x)t^{k+1}}{k!}=\sum \limits_{k=0}^\infty \frac{\partial^2 (U_k(x))}{\partial x^2 }\frac{t^k}{k!}$$

$$U_{n+1}(x)+(n+1)U_{n}(x)=\frac{\partial^2 (U_{n+1}(x))}{\partial x^2 } \tag 2$$ If we use the Equation (1) in Equation (2), we can get second order linear differential equation.

$$U_{n+1}(x)+2(n+1)\frac{\partial}{\partial x }(\frac{U_{n+1}(x)}{x})=\frac{\partial^2 (U_{n+1}(x))}{\partial x^2 } $$

$$\frac{\partial^2 (U_{n}(x))}{\partial x^2 }-2n\frac{\partial}{\partial x }(\frac{U_{n}(x)}{x})-U_{n}(x)=0 \tag3$$

$$\frac{\partial^2 (U_{n}(x))}{\partial x^2 }-2n\frac{\partial}{\partial x }(\frac{U_{n}(x)}{x})-U_{n}(x)=0 \tag4$$


$$U''_{n}(x)-\frac{2n}{x}U'_{n}(x)+(\frac{2n}{x^2}-1)U_{n}(x)=0 \tag5$$

$$x^2U''_{n}(x)-2nxU'_{n}(x)+(2n-x^2)U_{n}(x)=0 \tag6$$

It is clear that one one solution is

$$U_{n}(x)=\frac{\partial^n (e^{x\sqrt{1+t}})}{\partial t^n }|_{t=0}$$

Some examples, $$U_{0}(x)=e^{x}$$

$$U_{1}(x)=\cfrac{xe^{x}}{2}$$

Can I express the solution of the differential equation as known functions?

I also want to find orthogonal relation for $U_{n}(x)$ if possible.

Thanks a lot for answers.

EDIT: According to GEdgar's answer, I have used the transform
$$U_n(x)=x^{(n+\frac{1}{2})}T_n(x)$$

$$U'_n(x)=(n+\frac{1}{2})x^{(n-\frac{1}{2})}T_n(x)+x^{(n+\frac{1}{2})}T'_n(x)$$

$$U''_n(x)=(n^2-\frac{1}{4})x^{(n-\frac{3}{2})}T_n(x)+2(n+\frac{1}{2})x^{(n-\frac{1}{2})}T'_n(x)+x^{(n+\frac{1}{2})}T''_n(x)$$

$$U''_{n}(x)-\frac{2n}{x}U'_{n}(x)+(\frac{2n}{x^2}-1)U_{n}(x)=0 $$

$$(n^2-\frac{1}{4})x^{(n-\frac{3}{2})}T_n(x)+2(n+\frac{1}{2})x^{(n-\frac{1}{2})}T'_n(x)+x^{(n+\frac{1}{2})}T''_n(x)-\frac{2n}{x}((n+\frac{1}{2})x^{(n-\frac{1}{2})}T_n(x)+x^{(n+\frac{1}{2})}T'_n(x))+(\frac{2n}{x^2}-1)x^{(n+\frac{1}{2})}T_n(x)=0 $$

$$(n^2-\frac{1}{4})x^{(n-\frac{3}{2})}T_n(x)+2(n+\frac{1}{2})x^{(n-\frac{1}{2})}T'_n(x)+x^{(n+\frac{1}{2})}T''_n(x)-\frac{2n}{x}((n+\frac{1}{2})x^{(n-\frac{1}{2})}T_n(x)+x^{(n+\frac{1}{2})}T'_n(x))+(\frac{2n}{x^2}-1)x^{(n+\frac{1}{2})}T_n(x)=0 $$

$$x^{(n+\frac{1}{2})}T''_n(x) +\Big((2n+1)-2n\Big)x^{(n-\frac{1}{2})}T'_n(x)+\Big((n^2-\frac{1}{4})-2n(n+\frac{1}{2})+2n\Big)x^{(n-\frac{3}{2})}T_n(x)-x^{(n+\frac{1}{2})}T_n(x)=0 $$

$$x^2T''_n(x) +xT'_n(x)-((n-\frac{1}{2})^2+x^{2})T_n(x)=0 $$

Thus after transformation ,really We got Modified Bessel Differential Equation http://mathworld.wolfram.com/ModifiedBesselDifferentialEquation.html

The solutions are the modified Bessel functions of the first and second kinds, and can be written $$T_n(x)=C_1 I_{(n-\frac{1}{2})}( x)+C_2 K_{(n-\frac{1}{2})}( x)$$ $I_{n}( x)$ is a modified Bessel function of the first kind, and $K_{n}( x)$is modified Bessel function of the second kind.

$$U_n(x)=x^{(n+\frac{1}{2})}T_n(x)$$ $$U_n(x)=C_1 x^{(n+\frac{1}{2})}I_{(n-\frac{1}{2})}( x)+C_2 x^{(n+\frac{1}{2})}K_{(n-\frac{1}{2})}( x)$$ Thanks a lot to GEdgar for answer.

Now I need to find $C_1$ and $C_2$ that $$U_{n}(x)=\frac{\partial^n (e^{x\sqrt{1+t}})}{\partial t^n }|_{t=0}=C_1 x^{(n+\frac{1}{2})}I_{(n-\frac{1}{2})}( x)+C_2 x^{(n+\frac{1}{2})}K_{(n-\frac{1}{2})}( x)$$

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This is just Bessel's equation, right? –  mjqxxxx Oct 18 '12 at 15:14
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1 Answer 1

up vote 2 down vote accepted

Maple solves the DE $$ \frac{\partial^{2} f (x)}{\partial x^{2}} - \frac{2 n }{x}\;\frac{\partial f (x)}{\partial x} + \Biggl(\frac{2 n}{x^{2}} - 1\Biggr) f (x) = 0 $$ in terms of the Bessel I and K function: $$ f(x) = C_1 x^{1/2+n} I_{(n-1/2)}( x)+C_2 x^{1/2+n} K_{(n-1/2)}( x) $$ for $n>1/2$

added Oct 19

How about this $$ e^{x\sqrt{1-t}} =\sum_{n = 0}^{\infty} \frac{\sqrt{2}\;t^n\;x^{n+1/2}}{2^n n!} \Biggl((-1)^{n} \sqrt\pi\; I_{n-1/2}(x) - \frac{1}{\sqrt\pi}K_{n-1/2}(x)\Biggr) $$ (of course the half-integral order Bessel functions I and K are expressible in terms of hyperbolic functions.)

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What I said in meta (and received -6 :-) ). Before asking try to use computer. The answer produced by it can give ideas. (+1) –  vesszabo Oct 18 '12 at 18:28
    
@GEdgar What are $C_1$ and $C_2$ to satify $$e^{x\sqrt{1+t}}=\sum \limits_{k=0}^\infty \frac{U_k(x)t^k}{k!}$$ Thanks a lot for your help –  Mathlover Oct 18 '12 at 19:14
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