Answer to Question #188467 in Differential Equations for Alina

Question

Given the function

𝑔(𝑥) = 𝑥 𝐽1 (𝑥) − 𝐴 𝐽0(𝑥) with 𝐴 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ≥ 0,

determine 𝑔′, 𝑔 ′′ and 𝑔′′′ . Reduce all expressions to functions of 𝐽0(𝑥), 𝐽1(𝑥) and 𝐴 only.

Answer 1

$g(x)=xJ_1(x)-AJ_0(x)$ ....1)

Now differentiating the above equation we get-

$g^{'}(x)=J_1(x)-xJ_1^{'}(x)-AJ_0^{'}(x)$

Now by the recurrence relation ,we know that

$J_n^{'}(x)=\dfrac{J_n-1(x)-J_n+1(x)}{2}$ ......2)

now putting n=1 and n=0 we get the values of $J_1^{'}(x)\ and \ J_0^{'}(x).$

now putting the values we get g$^{'}(x)=J_1(x)-x \dfrac{J_0{(x)}-J_2(x)}{2}-A \dfrac{J_{-1}(x)-J_1(x)}{2}$ which is our required equation.

$g^{''}(x)=J_1^{'}(x)-J_1^{'}(x)-xJ_1^{''}(x)-AJ_o^{''}(x)$

Now differentiating recurrence relation .....2)

"J_n^{"}(x)= \\dfrac{J_n-1^{'}(x)-J_n+1{'}(x)}{2}" .....3)

Now in recurrence relation .....2), if we replace n$\to$ n-1,we get -$J_{n-1}^{'}(x)=\dfrac{J_{n-2}(x)-J_n(x)}{2}$

now replacing n$\to$ n+1, we get- $J_n+1{'}(x)=\dfrac{Jn(x)-J_{n+2}}{2}$

now puttting both these equation in .....3), we get-$\ Jn{''}(x)=$ $\dfrac{J_{n-2}(x)-2J_n(x)-J_{n+2}(x)}{2}$

now putting n=1,0 we get -

$g^{''}(x)=-x\dfrac {J_{-1}(x)-2J_1{(x)}}{2}$