Bessel Differential Equation
Solved Problems
Example 1.
Solve the differential equation
\[{x^2}y^{\prime\prime} + xy' + \left( {3{x^2} - 2} \right)y = 0.\]
Solution.
This equation has order \(\sqrt 2 \) and differs from the standard Bessel equation only by factor \(3\) before \({x^2}.\) Therefore, the general solution of the equation is expressed by the formula
\[y\left( x \right) = {C_1}{J_{\sqrt 2 }}\left( {\sqrt 3 x} \right) + {C_2}{Y_{\sqrt 2 }}\left( {\sqrt 3 x} \right),\]
where \({C_1},\) \({C_2}\) are constants, \({J_{\sqrt 2 }}\left( {\sqrt 3 x} \right)\) and \({Y_{\sqrt 2 }}\left( {\sqrt 3 x} \right)\) are Bessel functions of the \(1\)st and \(2\)nd kind, respectively.
Example 2.
Solve the equation
\[{x^2}y^{\prime\prime} + xy' - \left( {4{x^2} + {\frac{1}{2}}} \right)y = 0.\]
Solution.
This equation differs from the modified Bessel equation by factor \(4\) in front of \({x^2}.\) The order of the equation is \(v = {\frac{1}{{\sqrt 2 }}}.\) Then the general solution is written through the modified Bessel functions in the following way:
\[y\left( x \right) = {C_1}{I_{\frac{1}{{\sqrt 2 }}}}\left( {2x} \right) + {C_2}{K_{\frac{1}{{\sqrt 2 }}}}\left( {2x} \right),\]
where \({C_1}\) and \({C_2}\) are arbitrary constants.
Example 3.
Find the general solution of the differential equation
\[{x^2}y^{\prime\prime} + 2xy' + \left( {{x^2} - 1} \right)y = 0.\]
Solution.
We make the substitution:
\[y = {x^{\frac{{1 - 2}}{2}}}z = {x^{ - \frac{1}{2}}}z,\;\; \Rightarrow
y' = - \frac{1}{2}{x^{ - \frac{3}{2}}}z + {x^{ - \frac{1}{2}}}z',\;\; \Rightarrow
y^{\prime\prime} = \frac{3}{4}{x^{ - \frac{5}{2}}}z
- \frac{1}{2}{x^{ - \frac{3}{2}}}z' - \frac{1}{2}{x^{ - \frac{3}{2}}}z' + {x^{ - \frac{1}{2}}}z^{\prime\prime}
= \frac{3}{4}{x^{ - \frac{5}{2}}}z - {x^{ - \frac{3}{2}}}z' + {x^{ - \frac{1}{2}}}z^{\prime\prime}.\]
Put these expressions back into the equation:
\[{x^2}y^{\prime\prime} + 2xy' + \left( {{x^2} - 1} \right)y = 0,\]
\[\Rightarrow {x^2}\left( {\frac{3}{4}{x^{ - \frac{5}{2}}}z - {x^{ - \frac{3}{2}}}z' + {x^{ - \frac{1}{2}}}z^{\prime\prime}} \right)
+ 2x\left( { - \frac{1}{2}{x^{ - \frac{3}{2}}}z + {x^{ - \frac{1}{2}}}z'} \right)
+ \left( {{x^2} - 1} \right){x^{ - \frac{1}{2}}}z = 0,\]
\[ \Rightarrow \color{blue}{\frac{3}{4}{x^{ - \frac{1}{2}}}z} - \color{red}{{x^{ \frac{1}{2}}}z' + \color{black}{x^{\frac{3}{2}}}z^{\prime\prime}} - \color{blue}{{x^{ - \frac{1}{2}}}z} + \color{red}{2{x^{\frac{1}{2}}}z'} + \color{black}{{x^{\frac{3}{2}}}z} - \color{blue}{{x^{ - \frac{1}{2}}}z} = 0, \]
\[\Rightarrow {x^{\frac{3}{2}}}z^{\prime\prime} + \color{red}{{x^{ - \frac{1}{2}}}z'} + \color{blue}{\left( { - \frac{5}{4}{x^{ - \frac{1}{2}} + x^{\frac{3}{2}}}} \right)z} = 0,\]
\[ \Rightarrow {x^2}z^{\prime\prime} + xz' + \left( {{x^2} - \frac{5}{4}} \right)z = 0.\]
Indeed, we see that
\[{n^2} = {v^2} + \frac{1}{4}{\left( {a - 1} \right)^2} = 1 + \frac{1}{4}{\left( {2 - 1} \right)^2} = 1 + \frac{1}{4} = \frac{5}{4}.\]
Thus, the general solution for the function \(z\left( x \right)\) can be written in the form
\[z\left( x \right) = {C_1}{J_{\frac{{\sqrt 5 }}{2}}}\left( x \right) + {C_2}{Y_{\frac{{\sqrt 5 }}{2}}}\left( x \right).\]
Then the solution for the original function \(y\left( x \right)\) is given by
\[y\left( x \right) = {x^{ - \frac{1}{2}}}z\left( x \right) = \frac{1}{{\sqrt x }}\left[ {{C_1}{J_{\frac{{\sqrt 5 }}{2}}}\left( x \right) + {C_2}{Y_{\frac{{\sqrt 5 }}{2}}}\left( x \right)} \right],\]
where \({C_1}\) and \({C_2}\) are arbitrary constants.
