# Bessel Differential Equation

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the differential equation

${x^2}y^{\prime\prime} + xy' + \left( {3{x^2} - 2} \right)y = 0.$

### Example 2

Solve the equation

${x^2}y^{\prime\prime} + xy' - \left( {4{x^2} + {\frac{1}{2}}} \right)y = 0.$

### Example 3

Find the general solution of the differential equation

${x^2}y^{\prime\prime} + 2xy' + \left( {{x^2} - 1} \right)y = 0.$

### 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.