Question

In Problems 1 and 2 answer true or false without referring back to the text.

The general solution of \({x^2}y'' + xy' + ({x^2} - 1)y = 0\) is\(y = {c_1}{J_1}(x) + {c_2}{J_{ - 1}}(x)\).

Short answer

False. The general solutions of the given differential equation is \(y(x) = {C_1}{J_1}\left( x \right) + {C_2}{Y_1}\left( x \right)\).

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be \({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation has two linearly independent solutions for a fixed value of \(n\). A Bessel equation of the first kind, indicated by \({J_n}(x)\), is one of these solutions that may be derived using Frobinousapproach.

\(\begin{aligned}{}{y_1} &= {x^a}{J_p}\left( {b{x^c}} \right)\\{y_2} &= {x^a}{J_{ - p}}\left( {b{x^c}} \right)\end{aligned}\)

At\(x = 0\), this solution is regular. The second solution, which is singular at\(x = 0\), is represented by\({Y_n}(x)\)and is calleda Bessel function of the second kind.

\({y_3} = {x^a}\left( {\frac{{cosp\pi {J_p}\left( {b{x^c}} \right) - {J_{ - p}}\left( {b{x^c}} \right)}}{{sinp\pi }}} \right)\)

Determine the general form of the Bessel’s equation.

Let the given differential equation be \({x^2}y'' + xy' + ({x^2} - 1)y = 0\), that has a singular point at \(x = 0\).

The equation becomes in the following form:

\({x^2}y'' + xy' + ({x^2} - {v^2})y = 0\)

That yields,

\(\begin{aligned}{}{v^2} &= 1\\v &= 1, - 1\end{aligned}\)

Obtain the general solution.

There are two series which are linearly independent.

\(\begin{aligned}{}{y_1} &= {J_1}\left( x \right)\\{y_2} &= {Y_1}\left( x \right)\end{aligned}\)

The general solution by using superposition principle is,

\(y(x) = {C_1}{J_1}\left( x \right) + {C_2}{Y_1}\left( x \right)\)

That contradicts the given solution. Hence, the answer is false.