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In Problems \(13 - 20\) use \((20)\) to find the general solution of the given differential equation on \((0,\infty )\).

\(9{x^2}y'' + 9xy' + \left( {{x^6} - 36} \right)y = 0\)

Short Answer

Expert verified

The general solutions of the given differential equation are \(y = {C_1}{J_{2/3}}\left( {\frac{1}{9}{x^3}} \right) + {C_2}{J_{ - 2/3}}\left( {\frac{1}{9}{x^3}} \right)\) and \(y = {C_1}{J_{2/3}}\left( {\frac{1}{9}{x^3}} \right) + {C_2}{Y_{2/3}}\left( {\frac{1}{9}{x^3}} \right)\).

Step by step solution

01

Define Bessel’s equation.

Let the Bessel equation be\({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation hastwo linearly independent solutionsfor 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 usingFrobinous approach.

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

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.

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

02

Determine the general form of the Bessel’s equation.

Let the given differential equation be \(9{x^2}y'' + 9xy' + \left( {{x^6} - 36} \right)y = 0\), that has a singular point at \(x = 0\).

The equation becomes in the following form:

\(y'' + \frac{{1 - 2a}}{x}y' + \left( {{b^2}{c^2}{x^{2c - 2}} + \frac{{{a^2} - {p^2}{c^2}}}{{{x^2}}}} \right)y = 0\)鈥 (1)

That yields,

\(\frac{{9{x^2}}}{{9{x^2}}}y'' + \frac{{9x}}{{9{x^2}}}y' + \frac{{{x^6} - 36}}{{9{x^2}}}y = 0\)

\(y'' + \frac{1}{x}y' + \left( {\frac{1}{9}{x^4} - \frac{4}{{{x^2}}}} \right)y = 0\) 鈥 (2)

03

Find the value of constants.

Compare the equations (1) and (2).

Solve for \(a\):

\(\begin{aligned}{c}1 - 2a = 1\\a = 0\end{aligned}\)

Solve for \(c\):

\(\begin{aligned}{c}2c - 2 = 4\\c = 3\end{aligned}\)

Solve for \(b\):

\(\begin{aligned}{c}{b^2}{c^2} = \frac{1}{9}\\b = \quad \frac{1}{9}\end{aligned}\)

Solve for \(p\):

\(\begin{aligned}{c}{a^2} - {p^2}{c^2} = - 4\\{p_1} = \frac{2}{3},{p_2} = \frac{{ - 2}}{3}\end{aligned}\)

04

Obtain the general solution.

There are two series which are linearly independent.

\(\begin{aligned}{c}{y_1} = {x^0}{J_{2/3}}\left( {\frac{1}{9}{x^3}} \right)\\ = {J_{2/3}}\left( {\frac{1}{9}{x^3}} \right)\\{y_2} = {J_{ - 2/3}}\left( {\frac{1}{9}{x^3}} \right)\end{aligned}\)

The general solution by using superposition principle is,

\(y = {C_1}{J_{2/3}}\left( {\frac{1}{9}{x^3}} \right) + {C_2}{J_{ - 2/3}}\left( {\frac{1}{9}{x^3}} \right)\)

There is another general solution obtained from the Bessel鈥檚 equation of second order. (i.e.) \(y = {C_1}{J_{2/3}}\left( {\frac{1}{9}{x^3}} \right) + {C_2}{Y_{2/3}}\left( {\frac{1}{9}{x^3}} \right)\).

Hence, the general solutions are \(y = {C_1}{J_{2/3}}\left( {\frac{1}{9}{x^3}} \right) + {C_2}{J_{ - 2/3}}\left( {\frac{1}{9}{x^3}} \right)\) and \(y = {C_1}{J_{2/3}}\left( {\frac{1}{9}{x^3}} \right) + {C_2}{Y_{2/3}}\left( {\frac{1}{9}{x^3}} \right)\).

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