Differential Equations

Second Order Equations

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Chebyshev Differential Equation

Solved Problems

Example 1.

Find the general solution of the equation

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

for \(\left| x \right| \lt 1.\)

Solution.

The given equation is the Chebyshev differential equation with the fractional parameter \(n = \sqrt 2 .\) Its general solution can be written in the trigonometric form:

\[y\left( x \right) = {C_1}\cos \left( {\sqrt 2 \arccos x} \right) + {C_2}\sin\left( {\sqrt 2 \arccos x} \right),\]

where \({C_1},\) \({C_2}\) are constants. Note that the solution in this case is not expressed in terms of the Chebyshev polynomials due to the irrational number \(\sqrt 2.\)

Example 2.

Find the general solution of the differential equation

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

for \(\left| x \right| \lt 1.\)

Solution.

Here we deal with the Chebyshev equation with the parameter \(n = 2.\) Therefore, we can directly write the general solution in the form:

\[y\left( x \right) = {C_1}\cos \left( {2\arccos x} \right) + {C_2}\sin\left( {2\arccos x} \right),\]

where \({C_1},\) \({C_2}\) are arbitrary constants.

We can express this solution via the Chebyshev polynomials of the first kind. As

\[\cos \left( {2\arccos x} \right) = {T_2}\left( x \right) = 2{x^2} - 1,\]

we obtain the final answer:

\[y\left( x \right) = {C_1}\cos \left( {2\arccos x} \right) + {C_2}\sqrt {1 - {\cos^2}\left( {2\arccos x} \right)} = {C_1}{T_2}\left( x \right) + {C_2}\sqrt {1 - T_2^2\left( x \right)} = {C_1}\left( {2{x^2} - 1} \right) + {C_2}\sqrt {1 - {{\left( {2{x^2} - 1} \right)}^2}} = {C_1}\left( {2{x^2} - 1} \right) + 2{C_2}x\sqrt {1 - {x^2}} .\]
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