x^2y'' + xy' - 2y = 0Dfferential equation has a regular singular point at x = 0. Determine the indicial equation, the recursion relations, the roots of the indical equation, and the first three...

x^2y'' + xy' - 2y = 0

Dfferential equation has a regular singular point at x = 0. Determine the indicial equation, the recursion relations, the roots of the indical equation, and the first three terms of the series solutions.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should assume that the solution of the equation is of form `y(x) = x^r,`  hence `y'(x) = r*x^(r-1)`  ; `y''(x) = r*(r-1)x^(r-2).`

Substituting y' and y'' in the given equation yields:

`x^2*r*(r-1)x^(r-2) + x*r*x^(r-1) + 2x^r = 0`

`r*(r-1)x^r + r*x^r + 2x^r = 0`

Factoring out x^r yields:

`x^r (r^2 - r + r + 2) = 0`

Dividing by `x^r`  yields:

`r^2 + 2 = 0 =gt r_(1,2) = +-sqrt2`

Hence, evaluating the general solution to the differential equation  yields `y = c_1*x^(-sqrt2) + c_2*x^sqrt2.`

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