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Find values of m so that the function y xm is a solution of the given differential equation. x^2y''-7xy' 15y=0

Sagot :

LammettHash

If [tex]y=x^m[/tex] is a solution to the ODE

[tex]x^2 y'' - 7xy' + 15y = 0[/tex]

then substituting [tex]y[/tex] and its derivatives

[tex]y' = mx^{m-1} \text{ and } y'' = m(m-1)x^{m-2}[/tex]

gives

[tex]m(m-1)x^2x^{m-2} - 7mxx^{m-1} + 15x^m = 0[/tex]

[tex]m(m-1)x^{m} - 7mx^{m} + 15x^m = 0[/tex]

[tex](m(m-1) - 7m + 15)x^{m} = 0[/tex]

We ignore the trivial case of [tex]y=x^m=0[/tex]. Solve for [tex]m[/tex].

[tex]m(m-1) - 7m + 15 = 0[/tex]

[tex]m^2 - 8m + 15 = 0[/tex]

[tex](m - 3) (m - 5) = 0[/tex]

[tex]\implies \boxed{m=3} \text{ or } \boxed{m=5}[/tex]

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