Solve the boundary value problem
[tex]x^{2} \frac{d^{2}y}{dx^{2}} - 15x \frac{dy}{dx} + 64y=0[/tex]
where y(1)=0 and y(2)=1

Question

DramaQueen001 DramaQueen001

22-05-2017
Mathematics

contestada

Solve the boundary value problem
[tex]x^{2} \frac{d^{2}y}{dx^{2}} - 15x \frac{dy}{dx} + 64y=0[/tex]
where y(1)=0 and y(2)=1

Respuesta :

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LammettHash LammettHash · Answer 1 · 2017-05-23T00:37:37+02:00

Another Cauchy-Euler ODE. Take [tex]y=x^r[/tex] and you get

[tex]r(r-1)x^r-16rx^r+64x^r=0[/tex]
[tex]r^2-17r+64=0[/tex]
[tex]\implies r=\dfrac{17}2\pm\dfrac{\sqrt{33}}2[/tex]

so that the general solution is

[tex]y=C_1x^{(17+\sqrt{33})/2}+C_2x^{(17-\sqrt{33})/2}[/tex]

With the given boundary conditions, you have

[tex]y(1)=0\implies 0=C_1+C_2[/tex]
[tex]y(2)=1\implies 1=C_12^{(17+\sqrt{33})/2}+C_22^{(17-\sqrt{33})/2}[/tex]
[tex]\implies C_1=\dfrac{2^{(-17+\sqrt{33})/2}}{2^{\sqrt{33}}-1},C_2=-\dfrac{2^{(-17+\sqrt{33})/2}}{2^{\sqrt{33}}-1}[/tex]

Solve the boundary value problem [tex]x^{2} \frac{d^{2}y}{dx^{2}} - 15x \frac{dy}{dx} + 64y=0[/tex]where y(1)=0 and y(2)=1

Respuesta :

Otras preguntas

Solve the boundary value problem
[tex]x^{2} \frac{d^{2}y}{dx^{2}} - 15x \frac{dy}{dx} + 64y=0[/tex]
where y(1)=0 and y(2)=1