Question

Determine for which values of m the function variant phi (x )equals x Superscript m is a solution to the given equation.

Answer 1

Answer:

1) $m=-3\,;\,\, m=\frac{1}{3}$

2) $m=1\pm\sqrt{6}$

Step-by-step explanation:

As differential equation is not give so we consider both equations attached in figure below one by one.

$1) \,3x^{2}\frac{d^2y}{dx^2}+11x\frac{dy}{dx}-3y=0$

Writing in terms of ∅

$3x^{2}\frac{d^2\phi}{dx^2}+11x\frac{d\phi}{dx}-3\phi=0$

Substitute

$\phi=x^m\\\\\phi'=mx^{m-1}\\\\\phi''=m(m-1)x^{m-2}$

$3x^{2}(m(m-1)x^{m-2})+11x(mx^{m-1})-3x^m=0\\\\3x^{m}(m(m-1))+11mx^{m}-3x^m=0\\\\x^{m}(3m^2-3m+11m-3)=0\\\\3m^2+8m-3=0\\\\By \,\,factorization\\\\3m^2-m+9m-3=0\\\\m(3m-1)+3(3m-1)=0\\\\(m+3)(3m-1)=0\\\\\implies m=-3\,;\,\, m=\frac{1}{3}$

$2)\, x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}-5y=0---(2)$

Writing in terms of ∅

$x^2\frac{d^2\phi}{dx^2}-x\frac{d\phi}{dx}-5\phi=0$

Substitute

$\phi=x^m\\\\\phi'=mx^{m-1}\\\\\phi''=m(m-1)x^{m-2}$

$x^2(m(m-1)x^{m-2})-x(mx\phix^{m-1})-5x^{m}=0\\\\(m^2-m)x^m-mx^m-5x^m=0\\\\x^m(m^2-m-m-5)=0\\\\m^2-2m-5=0\\\\Using\,\,quadratic\,\,formula\\\\m=\frac{2\pm \sqrt{20+4}}{2}\\\\m=\frac{2\pm \sqrt{24}}{2}\\\\m=1\pm\sqrt{6}$