Respuesta :
Answer:
The functions satisfy the differential equation and linearly independent since W(x)≠0
Therefore the general solution is
[tex]y= c_1x^4+c_2x^6[/tex]
Step-by-step explanation:
Given equation is
[tex]x^2y'' - 9xy+24y=0[/tex]
This Euler Cauchy type differential equation.
So, we can let
[tex]y=x^m[/tex]
Differentiate with respect to x
[tex]y'= mx^{m-1}[/tex]
Again differentiate with respect to x
[tex]y''= m(m-1)x^{m-2}[/tex]
Putting the value of y, y' and y'' in the differential equation
[tex]x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0[/tex]
[tex]\Rightarrow m(m-1)x^m-9mx^m+24x^m=0[/tex]
[tex]\Rightarrow m^2-m-9m+24=0[/tex]
⇒m²-10m +24=0
⇒m²-6m -4m+24=0
⇒m(m-6)-4(m-6)=0
⇒(m-6)(m-4)=0
⇒m = 6,4
Therefore the auxiliary equation has two distinct and unequal root.
The general solution of this equation is
[tex]y_1(x)=x^4[/tex]
and
[tex]y_2(x)=x^6[/tex]
First we compute the Wronskian
[tex]W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|[/tex]
=x⁴×6x⁵- x⁶×4x³
=6x⁹-4x⁹
=2x⁹
≠0
The functions satisfy the differential equation and linearly independent since W(x)≠0
Therefore the general solution is
[tex]y= c_1x^4+c_2x^6[/tex]
