Question

Consider the differential equation x2y′′ − 9xy′ + 24y = 0; x4, x6, (0, [infinity]). Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. The functions satisfy the differential equation and are linearly independent since W(x4, x6) = ≠ 0 for 0 < x < [infinity]. Form the general solution. y =

Answer 1

Answer:

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

$y= c_1x^4+c_2x^6$

Step-by-step explanation:

Given equation is

$x^2y'' - 9xy+24y=0$

This Euler Cauchy type differential equation.

So, we can let

$y=x^m$

Differentiate with respect to x

$y'= mx^{m-1}$

Again differentiate with respect to x

$y''= m(m-1)x^{m-2}$

Putting the value of y, y' and y'' in the differential equation

$x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0$

$\Rightarrow m(m-1)x^m-9mx^m+24x^m=0$

$\Rightarrow m^2-m-9m+24=0$

⇒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

$y_1(x)=x^4$

and

$y_2(x)=x^6$

First we compute the Wronskian

$W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|$

         $= \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|$

         =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

$y= c_1x^4+c_2x^6$