ODE5.1

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$x^2y'' + 7xy' + 8y = 0\,$

The adjoint of an equation in the form

$a(x) y'' + b(x) y' + c(x) y = 0\,$ is

$a(x) y'' + [2 a'(x) - b(x)]y' + [a''(x) - b'(x) + c(x)]y = 0\,$

So the adjoint for this DE is:

$x^2y'' - 3xy' + 3y = 0\,$

Any solution of the adjoint equation is an integrating factor for the original problem.

$y=x\,$ is one solution. Multiply $x\,$ through the original DE.

$x^3y'' + 7x^2y' + 8xy = 0\,$

This is an exact equation, since

$\frac{d}{dx}[ a(x)y' + b(x)y ] = a'(x)y' + a(x)y'' + b'(x)y + b(x)y'\,$
$= a(x)y''+[a'(x)+b(x)]y' + b'(x)y\,$

In the case of an exact equation the last relation equals 0, so

$a(x) = x^3, a'(x) = 3x^2\,$
$b'(x) = 8x, b(x) = 4x^2\,$

$\frac{d}{dx} [ x^3y' + 4x^2y ] = 0\,$ . Integrating,

$x^3y' + 4x^2y = c_1\,$
$y' + 4x^{-1}y = c_1x^{-3}\,$

Use the integrating factor $e^{\int 4x^{-1} dx} = x^4\,$ .

$x^4y' + 4x^3y = c_1x\,$
$\frac{d}{dx}[x^4y] = c_1x\,$
$x^4y = \frac{c_1}{2}x^2 + c_2\,$

Finally,

$y = \frac{c_1}{2}x^{-2} + c_2x^{-4}\,$

==Alternate Solution==

$x^2y'' + 7xy' + 8y = 0\,$

This equation is recognizable as an Euler equation since the exponent on the $x$ is the same as the order of the derivative in each term. So try for a solution in the form $y = x r$ .

$y = x^r,\quad y' = rx^{r-1},\quad y''=r(r-1)x^{r-2}$

Substituting these in the equation gives:

$x^2r(r-1)x^{(r-2)} + 7xrx^{(r-1)} + 8x^r = 0\,$

$x^r(r^2 + 6r + 8) = 0\,$

$(r+2)(r+4) = 0,\quad r = -2,\ -4\,$

The solution is therefore $y = Ax^{-2} + Bx^{-4},\ x \ne 0\,$ .

Ordinary Differential Equations

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