# ODE5.1

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

The adjoint of an equation in the form

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

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

So the adjoint for this DE is:

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

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.

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

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

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

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

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

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

$x^{4}y'+4x^{3}y=c_{1}x\,$
${\frac {d}{dx}}[x^{4}y]=c_{1}x\,$
$x^{4}y={\frac {c_{1}}{2}}x^{2}+c_{2}\,$

Finally,

$y={\frac {c_{1}}{2}}x^{{-2}}+c_{2}x^{{-4}}\,$

==Alternate Solution==

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

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:

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

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

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

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