$$ xy'' - y' = 3x^{2} $$
$$ y' = p $$
$$ y'' = p' $$
$$ xp' - p =3x^{2} $$
$$ p' - \frac{1}{x}p = 3x $$
after multiplying by the integrating factor we get..
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p =3 $$
so $$ [\frac{1}{x}p]' = 3? $$
I know that these two below are equal, but can someone please show HOW
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p $$ equals $$[\frac{1}{x}p]' $$
Thank you!!
$$ y' = p $$
$$ y'' = p' $$
$$ xp' - p =3x^{2} $$
$$ p' - \frac{1}{x}p = 3x $$
after multiplying by the integrating factor we get..
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p =3 $$
so $$ [\frac{1}{x}p]' = 3? $$
I know that these two below are equal, but can someone please show HOW
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p $$ equals $$[\frac{1}{x}p]' $$
Thank you!!
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