Solving a Second Order Differential Equation: $$xy''-y'=3x^2$$

[vanceEE]

$$ xy'' - y' = 3x^{2} $$
$$ y' = p $$
$$ y'' = p' $$
$$ xp' - p =3x^{2} $$
$$ p' - \frac{1}{x}p = 3x $$
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!

 

[Dick]

$$ xy'' - y' = 3x^{2} $$
$$ y' = p $$
$$ y'' = p' $$
$$ xp' - p =3x^{2} $$
$$ p' - \frac{1}{x}p = 3x $$
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!

You want to find d/dx of the product (1/x)*p(x). Use the product rule.

 

[LCKurtz]

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!

Use the product rule on that last equation.

 

[vanceEE]

Awesome, I now understand that part. Now for any equation $$y' + a(x)y = b(x)$$ after multiplying the integrating factor, will my left side always be... $$ [e^{∫a(x) dx} * y]' ?$$

 

[LCKurtz]

I think you can answer your own question. is$$
\left(e^{∫a(x) dx} * y' + e^{∫a(x) dx} a(x)y\right)$$
the derivative of $$
e^{∫a(x) dx} * y$$

 

[vanceEE]

yes, thank you LCKurtz!