# Homework Help: DiffEq Question

1. Dec 29, 2013

### 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!!

2. Dec 29, 2013

### Dick

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

3. Dec 29, 2013

### LCKurtz

Use the product rule on that last equation.

4. Dec 29, 2013

### 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... $$[e^{∫a(x) dx} * y]' ?$$

Last edited: Dec 29, 2013
5. Dec 29, 2013

### 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 $$e^{∫a(x) dx} * y$$

6. Dec 29, 2013

### vanceEE

yes, thank you LCKurtz!!