1. [SOLVED] Linear DE

Hi, the question is

$y'=\frac{y}{x}+2x^2$
and y(1)=0

I put it to this form

$y' -y(x^{-1})=2x^2$

so $f(x)=x^{-1}$ and $h(x)=2x^2$

is that right?

Integrating Factor $I(x)=exp(\int f(x)dx)$
$=exp(\int x^{-1}dx)$

which means I(x) = e

So i mulptiply both sides by e?

$\frac{d}{dx}ey=2ex^2$

and then i'm lost and confused, any help will be marvellous

2. Originally Posted by Rapid_W
Hi, the question is

$y'=\frac{y}{x}+2x^2$
and y(1)=0

I put it to this form

$y' -y(x^{-1})=2x^2$

so $f(x)=x^{-1}$ and $h(x)=2x^2$

is that right?

Integrating Factor $I(x)=exp(\int f(x)dx)$
$=exp(\int x^{-1}dx)$

which means I(x) = e

So i mulptiply both sides by e?

$\frac{d}{dx}ey=2ex^2$

and then i'm lost and confused, any help will be marvellous
$I(x)=exp(\int - x^{-1}dx) = e^{- \ln x} = \frac{1}{x}$

Now continue.

3. Where is the dunce hat, I need it! Omg, if I can't remember x^(-1) integrates to ln(x) how am I going to pass my tests

4. Nah, i can't work it out still

I obvioulsy now know that I(x) = 1/x so i mulitply the equation by this...

$$y' \frac{y}{x} = \frac{2x^2}{x}$$

so now i'm meant to integrate both sides, i now get this

$ylnx = x^2 + c$

So now what do i do? The initial value is y(1)=0

$y' - \frac{1}{x} y = 2 x^2$ .

Multiplying by the integrating factor gives you:

$\frac{1}{x} y' - \frac{1}{x^2} y = 2 x$

$\Leftrightarrow \frac{d}{dx} \left( \frac{y}{x} \right) = 2 x$

so

$\frac{y}{x} = \int 2x \, \mathrm{d}x$.

You should be able to take it from there, right?

6. I found the answer doing it this way...

$f(x)=-x^{-1}$
$h(x)=2x^2$
$I(x)=\frac{1}{x}$
so
$\int I(x)h(x)dx$
$=\int (\frac{1}{x})(2x^2)dx$
$=\int 2x$
$x^2+c$

$z(x)=x^2+c$

general solution = y=\frac{z(x)}{I(x)}[/tex]
so
$y=\frac{x^2+c}{\frac{1}{x}}$
$=x^3+cx$

and the initial value was y(1)=0

so $0=1^3+c$
$c=-1$
so finaly
$y=x^3-x$