1. ## [SOLVED] Linear DE

Hi, the question is

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

I put it to this form

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

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

is that right?

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

which means I(x) = e

So i mulptiply both sides by e?

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

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

2. Originally Posted by Rapid_W
Hi, the question is

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

I put it to this form

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

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

is that right?

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

which means I(x) = e

So i mulptiply both sides by e?

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

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

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...

$$y' \frac{y}{x} = \frac{2x^2}{x}[/maTH] so now i'm meant to integrate both sides, i now get this \displaystyle ylnx = x^2 + c So now what do i do? The initial value is y(1)=0 5. Your equation was: \displaystyle y' - \frac{1}{x} y = 2 x^2 . Multiplying by the integrating factor gives you: \displaystyle \frac{1}{x} y' - \frac{1}{x^2} y = 2 x \displaystyle \Leftrightarrow \frac{d}{dx} \left( \frac{y}{x} \right) = 2 x so \displaystyle \frac{y}{x} = \int 2x \, \mathrm{d}x. You should be able to take it from there, right? 6. I found the answer doing it this way... \displaystyle f(x)=-x^{-1} \displaystyle h(x)=2x^2 \displaystyle I(x)=\frac{1}{x} so \displaystyle \int I(x)h(x)dx \displaystyle =\int (\frac{1}{x})(2x^2)dx \displaystyle =\int 2x \displaystyle x^2+c \displaystyle z(x)=x^2+c general solution = y=\frac{z(x)}{I(x)}$$
so
$\displaystyle y=\frac{x^2+c}{\frac{1}{x}}$
$\displaystyle =x^3+cx$

and the initial value was y(1)=0

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