solve

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

really helpful if someone could show me full work through i know they're not hard but need gd understanding thanks

Printable View

- May 23rd 2008, 08:01 AMskystarinitial value problem
solve

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

really helpful if someone could show me full work through i know they're not hard but need gd understanding thanks - May 23rd 2008, 08:14 AMKrizalid
$\displaystyle \begin{aligned}

y'&=\frac{y}{x}+2x^{2} \\

y'-\frac{y}{x}&=2x^{2} \\

\frac{y'}{x}-\frac{y}{x^{2}}&=2x \\

\left\{ y\cdot \frac{1}{x} \right\}'&=2x.

\end{aligned}$

In the third step I found the "integrating factor." Do you know what the I.F. is? - May 24th 2008, 03:56 AMskystar
for integrating factor would it be $\displaystyle e^{\int\frac{1}{x}} $ when is the integrating factor negative also,thanks.

- May 24th 2008, 05:58 AMKrizalid
It's actually $\displaystyle \exp \left\{ -\int{\frac{1}{x}\,dx} \right\}.$

- May 24th 2008, 07:17 AMskystar
hmmm why negative.

- May 24th 2008, 07:21 AMTheEmptySet
When you put your equation in standard form

$\displaystyle y'+a(x)y=f(x)$

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

you need to move the $\displaystyle \frac{y}{x}$ term to the left side of the equation

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

Good luck