initial value problem

• May 23rd 2008, 08:01 AM
skystar
initial value problem
solve

$y'=\frac{y}{x} +2x^{2}$, 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 AM
Krizalid
\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 AM
skystar
for integrating factor would it be $e^{\int\frac{1}{x}}$ when is the integrating factor negative also,thanks.
• May 24th 2008, 05:58 AM
Krizalid
It's actually $\exp \left\{ -\int{\frac{1}{x}\,dx} \right\}.$
• May 24th 2008, 07:17 AM
skystar
hmmm why negative.
• May 24th 2008, 07:21 AM
TheEmptySet
When you put your equation in standard form

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

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

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

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

Good luck