initial value problem

**skystar** · May 23rd 2008, 08:01 AM

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

**Krizalid** · May 23rd 2008, 08:14 AM

$\begin{aligned} y'&=\frac{y}{x}+2x^{2} \\ y'-\frac{y}{x}&=2x^{2} \\ \frac{y'}{x}-\frac{y}{x^{2}}&=2x \\ \left\{ y\cdot \frac{1}{x} \right\}'&=2x. \end{aligned}$

In the third step I found the "integrating factor." Do you know what the I.F. is?

**skystar** · May 24th 2008, 03:56 AM

for integrating factor would it be $e^{\int\frac{1}{x}}$ when is the integrating factor negative also,thanks.

**Krizalid** · May 24th 2008, 05:58 AM

It's actually $\exp \left\{ -\int{\frac{1}{x}\,dx} \right\}.$

**skystar** · May 24th 2008, 07:17 AM

hmmm why negative.

**TheEmptySet** · May 24th 2008, 07:21 AM

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

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