I can't find an easy way to use the integrating factor to solve this question, any ideas?

$\displaystyle \frac{dy}{dx}+x^3y=0$ The i.f is obviously exp{(x^4)/4} but I don't know where to go from there. thanks

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- Dec 23rd 2009, 03:31 AMi_zz_y_illintegrating factor
I can't find an easy way to use the integrating factor to solve this question, any ideas?

$\displaystyle \frac{dy}{dx}+x^3y=0$ The i.f is obviously exp{(x^4)/4} but I don't know where to go from there. thanks - Dec 23rd 2009, 04:03 AMAnonymitySquared
- Dec 23rd 2009, 07:52 AMSoroban
Hello, i_zz_y_ill!

If you found the integrating factor,

. . the next few steps are part of the procedure.

Exactlyis your difficulty?*where*

Quote:

$\displaystyle \frac{dy}{dx}+x^3y\:=\:0$

The I.F is obviously: .$\displaystyle e^{\frac{1}{4}x^4}$

but I don't know where to go from there. . You don't?

Multiply by the I.F. . $\displaystyle e^{\frac{1}{4}x^4}\frac{dy}{dx} + x^3e^{\frac{1}{4}x^4} y \:=\:0$

We have: .$\displaystyle \frac{d}{dx} \left(e^{\frac{1}{4}x^4}y\right) \;=\;0$

Integrate: .$\displaystyle e^{\frac{1}{4}x^4}y \;=\;C $

Therefore: .$\displaystyle y \;=\;Ce^{-\frac{1}{4}x^4} $