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
Hello, i_zz_y_ill!
If you found the integrating factor,
. . the next few steps are part of the procedure.
Exactly where is your difficulty?
$\displaystyle \frac{dy}{dx}+x^3y\:=\:0$
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} $