1. ## integrating factor

I can't find an easy way to use the integrating factor to solve this question, any ideas?
$\frac{dy}{dx}+x^3y=0$ The i.f is obviously exp{(x^4)/4} but I don't know where to go from there. thanks

2. Originally Posted by i_zz_y_ill
I can't find an easy way to use the integrating factor to solve this question, any ideas?
$\frac{dy}{dx}+x^3y=0$ The i.f is obviously exp{(x^4)/4} but I don't know where to go from there. thanks

This is a separable ODE, there is no need for integrating factors.

$\frac{dy}{y} = - x^3 dx$.

3. Hello, i_zz_y_ill!

If you found the integrating factor,
. . the next few steps are part of the procedure.

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

The I.F is obviously: . $e^{\frac{1}{4}x^4}$
but I don't know where to go from there. . You don't?

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

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

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

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