# Thread: Differential Equations - Intergrating by factors

1. ## Differential Equations - Intergrating by factors

Hi

Need help on the following question:

Find the general solution of the differential equation

$\frac{dy}{dx}-\frac{4y}{x}=e^{x^{-3}}$

$p(x)=\frac{-4}{x}$

$q(x)=e^{x^{-3}}$

$I=e^{\int \frac{-4}{x} dx$
$I=e^{ln|x^{-4}|}$
$I=x^{-4}$

$x^{-4}y=\int x^{-4}e^{x^{-3}} dx$

ok, i tried to integrate by parts however i never get the correct answer.

this is what i did

$u = e^{x^{-3}}$
$du = -3x^{x^{-3}}$

$dv = x^{-5}$
$v = \frac{x^{-4}}{-4}$

i have also tried the other way but still not correct.

P.S

2. $\frac{dy}{dx} - \frac{4}{x}y = e^{x^{-3}}$.

The integrating factor is $e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{x^{-4}}} = x^{-4}$.

Multiplying through by the integrating factor gives

$x^{-4}\frac{dy}{dx} - 4x^{-5}y = x^{-4}e^{x^{-3}}$

$\frac{d}{dx}(x^{-4}y) = x^{-4}e^{x^{-3}}$

$x^{-4}y = \int{x^{-4}e^{x^{-3}}\,dx}$

$x^{-4}y = -\frac{1}{3}\int{e^{x^{-3}}(-3x^{-4})\,dx}$.

Now let $u = x^{-3}$ so that $\frac{du}{dx} = -3x^{-4}$ and the DE becomes

$x^{-4}y = -\frac{1}{3}\int{e^u\,\frac{du}{dx}\,dx}$

$x^{-4}y = -\frac{1}{3}\int{e^u\,du}$

$x^{-4}y = -\frac{1}{3}e^u + C$

$x^{-4}y = -\frac{1}{3}e^{x^{-3}} + C$

$y = -\frac{1}{3}x^4e^{x^{-3}} + Cx^4$.

3. i don't understand how you got the following line?
$x^{-4}y = -\frac{1}{3}\int{e^{x^{-3}}(-3x^{-4})\,dx}$

4. I knew I needed to "create" the derivative of $x^{-3}$, which is $-3x^{-4}$.

I already had the $x^{-4}$, so all I needed to do was multiply by $-3$. But to keep the equation balanced, that means I had to multiply by $-\frac{1}{3}$ as well.