Thread: Differential Equations - Intergrating by factors

Hi

Need help on the following question:

Find the general solution of the differential equation

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

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

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

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

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

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

this is what i did

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

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

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

P.S

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

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


Multiplying through by the integrating factor gives

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

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

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

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


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

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

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

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

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

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

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

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

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