# First Order Linear Differential Equations Problems

• August 31st 2010, 04:05 AM
Paymemoney
First Order Linear Differential Equations Problems
Hi

Ok i have attempted the following question, but need help on what should be done next:

$y'+8y=sin(x)$

p(x)=8
$q(x)=sin(x)$
$e^{\int p(x)dx}$ $= e^{\int 8xdx}$ $= e^{4x^2}$

$e^{4x^2}y'+8ye^{4x^2}=e^{4x^2}sinx$

$\frac{d(e^{4x^2}y)}{dx}=e^{4x^2}sin(x)$

$e^{4x^2} = \int e^{4x^2} sin(x)$
i used integration by parts

$u=e^{4x^2} du=8xe^{4x^2}$
$dv=sin(x)$ $v=-cos(x)$

$=-e^{4x^2}cos(x) - \int -8xe^{4x^2}cos(x)$

what should i do now??
• August 31st 2010, 04:10 AM
Prove It
Integration by parts again.
• August 31st 2010, 04:45 AM
Paymemoney
after first integration
$=-e^{4x^2}cos(x)+8x-e^{4x^2}sin(x)-8[\int sin(x)8x^2e^{4x^2}-sin(x)e^{4x^2}]$
after two integrations
$=-e^{4x^2}cos(x)+8x-e^{4x^2}sin(x)+8e^{4x^2}sin(x)-64[-x^2cos(x)+2[xsin(x)-sin(x)]]$
• September 2nd 2010, 11:34 PM
Paymemoney
Can someone show me how this should be done.

I have tried numerous times and i seem to never stop integrating by parts, therefore it seems to never ends
• September 3rd 2010, 12:00 AM
Prove It
I found out that your original integrating factor is wrong...

Your DE is $\frac{dy}{dx} + 8y = \sin{x}$

The integrating factor is $e^{\int{8\,dx}} = e^{8x}$, so multiplying through gives

$e^{8x}\frac{dy}{dx} + 8e^{8x}y = e^{8x}\sin{x}$

$\frac{d}{dx}(e^{8x}y) = e^{8x}\sin{x}$

$e^{8x}y = \int{e^{8x}\sin{x}\,dx}$

Let $u = e^{8x}$ so that $du = 8e^{8x}$

and $dv = \sin{x}$ so that $v = -\cos{x}$ and the integral becomes

$\int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} - \int{-8e^{8x}\cos{x}\,dx}$

$\int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8\int{e^{8x}\cos{x}\,dx}$.

Now let $u = e^{8x}$ so that $du = 8e^{8x}$

and $dv = \cos{x}$ so that $v = \sin{x}$ and the integral becomes

$\int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8\left(e^{8x}\sin{x} - \int{8e^{8x}\sin{x}\,dx}\right)$

$\int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8e^{8x}\sin{x} - 64\int{e^{8x}\sin{x}\,dx}$

$65\int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8e^{8x}\sin{x}$

$\int{e^{8x}\sin{x}\,dx} = -\frac{1}{65}e^{8x}\cos{x} + \frac{8}{65}e^{8x}\sin{x}$.

So substituting back into the original DE gives

$e^{8x}y = - \frac{1}{65}e^{8x}\cos{x} + \frac{8}{65}e^{8x}\sin{x} + C$

$y = -\frac{1}{65}\cos{x} + \frac{8}{65}\sin{x} + Ce^{-8x}$.
• September 3rd 2010, 01:01 AM
Paymemoney
oh ic now thank you