# Thread: First Order Linear Differential Equations Problems

1. ## First Order Linear Differential Equations Problems

Hi

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

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

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

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

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

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

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

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

what should i do now??

2. Integration by parts again.

3. after first integration
$\displaystyle =-e^{4x^2}cos(x)+8x-e^{4x^2}sin(x)-8[\int sin(x)8x^2e^{4x^2}-sin(x)e^{4x^2}]$
after two integrations
$\displaystyle =-e^{4x^2}cos(x)+8x-e^{4x^2}sin(x)+8e^{4x^2}sin(x)-64[-x^2cos(x)+2[xsin(x)-sin(x)]]$

4. 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

5. I found out that your original integrating factor is wrong...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

So substituting back into the original DE gives

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

$\displaystyle y = -\frac{1}{65}\cos{x} + \frac{8}{65}\sin{x} + Ce^{-8x}$.

6. oh ic now thank you