Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)

Martingale · Postby **Martingale** » Fri Apr 17, 2009 1:53 am

Martingale wrote:
$y(\pi)=\frac{3\sin(\pi^5)}{\pi^4}+\frac{c}{\pi^4}=0$

solve for $c$

$\Leftrightarrow \frac{3\sin(\pi^5)}{\pi^4}=-\frac{c}{\pi^4}\Leftrightarrow -3\sin(\pi^5)=c$

doggy · Postby **doggy** » Fri Apr 17, 2009 2:02 am

I don't understand why it isn't c =0

Martingale · Postby **Martingale** » Fri Apr 17, 2009 2:04 am

doggy wrote:I don't understand why it isn't c =0

if c=0 then the answer would be what you had

doggy wrote:So then the final answer would be c=0

and y= 3 sin(x^5)/x^4?

but does $y(\pi)=0$ for this function? no it doesn't.

doggy · Postby **doggy** » Fri Apr 17, 2009 2:05 am

So then the final answer would be 3 sin(x^5)/x^4 -3 sin(pi^5)/x^4?

Martingale · Postby **Martingale** » Fri Apr 17, 2009 2:07 am

doggy wrote:So then the final answer would be 3 sin(x^5)/x^4 -3 sin(pi^5)/x^4?

yes

doggy · Postby **doggy** » Fri Apr 17, 2009 2:18 am

So the final answer would be y = 3 sin(x^5) - 3 sin (pi^5)

Martingale · Postby **Martingale** » Fri Apr 17, 2009 2:24 am

doggy wrote:So the final answer would be y = 3 sin(x^5) - 3 sin (pi^5)

no...what you had before is the correct answer...you don't remove the x^4 that is on the bottom.

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