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

Limits, differentiation, related rates, integration, trig integrals, etc.
doggy
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### Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)

I'm having a bit of trouble figuring out this differential equation. The question is:

Find the particular solution of the differential equation:
xy ' +4y = 15xcos(x^5) sastisfying the initial condition y(pi)=0.

Martingale
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### Re: Differential Equations

I'm having a bit of trouble figuring out this differential equation. The question is Find the particular solution of the differential equation:
xy ' +4y = 15xcos(x^5) sastisfying the initial condition y(pi)=0.

Use an Integrating factor.

doggy
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### Re: Differential Equations

I have tried using an integrating factor of x^4 but I'm still not getting the correct answer

Martingale
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### Re: Differential Equations

I have tried using an integrating factor of x^4 but I'm still not getting the correct answer

$xy ' +4y = 15x\cos(x^5)$

$y ' +\frac{4}{x}y = 15\cos(x^5)$

$x^4y ' +4x^3y = 15x^4\cos(x^5)$

$(x^4y)'=15x^4\cos(x^5)$

then $x^4y=3\sin(x^5)+c$

so $y=\frac{3\sin(x^5)}{x^4}+\frac{c}{x^4}$

now take care of the initial condition

doggy
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### Re: Differential Equations

I have gotten up to that point as well but I'm having trouble with the conditions when I sub in pi I got:
3 cos(pi)^5/(pi)^4 + c/(pi/4) =0
thus c= -3 cos((pi)^5)

So the final answer should be :
3 cos (x^5) - 3 cos(pi)^5 but that's not correct

Martingale
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### Re: Differential Equations

I have gotten up to that point as well but I'm having trouble with the conditions when I sub in pi I got:
3 cos(pi)^5/(pi)^4 + c/(pi/4) =0
thus c= -3 cos((pi)^5)

So the final answer should be :
3 cos (x^5) - 3 cos(pi)^5 but that's not correct

Why do you have $\cos(\pi^5)$?

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

Don't I need the cos (pi)^5?

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

Don't I need the cos (pi)^5?
no.
so $y=\frac{3\sin(x^5)}{x^4}+\frac{c}{x^4}$

now take care of the initial condition
$y(\pi)=\frac{3\sin(\pi^5)}{\pi^4}+\frac{c}{\pi^4}=0$

solve for $c$

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

So then the final answer would be c=0

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

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

So then the final answer would be c=0

and y= 3 sin(x^5)/x^4?
no