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.

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.

Find the particular solution of the differential equation:

xy ' +4y = 15xcos(x^5) sastisfying the initial condition y(pi)=0.

- Martingale
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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doggy wrote: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.

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

- Martingale
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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doggy wrote:I have tried using an integrating factor of x^4 but I'm still not getting the correct answer

then

so

now take care of the initial condition

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

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
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
**Contact:**

doggy wrote: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 ?

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

- Martingale
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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doggy wrote:Don't I need the cos (pi)^5?

no.

Martingale wrote:so

now take care of the initial condition

solve for

So then the final answer would be c=0

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

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

- Martingale
**Posts:**350**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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doggy wrote:So then the final answer would be c=0

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

no