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Stuck trying to integrate dy/dx-y=cos(x)-2

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dooogle
#1
Nov11-10, 04:19 PM
P: 21
1. The problem statement, all variables and given/known data

integrate

dy/dx-y=cos(x)-2

2. Relevant equations
3. The attempt at a solution

dy/dx-y=cos(x)-2

is in the form

dy/dx+p(x)=q(x)

take the I.F as e^int(p(x))dx=e^-x

multiplying throughout by e^-x

d(e^-x)z/dx=-(cos(x)e^-x)-(2e^-x)

(e^-x)z=-int((cos(x)e^-x)-(2e^-x))dx

so -int((cos(x)e^-x)-(2e^-x))dx=-int(cos(x)(e^-x)dx-2int(e^-x)dx

2int(e^-x)dx=-2(e^-x)+c

-int(cos(x)(e^-x)) this is where i have trouble im not sure if my method is correct using integration by parts multiple times

1st integration
let u=e^(-x) du/dx=-e^(-x) dv/dx=cos(x) v=-sin(x)

-int(cos(x)(e^-x))dx = -e^(-x)sin(x)+int(-sin(x)e^(-x))dx

2nd integration
int(-sin(x)e^(-x))dx
let u=e^(-x) du/dx=-e^(-x) dv/dx=-sin(x) v=cos(x)

e^(-x)cos(x)+int(cos(x)(e^-x))dx

so

-int(cos(x)(e^-x)dx=-e^(-x)sin(x)+e^(-x)cos(x)+int(cos(x)(e^-x))dx

this is where im not sure i take away from both sides int(cos(x)(e^-x))dx giving

-2int(cos(x)(e^-x)dx=-e^(-x)sin(x)+e^(-x)cos(x)

int(cos(x)(e^-x)dx=e^(-x)/2(sin(x)+cos(x))

so
(e^-x)z=-2(e^-x)+c+e^(-x)/2(sin(x)+cos(x))+c

z=-2+(sin(x)+cos(x))/2+c/(e^-x)

1/y=-2+(sin(x)+cos(x))/2+c/(e^-x)

y=1/(-2+(sin(x)+cos(x))/2+c/(e^-x))

thank you for your time to recap my problems is:

im not sure if i am allowed to add integrals

cheers

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fzero
#2
Nov11-10, 07:49 PM
Sci Advisor
HW Helper
PF Gold
P: 2,603
It's easier to do that integral by using the Euler formula

[tex]e^{\pm ix} = \cos x \pm i \sin x[/tex]

to obtain an expression for [tex]\cos x[/tex] in terms of exponentials. However, your integration is correct if you mean

int(cos(x)(e^-x)dx= (e^(-x)/2 ) (sin(x)+cos(x)) +c
HallsofIvy
#3
Nov11-10, 09:40 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,348
Sorry but the use of two "/"s in
y=1/(-2+(sin(x)+cos(x))/2
is meaning less to me.


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