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

1. Nov 11, 2010

### dooogle

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

dooogle

2. Nov 11, 2010

### fzero

It's easier to do that integral by using the Euler formula

$$e^{\pm ix} = \cos x \pm i \sin x$$

to obtain an expression for $$\cos x$$ 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

3. Nov 11, 2010

### HallsofIvy

Staff Emeritus
Sorry but the use of two "/"s in
is meaning less to me.