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## Homework Statement

integrate

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

## Homework Equations

## 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