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

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 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 dooogle
 Sci Advisor HW Helper PF Gold P: 2,603 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
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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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