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## Homework Statement

Find the particular solution of the differential equation

[tex]\frac{dy}{dx}[/tex]+ycos(x)=3cos(x)

## Homework Equations

y(0)=5

## The Attempt at a Solution

[tex]\frac{dy}{dx}[/tex]=3cos(x)-ycos(x)

[tex]\frac{dy}{dx}[/tex]=(cos(x))(3-y)

[tex]\int \frac{dy}{3-y}[/tex]=[tex]\int cos(x) dx[/tex]

-ln|3-y|=sin(x)+C

ln|3-y|=-sin(x)-C

to find C, I did

ln|3-5|=-sin(0)-C

ln|-2|=-C

C=-ln|2|

So the equation is

3-y=e

^{(-sin(x)+ln|2|)}

-y=e

^{(-sin(x)+ln|2|)}-3

y=-e

^{(-sin(x)+ln|2|)}+3

But this is incorrect. I'm not sure if my setup is right or if I made a mistake. I would greatly appreciate if someone could see were I went wrong.