# Solving a differential equation

## Homework Statement

Find the particular solution of the differential equation
$$\frac{dy}{dx}$$+ycos(x)=3cos(x)

y(0)=5

## The Attempt at a Solution

$$\frac{dy}{dx}$$=3cos(x)-ycos(x)
$$\frac{dy}{dx}$$=(cos(x))(3-y)
$$\int \frac{dy}{3-y}$$=$$\int cos(x) dx$$
-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.

Your steps and your solution have no flaws. However, you can simplify your solution further. Also, it is not necessary to continue using the absolute value signs when the argument is positive.

How would I simplify futher?

How would I simplify futher?

Remember that $a^{b+c} = a^ba^c$ and that the exponential function is the inverse function of the natural logarithm.

Oh! Okay that makes. Thank You so much!