# Solving a differential equation

1. Sep 8, 2009

### hardatwork

1. The problem statement, all variables and given/known data
Find the particular solution of the differential equation
$$\frac{dy}{dx}$$+ycos(x)=3cos(x)

2. Relevant equations

y(0)=5

3. 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.

2. Sep 8, 2009

### slider142

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.

3. Sep 8, 2009

### hardatwork

How would I simplify futher?

4. Sep 8, 2009

### slider142

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

5. Sep 8, 2009

### hardatwork

Oh! Okay that makes. Thank You so much!