Im having some trouble to solve this differential equation. If somebody knows the solution of this please reply.

So the equation is:

y'+y*sinX=0

I started to solve it but im stucked with a transformation problem. See below. If my method is incorrect please fix it

2. Originally Posted by gergofoto
Im having some trouble to solve this differential equation. If somebody knows the solution of this please reply.

So the equation is:

y'+y*sinX=0

I started to solve it but im stucked with a transformation problem. See below. If my method is incorrect please fix it
then $\ln y = \cos x - C$

$y = e^{\cos x - C}$, let $e^{-C} = A$

then

$y = Ae^{\cos x}$

3. Originally Posted by gergofoto
Im having some trouble to solve this differential equation. If somebody knows the solution of this please reply.

So the equation is:

y'+y*sinX=0

I started to solve it but im stucked with a transformation problem. See below. If my method is incorrect please fix it
It looks good so far

$-\ln(y)=-\cos(x)+C$

First mult both sides by $-1$ to get

$\ln(y)=\cos(x)+C$ Since C is arbitary we don't need the negative sign.

Since the inverse of the log function is the exponential we get

$e^{\ln(y)}=e^{\cos(x)+C}$ Now using exponential properties we get

$y=e^{\cos(x)}\cdot e^{C}$

Once again since $C$ is arbitary we can rename $e^{C}=A$ to get

$y=Ae^{cos(x)}$