• Jan 10th 2010, 09:18 AM
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(Happy)
• Jan 10th 2010, 09:25 AM
dedust
Quote:

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(Happy)

then $\displaystyle \ln y = \cos x - C$

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

then

$\displaystyle y = Ae^{\cos x}$
• Jan 10th 2010, 09:28 AM
TheEmptySet
Quote:

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(Happy)

It looks good so far

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

First mult both sides by $\displaystyle -1$ to get

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

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

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

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

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

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