solving exact differential eqn: (1-ysinx)dx + cosx dy = 0

[mathstresser]

Solve the exact equation

(1-ysinx)dx + cosx dy = 0

We know that it's an exact equation because Pdy = (-sin x) and Qdx= -sin x.

But, I don't know what to do with an exact equation.

dy/dx= (ysinx-1)/cosx

dy/dx= ysinx/cosx - 1/cosx

dy/dx= (1/cosx)(yxsinx-1)

I don't know what to do. I can't separate it. I know it's linear, but I don't know what to do.

 

[mark07]

When the DE $\displaystyle Q dx + P dy = 0$ is exact, solution is given by an implicit function $\displaystyle F(x,y)=0$, where $\displaystyle \frac{\partial F}{\partial x}=Q$ and $\displaystyle \frac{\partial F}{\partial y}=P$.

You already know it's exact. So you need to solve the following equations to find F:

1) \(\displaystyle \L \frac{\partial F}{\partial x}= 1- y \sin x\)

2) \(\displaystyle \L \frac{\partial F}{\partial y}= \cos x\)

The second one seems to be easier to start with. Just integrate with respect to y:

\(\displaystyle \L F(x,y) = \int \cos x dy = y \cos x + C(x)\)

Now differentiate this with respect to x and compare with the first equation above to get the function $\displaystyle C(x)$. As I said above, once you find $\displaystyle C(x)$, solution is given implicity by $\displaystyle F(x,y)=0$.