Solving a First Order DE: y'+ycot(x)=cos(x)

[iRaid]

Homework Statement

y'+ycot(x)=cos(x)

Homework Equations

The Attempt at a Solution

First I found the integrating factor:
\rho (x)=e^{\int cot(x)dx}=e^{ln(sinx)}=sinx
Plug into the equation for first order DE...
\int \frac{d}{dx} ysinx=\int cosxsinx dx
End up with:
ysinx=\frac{-cos^{2}x}{2}+C\\y=\frac{-cosxcotx}{2}+\frac{C}{sinx}

I think that's wrong tho..

 

[pasmith]

Homework Statement

y'+ycot(x)=cos(x)

Homework Equations

The Attempt at a Solution

First I found the integrating factor:
\rho (x)=e^{\int cot(x)dx}=e^{ln(sinx)}=sinx
Plug into the equation for first order DE...
\int \frac{d}{dx} ysinx=\int cosxsinx dx
End up with:
ysinx=\frac{-cos^{2}x}{2}+C\\y=\frac{-cosxcotx}{2}+\frac{C}{sinx}


I think that's wrong tho..

You have the right answer. But since \cos^2 x + \sin^2 x = 1, it follows that (\sin^2 x)' = -(\cos^2 x)'. Perhaps it would have been better to take \sin x \cos x = \frac12(\sin^2 x)', since you have (y\sin x)' on the other side of the equation.

You can in any event use \cos^2 x + \sin^2 x = 1 to simplify your answer:
<br /> \frac{\cos^2 x}{\sin x} = \frac{1 - \sin^2 x}{\sin x} = \frac{1}{\sin x} - \sin x