First order of equation: (dy/dx) + (cot x)y = csc x

I'm brand new to this stuff, and got stuck on this problem. Anyone has a minute? Thanks

Find the general solution of this equation:

(dy/dx)+ (cot x)y = csc x

This actually falls in place pretty easily.

You have a linear first order DE.

\(\displaystyle \L\\\frac{dy}{dx}+ycot(x)=csc(x)\)

Find the integrating factor:

\(\displaystyle \L\\e^{\int{cot(x)}dx}=e^{ln(sin(x))}=sin(x)\)

\(\displaystyle \L\\\frac{d}{dx}[ysin(x)]=sin(x)csc(x)\)

Integrate:

\(\displaystyle \L\\ysin(x)=x+C\)

Solve for y:

\(\displaystyle \L\\y=\frac{x+C}{sin(x)}\)

thank you sir, actually, i got stuck at e^ln(sinx), i forgot all the properties of exponent. Perhaps you know a site for me to refesh that? Thanks

Try purplemath.com

okay, working on the last one tonight, here what i have so far

y' +ay = be ^-kt

\(\displaystyle \L\\\frac{dy}{dx}+ay=be^{-kt}\)
\(\displaystyle \L\\e^{\int{a}dt}=e^{t}\)

\(\displaystyle \L\\\frac{d}{dx}[ye^{t}]=b e^{t} e^{-kt]\)


how's that look so far? what i'm i gonna do with those two e on the right side? btw, -k = - lamda ( i dont know how to get the symbol). Thanks

Your integrating factor is wrong. It should be \(\displaystyle e^{at}\)

Skipping ahead to the end after integration

\(\displaystyle \L\\ye^{at}=\frac{be^{(a-\lambda)t}}{a-k}+C\)

Divide through to isolate y:

\(\displaystyle \L\\y=\frac{be^{{-\lambda}t}}{a-k}+Ce^{-at}\)

Incidentally, to get a Greek letter, type {\lambda}. If you want a capital just capitalize the first letter.