Differential Equation

Can anyone agree/disagree?

I think, best to swiftly draw a line under that lot.

$y'+\cot{x}=e^{-x}$

So

$y'=e^{-x}-\cot{x}$


$y=\int e^{-x}-\cot{x}\;dx$

oops its meant to be $\cot{x} y$

How do you solve

$y'+y \cot{x}=e^{-x}$

it's been years since i did this stuff.

is it the old complementary function and particular integral?

if so, i get

$\frac{dy}{dx}=-\cot{x}$
$\frac{dy}{y}=-\cot{x} dx$
$\ln{y}=-\ln{\sin{x}}+C$
$y=Be^{-\ln{\sin{x}}}$
$y=\frac{C}{\sin{x}}$

and then for the particular integral if we take $y=ke^{-x}$

the equation implies

$ke^{-x} + k \cot{x} e^{-x} = e^{-x}$
$k+k \cot{x} = 1$
$k (1 + \cot{x} ) = 1$
$k = \frac{1}{(1 + \cot{x})}$


and so the general solution is

$y= \frac{C}{\sin{x}} + \frac{1}{(1 + \cot{x})} e^{-x}$

Can anyone agree/disagree?

How do you solve

$y'+y \cot{x}=e^{-x}$

it's been years since i did this stuff.

is it the old complementary function and particular integral?

if so, i get

$\frac{dy}{dx}=-\cot{x}$
$\frac{dy}{y}=-\cot{x} dx$
$\ln{y}=-\ln{\sin{x}}+C$
$y=Be^{-\ln{\sin{x}}}$
$y=\frac{C}{\sin{x}}$

and then for the particular integral if we take $y=ke^{-x}$

the equation implies

$ke^{-x} + k \cot{x} e^{-x} = e^{-x}$
$k+k \cot{x} = 1$
$k (1 + \cot{x} ) = 1$
$k = \frac{1}{(1 + \cot{x})}$


and so the general solution is

$y= \frac{C}{\sin{x}} + \frac{1}{(1 + \cot{x})} e^{-x}$

Can anyone agree/disagree?