1. ## Integrating Factor Differentiation

Hello. I need help on the following questions. I've done some stages but then I'm unaware how to go further so can you please check the work and show the steps ahead. If there are mistakes, please amend them. Thanks in advance.
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1.

$\frac{\mathrm{d}y}{\mathrm{d}x} + ycotx = cos3x$
$\mathrm{Integrating factor} = tanx$
$(. tanx)$
$tanx\frac{\mathrm{d}y}{\mathrm{d}x} + y = tanxcosx$
$tanx = \int(tanxcos3x)dx$

2.

$\frac{\mathrm{d}y}{\mathrm{d}x} + 2ytanx = sinx$
$\mathrm{Integrating factor} = sec^2x$
$(.sec^2x)$
$sec^2x\frac{\mathrm{d}y}{\mathrm{d}x} + 2ytanxsec^2x = sinxsec^2x$
$ysec^2x = \int(sinxsec^2x)dx$

3.

$\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{y}{x+1} = x$
$\mathrm{Integrating factor} = \frac{1}{x+1}$
$(. \frac{1}{x+1})$
$\frac{1}{x+1} \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{y}{(x+1)^2} = \frac {x}{x+1}$
$\frac {y}{x+1} = \int(\frac{x}{x+1})dx$

2. [quote=Air;76656]Hello. I need help on the following questions. I've done some stages but then I'm unaware how to go further so can you please check the work and show the steps ahead. If there are mistakes, please amend them. Thanks in advance.
__________________________________________________ _______________

1.

$\frac{\mathrm{d}y}{\mathrm{d}x} + ycotx = cos3x$
$\mathrm{Integrating factor} = tanx$
$(. tanx)$
$tanx\frac{\mathrm{d}y}{\mathrm{d}x} + y = tanxcosx$
$tanx = \int(tanxcos3x)dx$
The integrating factor is cot(x),

So we have $e^{\int{cot(x)}dx}=sin(x)$

3. Originally Posted by galactus
The integrating factor is cot(x),

So we have $e^{\int{cot(x)}dx}=sin(x)$
Oh yeah. I had done it by changing $cotx$ to $\frac{1}{tanx}$ and thinking it would be $tanx$.

4. Hello, Air!

$1)\;\;\frac{dy}{dx} + y\!\cdot\cot x \:= \:\cos3x$
$\text{Integrating factor: }\: I \;=\;e^{\int\cot x\,dx} \;=\;e^{\ln|\sin x|} \;=\;\sin x$

Then we have: . $\sin x\cdot\frac{dy}{dx} + \sin x\cdot\cot x\cdot y \;=\;\sin x\cos3x$

. . $\frac{d}{dx}(y\cdot\sin x) \;=\;\sin x\cos3x$

. . $y\cdot\sin x \;=\;\int\sin x\cos3x\,dx$ . . . etc.

$2)\;\;\frac{dy}{dx} + 2y\!\cdot\tan x \:= \:\sin x$

$\text{Integrating factor} = \sec^2x$ . . . . Right!

$\sec^2x\!\cdot\!\frac{dy}{dx} + 2\sec^2\!x\tan x\cdot y \:= \;\sin x\sec^2x$

$y\!\cdot\!\sec^2x \:= \:\int \sin x\sec^2\!x\,dx$
Note that: . $\sin x\sec^2x \:=\:\sin x\cdot\frac{1}{\cos^2x} \:=\:\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} \:=\:\sec x\tan x$

$3)\;\;\frac{dy}{dx} - \frac{y}{x+1} = x$

$\text{Integrating factor} = \frac{1}{x+1}$ . . . . Yes!

$\frac{1}{x+1} \frac{dy}{dx} - \frac{y}{(x+1)^2} \:= \:\frac {x}{x+1}$

$\frac {y}{x+1} \:= \:\int\left(\frac{x}{x+1}\right)\,dx$ . . . . Good!
Note that: . $\frac{x}{x+1} \:=\:\frac{x + 1 - 1}{x+1} \:=\:\frac{x+1}{x+1} - \frac{1}{x+1} \:=\:1 - \frac{1}{x+1}$

5. Originally Posted by Soroban
Hello, Air!

. . $y\cdot\sin x \;=\;\int\sin x\cos3x\,dx$ . . . etc.
How would I do about integrating this? Do I used substitution of $u=cos3x$?

Originally Posted by Soroban
Hello, Air!
Note that: . $\sin x\sec^2x \:=\:\sin x\cdot\frac{1}{\cos^2x} \:=\:\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} \:=\:\sec x\tan x$

Note that: . $\frac{x}{x+1} \:=\:\frac{x + 1 - 1}{x+1} \:=\:\frac{x+1}{x+1} - \frac{1}{x+1} \:=\:1 - \frac{1}{x+1}$

For these, Do I integrate the equivalent values?
E.g:
$\int(\sec x \tan x)dx$ & $\int(1 - \frac{1}{x+1})dx$

6. Rewrite:

$\int{sin(x)cos(3x)}dx=\frac{1}{2}\int{sin(4x)}dx-\frac{1}{2}\int{sin(2x)}dx$