Integrating Factor Differentiation

**Simplicity** · October 13th 2007, 02:05 PM

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.

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

**galactus** · October 13th 2007, 02:25 PM

[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 $e^{\int{cot(x)}dx}=sin(x)$

**Simplicity** · October 13th 2007, 02:28 PM

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 $tanx$ .

**Soroban** · October 13th 2007, 02:52 PM

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

**Simplicity** · October 13th 2007, 03:07 PM

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