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Math Help - Integrating Factor Differentiation

  1. #1
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    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
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  2. #2
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    [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)
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  3. #3
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    Quote Originally Posted by galactus View Post
    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.
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  4. #4
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    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}

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  5. #5
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    Quote Originally Posted by Soroban View Post
    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?

    Quote Originally Posted by Soroban View Post
    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
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  6. #6
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    Rewrite:

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