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Math Help - Can someone check my answers?

  1. #1
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    Can someone check my answers?

    Hey people,

    I would just like a few of my answers to be checked, if anyone wants to,

    1) y'=e(-y)-1

    I got ln(1-e(y))+x=c, but i think this is wrong(it doesn't match the given answer)

    My working was:

    dy/dx=1/e(y)-1
    dy/dx=(1-e(y))/e(y)
    int (1-e(y))/e(y) dy -int 1 dx=0
    ln(1-e(y))+x=c

    2) (dy/dx)tanx+tany=0

    I got sinysinx=c

    3) Find the general solution of:

    d2y/dx2-5dy/dx+4y=2e(3x)

    I got: Ae(4x)+Be(x)-e(3x)

    4) d2y/dt2+4y=12cos(2t)

    i) find Yc:

    I got Yc=Acos(2t)+Bsin(2t)

    ii) find the full general solution:

    I got y=Acos(2t)+Bsin(2t)+3tsin(2t)

    iii) given y'(0)=0 and y(0)=0 find the full solution

    I got y=3tsin(2t)

    5) evaluate int e(2x)sinx dx via trap rule, simpsons 1/3rd and gauss integration. With 4 strips for trapezoidal/simpsons.

    I got:

    Trap: 23.02829

    Simp: 22.70988612

    Gauss: 22.768

    Any answer checks would be greatly appreciated, especially for the first question.
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  2. #2
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    Quote Originally Posted by deragon999 View Post
    Hey people,

    I would just like a few of my answers to be checked, if anyone wants to,

    1) y'=e(-y)-1

    I got ln(1-e(y))+x=c, but i think this is wrong(it doesn't match the given answer)

    My working was:

    dy/dx=1/e(y)-1
    dy/dx=(1-e(y))/e(y)
    int (1-e(y))/e(y) dy -int 1 dx=0
    ln(1-e(y))+x=c

    2) (dy/dx)tanx+tany=0

    I got sinysinx=c

    3) Find the general solution of:

    d2y/dx2-5dy/dx+4y=2e(3x)

    I got: Ae(4x)+Be(x)-e(3x)

    4) d2y/dt2+4y=12cos(2t)

    i) find Yc:

    I got Yc=Acos(2t)+Bsin(2t)

    ii) find the full general solution:

    I got y=Acos(2t)+Bsin(2t)+3tsin(2t)

    iii) given y'(0)=0 and y(0)=0 find the full solution

    I got y=3tsin(2t)

    5) evaluate int e(2x)sinx dx via trap rule, simpsons 1/3rd and gauss integration. With 4 strips for trapezoidal/simpsons.

    I got:

    Trap: 23.02829

    Simp: 22.70988612

    Gauss: 22.768

    Any answer checks would be greatly appreciated, especially for the first question.
    Dear deragon999,

    I suppose you mean y'=e^{-y}-1 in the first question. Is'nt??

    If that is so,

    \frac{dy}{dx}=e^{-y}-1

    \frac{dy}{dx}=\frac{1-e^{y}}{e^y}

    \int{\frac{e^y}{1-e^y}}dy=\int{dx}

    -\int{\frac{e^y}{e^y-1}}dy=\int{dx}

    -\int{\frac{d}{dy}(e^{y}-1)\frac{d}{d(e^y-1)}ln(e^y-1)}dy=\int{dx}

    -\int{d(ln(e^y-1))}=\int{dx}

    -ln(e^y-1)=x+C ; Where C is an arbitary constant.

    Hope this will help you.
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  3. #3
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    Sudharaka

    Thanks, can you show me how you got



    Thanks,
    Simon
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  4. #4
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    Quote Originally Posted by deragon999 View Post
    Sudharaka

    Thanks, can you show me how you got



    Thanks,
    Simon
    Dear deragon999,

    \frac{dy}{dx}=e^{-y}-1=\frac{1}{e^y}-1=\frac{1-e^y}{e^y}

    Was it helpful?
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  5. #5
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    Haha not relly, that bit i had already done in my working, the bit that you did different to me was taking -1 out of the dy integral

    Thanks you for all your time spent anyway
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