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Math Help - Two First Order Differentia Questions

  1. #1
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    Two First Order Differentia Questions

    Hi

    I am having trouble getting the correct answer according to my textbook
    This is what i have done:
    1) y'+3y=2cosx
    I=e^{3x}
    e^{3x}(y'+3y)=2e^{3x}cos(x)
    e^{3x}y=2\int e^{3x}cos(x) dx

    u=e^{3x} du=3e^{3x}
    dv=cos(x) v=sin(x)

    2\int e^{3x}cos(x)=e^{3x}sin(x)-3\int e^{3x}sin(x) dx

    u=e^{3x} du=3e^{3x}
    dv=sin(x)  v=-cos(x)

    2\int e^{3x}cos(x)=e^{3x}sin(x)-3[-e^{3x}cos(x) -\int -3e^{3x}cos(x) dx]

    2\int e^{3x}cos(x)=e^{3x}sin(x)+3e^{3x}cos(x) -3[3\int e^{3x}cos(x) dx]

    2\int e^{3x}cos(x)=e^{3x}sin(x)+3e^{3x}cos(x) +9\int e^{3x}cos(x) dx
    11\int e^{3x}cos(x) = e^{3x}sin(x)+3e^{3x}cos(x)
    \int e^{3x}cos(x) = \frac{e^{3x}sin(x)+3e^{3x}cos(x)}{11}

    go back to original equation:
    e^{3x}y=\frac{e^{3x}sin(x)+3e^{3x}cos(x)}{11}
    y=\frac{1}{11}sin(x)+\frac{3}{11}3cos(x)

    2) y' +3y=5cos(2x)
    I=e^{3x}

    e^{3x}(y'+3y)=e^{3x}(5cos(2x))
    \frac{d(e^{3x})}{dx}=5e^{3x}cos(2x)
    e^{3x}y=5\inte^{3x}cos(2x)

    u=e^{3x}  du=3e^{3x}
    dv=cos(2x) v=\frac{sin(2x)}{2}

    5\int e^{3x}cos(2x)=\frac{1}{2}e^{3x}sin(2x)-\frac{3}{2}\int e^{3x}sin(2x)

    u=e^{3x} du=3e^{3x}
    dv=sin(2x) v=\frac{-cos(2x)}{2}

    5\int e^{3x}cos(2x)=\frac{1}{2}e^{3x}sin(2x)-\frac{3}{2}[e^{3x}*\frac{-cos(2x)}{2}]+\frac{3}{2}[\frac{-3e^{3x}cos(2x)}{2}]<br />

    5\int e^{3x}cos(2x)=\frac{1}{2}e^{3x}sin(2x)-\frac{3}{4}e^{3x}-cos(2x)-\frac{9}{4}e^{3x}cos(2x)<br />

    \frac{29}{4}\int e^{3x}cos(2x)=\frac{1}{2}e^{3x}sin(2x)-\frac{3}{4}e^{3x}

    \int e^{3x}cos(2x)=\frac{\frac{1}{2}e^{3x}sin(2x)-\frac{3}{4}e^{3x}}{\frac{29}{4}}

    sub back into original equation:
    y=\frac{2}{29}sin(2x)+\frac{3}{29}cos(2x)+Ce^{-3x}

    P.S
    Last edited by Paymemoney; September 3rd 2010 at 08:21 PM.
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  2. #2
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    With regard to your solution of 1): a. You forgot the constant of integration on the RHS. b. I think your constants might be a little bit off. Check them again, especially with respect to your integration by parts.

    With regard to your solution of 2): I think your constants are off again. Check the same thing as with your solution of #1.

    In general, you've got the right method of solution. Just make sure the details are correct!
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  3. #3
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    after doing the questions again i get a different answer.

    This time i have integrated with the constant values inside the integral.

    1)
    e^{3x}=\int 2e^{3x}cos(x)
    u=2e^{3x} du=6e^{3x}

    dv=cos(x) v=sin(x)

    \int 2e^{3x}cos(x) = 2e^{3x}sin(x) - 6\int e^{3x}sin(x)
    u=e^{3x}  du=3e^{3x}

    dv=sin(x) v=-cos(x)

    \int 2e^{3x}cos(x) = 2e^{3x}sin(x) + 6e^{3x}cos(x)-18e^{3x}cos(x)
    \int 20e^{3x}cos(x) = 2e^{3x}sin(x) + 6e^{3x}cos(x)
    \int e^{3x}cos(x)=\frac{2}{20}e^{3x}sin(x)+\frac{6}{20}  e^{3x}cos(x)
    \int e^{3x}cos(x)=\frac{1}{10}e^{3x}sin(x)+\frac{3}{10}  e^{3x}cos(x)

    e^{3x}y=\frac{1}{10}e^{3x}sin(x)+\frac{3}{10}e^{3x  }cos(x) + C

    y=\frac{1}{10}sin(x)+\frac{3}{10}cos(x) +Ce^{-3x}

    2)
    e^{3x}y=\int 5e^{3x}cos(2x)

    u=5e^{3x} du=15e^{3x}
    dv=cos(2x) v=sin(2x)
    <br />
/int 5e^{3x}cos(2x)=/frac{5}{2}e^{3x}sin(2x)-\int \frac{15}{2}e^{3x}sin(2x)

    u=\frac{15}{2}e^{3x} du=\frac{45}{2}e^{3x}
    dv=cos(2x) v=\frac{-cos(2x)}{2}

    \int 5e^{3x}cos(2x)=\frac{5}{2}e^{3x}sin(2x)+\frac{15}{  4}e^{3x}cos(2x)+\frac{45}{4}e^{3x}cos(2x)

    \frac{-25}{4}/int e^{3x}cos(2x)=\frac{5}{2}e^{3x}sin(2x)+\frac{15}{4  }e^{3x}cos(2x)<br />

    \int e^{3x}cos(2x)=\frac{-20}{50}e^{3x}sin(2x)+\frac{60}{100}e^{3x}cos(2x)<br />

    \int e^{3x}cos(2x)=\frac{-2}{5}e^{3x}sin(2x)+\frac{3}{2}e^{3x}cos(2x)<br />

    e^{3x}y=\frac{-2}{5}e^{3x}sin(2x)+\frac{3}{2}e^{3x}cos(2x)+C<br />

    y=\frac{-2}{5}sin(2x)+\frac{3}{5}cos(2x)+Ce^{-3x}<br />
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  4. #4
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    I still don't buy either solution. You can always check your work by plugging back into the DE, you know.

    In the first problem, a 0 appears mysteriously inside an integral. In the second problem, you have a bit of a messed up notation there. Try cleaning things up, and making sure extra digits don't appear where they're not wanted.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    In the first problem, a 0 appears mysteriously inside an integral.

    Would you tell me which line this is.
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  6. #6
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    Hmm. Actually, the first issue I raised wasn't the problem. Since you obviously know all the steps required to solve the problem, I'll just post a complete solution here for both problems. Then you can check your work against mine.

    1. y'+3y=2\cos(x).

    \displaystyle{e^{3x}y'+3e^{3x}y=2e^{3x}\cos(x)}

    \displaystyle{(e^{3x}y)'=2e^{3x}\cos(x)}

    \displaystyle{e^{3x}y-C=2\int e^{3x}\cos(x)\,dx}

    Here I have used the -C for the constant of integration on the LHS so I don't have to have one on the RHS.

    We will investigate the remaining integral. Let u=e^{3x}. Then du=3\,e^{3x}\,dx. Let dv=\cos(x)\,dx. Then v=\sin(x). It follows, then, that

    \displaystyle{\int e^{3x}\cos(x)\,dx=e^{3x}\sin(x)-3\int e^{3x}\sin(x)\,dx.}

    Again, let u=e^{3x}, from which we get that du=3e^{3x}\,dx. This time, let dv=\sin(x)\,dx, and thus it is that v=-\cos(x). Our equation becomes

    \displaystyle{\int e^{3x}\cos(x)\,dx=e^{3x}\sin(x)-3\left(-e^{3x}\cos(x)+3\int e^{3x}\cos(x)\,dx\right).}

    Therefore,

    \displaystyle{\int e^{3x}\cos(x)\,dx=e^{3x}\sin(x)+3 e^{3x}\cos(x)-9\int e^{3x}\cos(x)\,dx.} Hence,

    \displaystyle{10\int e^{3x}\cos(x)\,dx=e^{3x}\sin(x)+3 e^{3x}\cos(x),} and thus

    \displaystyle{\int e^{3x}\cos(x)\,dx=\frac{1}{10}\,e^{3x}\sin(x)+\fra  c{3}{10}\, e^{3x}\cos(x).}

    Going back to the equation from whence we needed this integral, we have the following:

    \displaystyle{e^{3x}y-C=2\int e^{3x}\cos(x)\,dx=2\left(\frac{1}{10}\,e^{3x}\sin(  x)+\frac{3}{10}\, e^{3x}\cos(x)\right)=\frac{1}{5}\,e^{3x}\sin(x)+\f  rac{3}{5}\, e^{3x}\cos(x)}.

    It follows, then, that

    \displaystyle{e^{3x}y=\frac{1}{5}\,e^{3x}\sin(x)+\  frac{3}{5}\, e^{3x}\cos(x)+C}, and thus

    \displaystyle{y(x)=\frac{1}{5}\,\sin(x)+\frac{3}{5  }\,\cos(x)+Ce^{-3x}.}

    Check:

    \displaystyle{y'+3y=\frac{1}{5}\,\cos(x)-\frac{3}{5}\,\sin(x)-3Ce^{-3x}+\frac{3}{5}\,\sin(x)+\frac{9}{5}\,\cos(x)+3Ce^  {-3x}=2\cos(x),}

    as required. In comparing this with your solution, I think you dropped the original RHS 2 somewhere.

    Suggestion: don't skip lots of steps! You can see in my solution that I don't skip a lot of steps.

    2. Instead of posting the entire solution, I think I know where part of your problem is: in your integration by parts the first time: integrating dv=\cos(2x)\,dx. You should get

    \displaystyle{v=\frac{\sin(2x)}{2}}. You've done this correctly before, but not the second time you posted a solution to this problem.

    So carry that through, and check your answer by differentiating to make sure it satisfies the original DE! That's very important.
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  7. #7
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    oh ic now

    My problem was that in both question i included the constant in my integration by parts. This is what i mean

    e^{3x}y=2\int e^{3x}cos(x)
    meant to be
    <br />
\int e^{3x}cos(x) = e^{3x}sin(x)-3\int e^{3x}sin(x)
    not
    2\int e^{3x}cos(x) =e^{3x}sin(x)-3\int e^{3x}sin(x)

    i included the two which meant the next integration by parts i was going to 9+2 on the LHS.

    This was the same problem with the other question as well.
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  8. #8
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    So, all clear now?
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  9. #9
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    yep
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  10. #10
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    Great! Have a good one.
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