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Solving y'=y-sin(2x) |
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| Jan22-07, 11:15 PM | #1 |
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Solving y'=y-sin(2x)
I found the integrating factor to be e^x for DE y'=y-sin(2x)
Now i am stuck at integral of -sin(2x)e^x Can you help me with it? I tried using integration by parts and i get to following integral(sin(2x)(e^x)=sin(2x)e^x-2cos(2x)e^x-4*integral(sin(2x)(e^x) I am stuck. Help! Thanks |
| Jan22-07, 11:33 PM | #2 |
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Notice that you have the expression "integral(sin(2x)(e^x))" on both sides of your equation. You can therefore solve for it directly, and be done.
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| Jan23-07, 12:28 AM | #3 |
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directly? I think i am not seeing something
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| Jan23-07, 12:33 AM | #4 |
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Solving y'=y-sin(2x)
The answer to this DE is 1/5(sin(2x)+2cos(2x) and i am not seeing how to get to this from what i have
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| Jan23-07, 12:35 AM | #5 |
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OH ok i got it....wasnt seeing that lol...THanks
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| Jan23-07, 04:21 AM | #6 |
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also careful with your results. looks like you missed a couple small details. your integrating factor is actually $e^{-x}$ and you should have the boundaries of the LHS of your equation(unless they're supposed to be =0)
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