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Thread: First order DE

  1. #1
    Member Jones's Avatar
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    First order DE

    Hi,

    I have the following DE:

    $\displaystyle y' + ycos~x = sin2x$

    So i have $\displaystyle e^{\int cos~x} = e^{sinx}$

    Multiply both sides by this factor and integrate. Gives me.

    $\displaystyle ye^{sinx} = \int e^{sinx} \cdot 2sin~x\cdot cos~x$

    This is where im stuck, how to i integrate the right hand side?
    Last edited by Jones; Feb 13th 2010 at 12:30 PM.
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi,

    I have the following DE:

    $\displaystyle y' + ycos~x = sin2x$

    So i have $\displaystyle e^{\int cos~x} = e^sinx$

    Multiply both sides by this factor and integrate. Gives me.

    $\displaystyle ye^{sinx} = \int e^{sinx} \cdot 2sin~x\cdot cos~x$

    This is where im stuck, how to i integrate the right hand side?
    Dear Jones,

    $\displaystyle \int{e^{sinx}sin2x~dx}$

    $\displaystyle Substitute,~u=e^{sinx}\Rightarrow{sinx=lnu}$

    $\displaystyle du=e^{sinx}cosx~dx$

    $\displaystyle =\int{u\times{2}\times{sinx}\times{cosx}\frac{du}{ ucosx}}$

    $\displaystyle =\int{2\ln{u}~du}$

    $\displaystyle =2\int{lnu~du}$

    $\displaystyle =2\left[ulnu-u\right]+C~;~Where~C~is~an~arbitary~constant.$

    $\displaystyle =2\left[e^{sinx}sinx-e^{sinx}\right]+C$

    Hope this will help you.
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  3. #3
    Member Jones's Avatar
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    Thank you very much Sudharaka.

    I hate myself for not seeing that obvious substitution
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