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Math Help - Another Differential Equation Question

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    Another Differential Equation Question

    Hi again, here's another question that I had a go at but haven't got any answers for, does the working and answer seem ok?

    Solve y' + y = y^2e^x

    The method I use is outlined in my lecture notes, however I am unsure of it's actual name.

    First thing I did was to isolate the x term on the right hand side by multiplying everything by y^{-2}, giving the equation y'y^{-2} + y^{-1} = e^x

    I then use the substitution of w = y^{-1}, then differentiate this equation with respect to x and you get w' = -y^{-2}y'.

    Placing these two substitutions into my original equation this simplifies to be w' - w = -e^x.

    I wasn't too sure what to do at this point, I took a guess at using an integrating factor of e^{\int{-1}}, which when I put back into the equation gave me e^{-x}w' - we^{-x} = -1.

    This then integrates to we^{-x} = x, which after putting back in my substitution gave the answer of xye^{x} = 1.

    Is any of this correct, or know me have I made an early error and then just answered a completely different question?

    Thanks again

    Craig
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  2. #2
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    Quote Originally Posted by craig View Post
    Hi again, here's another question that I had a go at but haven't got any answers for, does the working and answer seem ok?

    Solve y' + y = y^2e^x

    The method I use is outlined in my lecture notes, however I am unsure of it's actual name.

    First thing I did was to isolate the x term on the right hand side by multiplying everything by y^{-2}, giving the equation y'y^{-2} + y^{-1} = e^x

    I then use the substitution of w = y^{-1}, then differentiate this equation with respect to x and you get w' = -y^{-2}y'.

    Placing these two substitutions into my original equation this simplifies to be w' - w = -e^x.

    I wasn't too sure what to do at this point, I took a guess at using an integrating factor of e^{\int{-1}}, which when I put back into the equation gave me e^{-x}w' - we^{-x} = -1.

    This then integrates to we^{-x} = x, which after putting back in my substitution gave the answer of xye^{x} = 1.

    Is any of this correct, or know me have I made an early error and then just answered a completely different question?

    Thanks again

    Craig
    I think you'll find that w = -x e^{x} + C e^x = e^x (C - x).

    Then y = \frac{e^{-x}}{C - x}.
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