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Math Help - 1st order non-homogeneous - solve by substitution?

  1. #1
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    1st order non-homogeneous - solve by substitution?

    Hi everyone,

    I'm having trouble solving this problem. I think substitution is the way to go but I haven't had a lot of luck.

    y^2/2+2ye^x+(y+e^x)dy/dx = e^-x
    using u=y+e^x
    dy/dx=-y^2/2u-2ye^x/u+e^-x

    If this is correct so far, can someone help me with the next few steps? I'm pretty lost on this one and I've been working on it for a few days

    Thanks!
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  2. #2
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    You donít want a mixed equation with u and y. The substitution means that replace all instances of y with something involving u.

    So if you make the substitution u=y+e^x, then you will note that y=u-e^x, so y^2=u^2-2ue^x+e^{2x}, and y'=u'-e^x.

    Then, the equation \frac{1}{2}y^2+2ye^x+(y+e^x)y'=e^{-x} becomes \frac{1}{2}u^2-ue^x+\frac{1}{2}e^{2x}+2ue^x-2e^{2x}+uu'-ue^x=e^{-x}, which simplifies to u^2+2uu'=3e^{2x}+2e^{-x}. If we now make another substitution, v=u^2, then v'=2uu', so we have v+v'=3e^{2x}+2e^{-x}, which is linear.
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  3. #3
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    Thank you so much for the help!!

    I proceeded to solve as I would for a normal 1st order linear non-homogeneous equation. I got the following result:

    y=(e^3x+2x+C)e^-x

    Does this seem about right?
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  4. #4
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    That seems like the solution for v. In order to get y, you need to take y=u-e^x along with u^2=v to get y=\pm\sqrt{v}-e^x.

    You can always check your final answer by substituting back into the original equation.
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