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    Could anyone tell me where the x^2/2 part comes from in the answer? Since I end up with 4Ae^x = e^x + e^-x when solving the particular solution which only gives the e^x/4 part so I'm not sure where the (x^2 e^-x)/2 is from..?

    My general solution is Ae^-x + Bxe^-x
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    (Original post by As_Dust_Dances_)
    Name:  differential.png
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    Could anyone tell me where the x^2/2 part comes from in the answer? Since I end up with 4Ae^x = e^x + e^-x when solving the particular solution which only gives the e^x/4 part so I'm not sure where the (x^2 e^-x)/2 is from..?

    My general solution is Ae^-x + Bxe^-x
    The RHS of the ODE is part of your solution, so you have to multiply your "guess" for the particular solution by powers of x until you get something that works
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    (Original post by As_Dust_Dances_)
    Name:  differential.png
Views: 368
Size:  66.8 KB

    Could anyone tell me where the x^2/2 part comes from in the answer? Since I end up with 4Ae^x = e^x + e^-x when solving the particular solution which only gives the e^x/4 part so I'm not sure where the (x^2 e^-x)/2 is from..?

    My general solution is Ae^-x + Bxe^-x
    1. The characteristic polinomial is complete square, so the multiciplity of
    the root is 2.
    The two independent solutions for homogeneous part is
    u_1=e^{-x} and u_2=x\cdot e^{-x}
    THe Wronskian determinant is W=e^{-2x}
    So your general solution for non-homogeneous equation is
    C_1(x)\cdot e^{-x}+C_2(x)\cdot x\cdot e^{-x}
    2.
    Calculating C_1(x) and C_2)x) from the Wronskian with Cramer rule
    and with integration you will get that functions with constant A and B respectively, and substituting back that in the general solution and arranging you will get x^2/2*e^(-x) as one of the terms.
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    (Original post by Indeterminate)
    The RHS of the ODE is part of your solution, so you have to multiply your "guess" for the particular solution by powers of x until you get something that works
    Oh yeah, I forgot about those laws. Yeah just realised you would get Ae^x + Bx^2e^-x as opposed to Ae^x + Be^-x :L

    Thanks anyway!
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    (Original post by ztibor)
    1. The characteristic polinomial is complete square, so the multiciplity of
    the root is 2.
    The two independent solutions for homogeneous part is
    u_1=e^{-x} and u_2=x\cdot e^{-x}
    THe Wronskian determinant is W=e^{-2x}
    So your general solution for non-homogeneous equation is
    C_1(x)\cdot e^{-x}+C_2(x)\cdot x\cdot e^{-x}
    2.
    Calculating C_1(x) and C_2)x) from the Wronskian with Cramer rule
    and with integration you will get that functions with constant A and B respectively, and substituting back that in the general solution and arranging you will get x^2/2*e^(-x) as one of the terms.
    I haven't studied the Wronskian method yet and we haven't done Cramer's Rule in Calculus, only Algebra. I just used the fact that the solutions for the characteristic equation were -1 for both and compared them with the e^-x, then realised you would need an x^2 to make the guess solution Ae^x + Bx^2 e^-x

    Thanks Anyway!
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    you can, if you want, split the equation into 3 different ones, two in-homogeneous and one homogeneous, with generating functions of:

    0, e^{x}, e^{-x}

    respectively, and solve each one - bearing in mind that "each" of the particular solutions cannot be a multiple of the homogeneous
 
 
 
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