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# Nonhomogeneous differential equation! watch

1. 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
2. (Original post by As_Dust_Dances_)

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
3. (Original post by As_Dust_Dances_)

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
and
THe Wronskian determinant is
So your general solution for non-homogeneous equation is

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.
4. (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!
5. (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
and
THe Wronskian determinant is
So your general solution for non-homogeneous equation is

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

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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