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About LinkBacksI want to use A(D)= (D+2)^2 to annihilate F(X)=5xe^(-2x).
However, I also need to compute A(D) operating on y''+4y'+4y in order to find Yp.
Thus, my question is this.
How do I compute A(D) operating on y''+4y'+4y?
I know it's simple but I'm stuck.
I want to use A(D)= (D+2)^2 to annihilate F(X)=5xe^(-2x).
However, I also need to compute A(D) operating on y''+4y'+4y in order to find Yp.
Thus, my question is this.
How do I compute A(D) operating on y''+4y'+4y?
I know it's simple but I'm stuck.
The equation that we want to solve isOriginally Posted by divinelogos
If we write this in terms of the operator D we get
From here we see that the complimentary solution is
Now find the particular solution we need to annihilate the right hand side.
Using the Annihilator that you have above gives
This "new" equation have solution
Now we need to remove any terms that are in the complimentary solution. If we relabel the constants we get
From here it is just finding A and B.
If that is what you meant to say, you don't! y''+ 4y'+ 4y is already (D+ 2)^2y.
(D+ 2)^2y means (D+ 2)(D+ 2)y= (D+ 2)(Dy+ 2y)= (D+ 2)(y'+ 2y)= D(y')+ D(2y)+ 2(y'+ 2y)= y''+ 2y'+ 2y'+ 4y= y''+ 4y'+ 4y.
In order to find yp you need to find a function y(x) such that (D+ 2)^2y= 5xe^{-2x}I know it's simple but I'm stuck.
I recommend trying something of the form y= (Ax^3+ Bx^2)e^{-2x}. Then Dy= [(Ax^3+ Bx^2)e^{-2x}]'= (3Ax^2+ 2Bx)e^{-2x}- 2(Ax^3+ Bx^2)e^{-2x}= (-2Ax^3+ (3A- 2B)x^2+ 2Bx)e^{-2x} and D^2y= [(-2Ax^3+ (3A- 2B)x^2+ 2Bx)e^{-2x}]'= (-6Ax^2+ (6A- 4B)x+ 2B)e^{-2x}- 2(-2Ax^3+ (3A- 2B)x^2+ 2Bx)e^{-2x}= (4Ax^3+ (-12A+ 4B)x^2+ (-6A- 4B)x+ 2B)e^{-2x}. Put those into (D^2+ 4D+ 4)y= 5xe^{-2x} and try to find A and B.
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