(d2+1)y=e-x+cosx+x3+exsinx

Question

plz send me answer

Answer 1

This is the same as:

y"+y=e^-x+cos(x)+x³+e^xsin(x).

First consider the general (characteristic) solution of y"+y=0, which is y_c=Acos(x)+Bsin(x).

To find the particular solution we need to work out how each term on the RHS can be produced.

For example, if y₁=ae^xcos(x)+be^xsin(x),

y₁'=-ae^xsin(x)+ae^xcos(x)+be^xcos(x)+be^xsin(x)=e^x(cos(x)(a+b)+sin(x)(b-a)),

y₁"=e^x(-sin(x)(a+b)+cos(x)(b-a))+e^x(cos(x)(a+b)+sin(x)(b-a))=

e^x(cos(x)(b-a+a+b)+sin(x)(-a-b+b-a))=e^x(2bcos(x)-2asin(x))

But y₁"+y₁=e^x(2bcos(x)-2asin(x))+ae^xcos(x)+be^xsin(x)=e^x(cos(x)(2b+a)+sin(x)(-2a+b)).

To arrive at e^xsin(x) in the DE, -2a+b=1 and 2b+a=0, which can be written 2a+4b=0. Add the two equations: 5b=1, b=⅕, a=-2b=-⅖.

This makes y₁=-⅖e^xcos(x)+⅕e^xsin(x).

We now have to consider three other terms e^-x+cos(x)+x³. This is a heterogeneous set of terms.

Let y₂=ce^-x, y₂'=-ce^-x, y₂"=ce^-x; y₂"+y₂=2ce^-x, so 2c=1, c=½, y₂=½e^-x.

Let y₃=a₁x+a₂x²+a₃x³, y₃'=a₁+2a₂x+3a₃x², y₃"=2a₂+6a₃x; y₃"+y₃=2a₂+6a₃x+a₁x+a₂x²+a₃x³⇒a₃=1; a₂=0; 6a₃+a₁=0⇒a₁=-6.

Therefore y₃=-6x+x³.

This leaves cos(x). Let y₄=dxsin(x), y₄'=dxcos(x)+dsin(x), y₄"=-dxsin(x)+dcos(x)+dcos(x)=-dxsin(x)+2dcos(x);

y₄"+y₄'=-dxsin(x)+2dcos(x)+dxsin(x)=2dcos(x). When d=½, y₄"+y₄'=cos(x), and y₄=½xsin(x)

Therefore we have the (verified) solution, where A and B are constants:

y=y_c+y₁+y₂+y₃+y₄=Acos(x)+Bsin(x)-⅖e^xcos(x)+⅕e^xsin(x)+½e^-x-6x+x³+½xsin(x).