Question #322338

-3x + (D+2)y=e-t

1
Expert's answer
2022-04-05T13:29:54-0400

y+2y=ex+3xy'+2y=e^{-x}+3x

The solutions to a nonhomogeneous equation are of the form

y(x)=yc(x)+yp(x)y(x) = y_c(x) + y_p(x) ,

where ycy_c is the general solution to the associated homogeneous equation and ypy_p is a particular solution.

The associated homogeneous equation:

y+2y=0y'+2y=0

The general solution of this equation is determined by the roots of the characteristic equation:

λ+2=0\lambda+2=0

λ=2\lambda=-2

yc(x)=Ce2xy_c(x)=Ce^{-2x} .

The particular solution of the differential equation:

yp(x)=Aex+(Bx+C)y_p(x)=Ae^{-x}+(Bx+C)

yp(x)=Aex+By_p'(x)=-Ae^{-x}+B

Now put these into the original differential equation to get:

Aex+B+2(Aex+(Bx+C))=ex+3x-Ae^{-x}+B+2(Ae^{-x}+(Bx+C))=e^{-x}+3x

Equating coefficients, we get

A=1A=1;   B=32B=\frac32C=B2=34C=-\frac{B}{2}=-\frac34.

y(x)=yc(x)+yp(x)=Ce2x+ex+32x34y(x)=y_c(x)+y_p(x)=Ce^{-2x}+e^{-x}+\frac32x-\frac34

Answer: y(x)=Ce2x+ex+32x34y(x)=Ce^{-2x}+e^{-x}+\frac32x-\frac34.


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