y′+2y=e−x+3x
The solutions to a nonhomogeneous equation are of the form
y(x)=yc(x)+yp(x) ,
where yc is the general solution to the associated homogeneous equation and yp is a particular solution.
The associated homogeneous equation:
y′+2y=0
The general solution of this equation is determined by the roots of the characteristic equation:
λ+2=0
λ=−2
yc(x)=Ce−2x .
The particular solution of the differential equation:
yp(x)=Ae−x+(Bx+C)
yp′(x)=−Ae−x+B
Now put these into the original differential equation to get:
−Ae−x+B+2(Ae−x+(Bx+C))=e−x+3x
Equating coefficients, we get
A=1; B=23; C=−2B=−43.
y(x)=yc(x)+yp(x)=Ce−2x+e−x+23x−43
Answer: y(x)=Ce−2x+e−x+23x−43.
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