Answer to Question #322338 in Differential Equations for MNS

Question

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

Answer 1

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

The solutions to a nonhomogeneous equation are of the form

$y(x) = y_c(x) + y_p(x)$ ,

where $y_c$ is the general solution to the associated homogeneous equation and $y_p$ 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:

$\lambda+2=0$

$\lambda=-2$

$y_c(x)=Ce^{-2x}$ .

The particular solution of the differential equation:

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

$y_p'(x)=-Ae^{-x}+B$

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

$A=1$;   $B=\frac32$; $C=-\frac{B}{2}=-\frac34$.

$y(x)=y_c(x)+y_p(x)=Ce^{-2x}+e^{-x}+\frac32x-\frac34$

Answer: $y(x)=Ce^{-2x}+e^{-x}+\frac32x-\frac34$.