Question

(D^3+1) y= 3+e^(-x) +5e^2x?

Answer 1

$y'''+y=3+e^{-x}+5e^{2x}$

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:

$y'''+y=0$

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

$\lambda^3+1=0$

$(\lambda+1)(\lambda^2-\lambda+1)=0$

$\lambda_1=-1$

$(\lambda^2-\lambda+1)=0$

$D=b^2-4ac=1-4=-3$

$\lambda_{2,3}=\frac{1\pm i\sqrt 3}{2}$ , where $i=\sqrt{-1}$ .

$y_c(x)=c_1e^{-x}+e^{\frac x2}(c_2\cos{\frac{\sqrt3}{2}x}+c_3\sin{\frac{\sqrt3}{2}x})$ .

The particular solution of the differential equation:

$y_p(x)=A+xBe^{-x}+Ce^{2x}$

$y_p'=Be^{-x}-xBe^{-x}+2Ce^{2x}$

$y_p''=-2Be^{-x}+xBe^{-x}+4Ce^{2x}$

$y_p'''=3Be^{-x}-xBe^{-x}+8Ce^{2x}$

Now put these into the original differential equation to get:

$3Be^{-x}-xBe^{-x}+8Ce^{2x}+A+xBe^{-x}+Ce^{2x}=$$3+e^{-x}+5e^{2x}$

Equating coefficients, we get

$A=3$;   $B=\frac13$; $C=\frac59$.

$y_p(x)=3+\frac x3e^{-x}+\frac59e^{2x}$

Answer: $y=c_1e^{-x}+e^{\frac x2}(c_2\cos{\frac{\sqrt3}{2}x}+c_3\sin{\frac{\sqrt3}{2}x})+$$3+\frac x3e^{-x}+\frac59e^{2x}$ .