Question

In Problems 1–10 write the given differential equation in the form where L(y) = g(x), is a linear differential operator with constant coefficients. If possible, factor L.

$\mathit{y}\mathbf{"}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{"}\mathbf{-}\mathbf{13}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

Short answer

$\left(D-1\right)\left(D+5\right)\left(D-2\right)y=x{e}^{-x}.$

Step-by-step solution

Use the Differential linear operators to find the solution

We have the non-homogeneous second order differential equation,

$y"\text{'}+2y"-13y\text{'}+10y=x{e}^{-x}$

And we have to write it in the form using the differential linear operators as

$$\begin{gathered} D^{3}y+2D^{2}y-13Dy+10y=x{e}^{-x} \\ \left(D^3+2D^2-13D+10\right)y=x{e}^{-x} \end{gathered}$$

After that if we put (D = 1), we find that the equation equals to 0 and (D - 1) is a factor of these equation, after that we can make a long division, we can factorize the third equation as,

role="math" $$\begin{gathered} \left(D-1\right)\left(D^2+3D-10\right)y=x{e}^{-x} \\ \left(D-1\right)\left(D+5\right)\left(D-2\right)y=x{e}^{-x} \end{gathered}$$

Where g(x) = xe^-x

Hence the solution is $\left(D-1\right)\left(D+5\right)\left(D-2\right)\mathit{y}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$