Question

Solve the given differential equation by undetermined coefficients.

_{${\mathbf{y}}^{\text{'}\text{'}\text{'}}\mathbf{-}\mathbf{3}{\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{3}{\mathbf{y}}^{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{16}$}

Short answer

$y=C_1{e}^x+C_{2}x{e}^x+C_3{x}^2{e}^x+\frac{1}{6}{x}^3{e}^x+x-13$

Step-by-step solution

Definition of complementary function

A complementary function is the general solution of a homogenous, linear differential equation.

Given data

Consider the following differential equation:

$y^{\text{'}\text{'}\text{'}}-3y^{\text{'}\text{'}}+3y^{\text{'}}-y=e^x-x+16$

The objective is to find the general solution to the differential equation.

First find the solution to the homogeneous differential equation$y^{\text{'}\text{'}\text{'}}-3y^{\text{'}\text{'}}+3y^{\text{'}}-y=0$ .

The auxiliary equation is

$m^3-3m^2+3m-1=0$

Then,$(m-1{)}^3=0$

$m_1=m_2=m_3=1$

So, the complementary function is$y_c=\left(C_1+C_{2}x+C_3{x}^2\right)e^x$ .

Apply differential operator

Since$(D-1)e^x=0$and$D^2\left(x\right)=0$ , apply$D^2(D-1)$to the both sides of the given differential equation.

$D^2(D-1)\left(D^3-3D^2+3D-1\right)y=D^2(D-1)\left[e^x-x+16\right]$

$D^2(D-1{)}^{4}y=0$

The auxiliary equation corresponding to (2) is

$m^2(m-1{)}^4=0$

Then,

$m_1=m_2=m_3=m_4=1,m_5=m_6=0$

So, the general solution must be

$$\begin{gathered} =\left(C_1+C_{2}x+C_3{x}^2+C_4{x}^3\right)e^x+\left(C_5+C_{6}x\right)e^{0x} \\ =C_1{e}^x+C_{2}x{e}^x+C_3{x}^2{e}^x+C_4{x}^3{e}^x+C_5+C_{6}x \\ =\underset{y_c}{\underbrace{C_1{e}^x+C_{2}x{e}^x+C_3{x}^2{e}^x}}+C_4{x}^3{e}^x+C_5+C_{6}x \end{gathered}$$

Solve for yp

The first three-terms in y constitute the complementary function$y_c$ .

Since$y_p$particular solution of the given differential equation should also satisfy (2)

So, the remaining terms in y must be the basic form of$y_p$:

Let$y_p=A{x}^3{e}^x+B+Cx$

Then$y_p^{\text{'}}=3A{x}^2{e}^x+A{x}^3{e}^x+C$

$\begin{array}{rcl}{y}_p^{*}& =& 6Ax{e}^x+3A{x}^2{e}^x+3A{x}^2{e}^x+A{x}^3{e}^x\\ & =& 6A{x}^2{e}^x+6Ax{e}^x+A{x}^3{e}^x{y}_p^{\text{'}}\\ & =& 12Ax{e}^x+6A{x}^2{e}^x+6A{e}^x+6Ax{e}^x+3A{x}^2{e}^x+A{x}^3{e}^x\\ & =& 18Ax{e}^x+9A{x}^2{e}^x+A{x}^3{e}^x+6A{e}^x\end{array}$

Substitution

Substitute the values of$y_p^{\text{'}},y_p^{\text{'}\text{'}},y_p^{\text{m}}$and$y_p$in differential equation (1).

$$\begin{gathered} 18Ax{e}^x+9A{x}^2{e}^x+A{x}^3{e}^x+6A{e}^x-3\left(6A{x}^2{e}^x+6Ax{e}^x+A{x}^3{e}^x\right) \\ +3\left(3A{x}^2{e}^x+A{x}^3{e}^x+C\right)-\left(A{x}^3{e}^x+B+Cx\right)=e^x-x+16 \\ 6A{e}^x+3C-B-Cx=e^x-x+16 \end{gathered}$$

i.e,$6A{e}^x+(-C)x+(3C-B)=e^x-x+16$

Equate the coefficient

On equating the coefficients of like terms,

$\begin{array}{rcl}6A& =& 1\\ A& =& \frac{1}{6}\\ -C& =& -1\\ C& =& 1\\ 3C-B& =& 16\\ B& =& 3C-16\\ & =& 3-16\\ & =& -13\\ & & \end{array}$

So, the particular solution is .$y_p=\frac{1}{6}{x}^3{e}^x-13+x$

Thus, the required solution is .$y=C_1{e}^x+C_{2}x{e}^x+C_3{x}^2{e}^x+\frac{1}{6}{x}^3{e}^x+x-13$