Question

Solve the given differential equation by undetermined coefficients.

${\mathbf{y}}^{\text{'}\text{'}\text{'}}\mathbf{-}{\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}{\mathbf{y}}^{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{xe}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}\mathbf{7}$

Short answer

$y=C_1\cos x+C_2\sin x+C_3{e}^x-\frac{1}{2}x{e}^x+\frac{1}{4}{x}^2{e}^x+\frac{1}{4}{e}^{-x}-7$

Step-by-step solution

Definition of complementary function

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

Given data

From $m^3-m^2+m-1=0$we have$(m-1)\left(m^2+1\right)=0\Rightarrow m=1,\pm i$

So, the complementary function is .$y_c=C_1\cos x+C_2\sin x+C_3{e}^x$

We know$D\left(7\right)=0$

And from (5) section 4.5

With $\alpha =1,n=2$we have$(D-1{)}^2\left[x{e}^x\right]=0$

And with$\alpha =-1,n=1$ we have$(D+1)\left[e^{-x}\right]=0$

So,$D(D-1{)}^2(D+1)\left[x{e}^x-e^{-x}+7\right]=0$

Apply differential operator

Applying$D(D-1{)}^2(D+1)$ to the both sides of the given differential equation gives

$D(D-1{)}^2(D+1)\left(D^3-D^2+D-1\right)y=D(D-1{)}^2(D+1)\left[x{e}^x-e^{-x}+7\right]$

i.e,$D(D-1{)}^3(D+1)\left(D^2+1\right)y=0\cdots (*)$

Now, the auxiliary equation corresponding to (*) is

$m(m-1{)}^3(m+1)\left(m^2+1\right)=0$

$\Rightarrow m_1=0,m_2=1=m_3=m_4,m_5=-1,m_6=i\text{and}{m}_7=-i$

Thus, the general solution must be .

$y=\underset{y_c}{\underbrace{C_1\cos x+C_2\sin x+C_3{e}^x}}+C_{4}x{e}^x+C_5{x}^2{e}^x+C_6{e}^{-x}+C_7$

Find particular solution

Now, $y_p$particular solution of the given differential equation should also satisfy (*) and the first three-terms in y constitute$y_c$ .

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

$y_p=Ax{e}^x+B{x}^2{e}^x+C{e}^{-x}+D$

Then$$\begin{gathered} y_p^{\text{'}}=Ax{e}^x+B{x}^2{e}^x-C{e}^{-x}+A{e}^x+2Bx{e}^x \\ {y}_p^{*}=Ax{e}^x+B{x}^2{e}^x+C{e}^{-x}+2A{e}^x+4Bx{e}^x+2B{e}^x \\ {y}_p^{\text{'}}=Ax{e}^x+B{x}^2{e}^x-C{e}^{-x}+3A{e}^x+6Bx{e}^x+6B{e}^x \end{gathered}$$

Substitution

Substituting$y_p$ into the given differential equation gives

$4Bx{e}^x-4C{e}^{-x}+(2A+4B)e^x-D=x{e}^x-e^{-x}+7$

On equating the coefficient of like terms, we get

$$\begin{gathered} 4B=1 \\ -4C=-1 \\ -D=7 \\ 2A+4B=0 \\ \therefore A=-\frac{1}{2},B=\frac{1}{4},C=\frac{1}{4},\text{and}D=-7 \end{gathered}$$

So,$y_p=-\frac{1}{2}x{e}^x+\frac{1}{4}{x}^2{e}^x+\frac{1}{4}{e}^{-x}-7$

Thus, the required solution is .$y=C_1\cos x+C_2\sin x+C_3{e}^x-\frac{1}{2}x{e}^x+\frac{1}{4}{x}^2{e}^x+\frac{1}{4}{e}^{-x}-7$