Question

In Problemssolve the given initial-value problem.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathbf{5}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

$y=C_1+C_2{e}^x+C_{3}x{e}^x+5x-\frac{1}{2}{x}^2{e}^x+\frac{1}{6}{x}^3{e}^x$

Step-by-step solution

Definition of initial value problem

An ordinary differential equation in which an initial condition is provided that identifies the unknown function within the domain is known as the initial value problem.

Complementary function

Consider the differential equation

$y^{\text{'}\text{'}\text{'}}-2y^{\text{'}\text{'}}+y^{\text{'}}=x{e}^x+5,y\left(0\right)=2,y^{\text{'}}\left(0\right)=2,y^{\text{'}\text{'}}\left(0\right)=-1$

The auxiliary equation corresponding to homogenous equation gives:

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

we find $m_1=0$and$m_2=1=m_3$

So, the complementary function is .$y_c=C_1+C_2{e}^x+C_{3}x{e}^x$

General solution

Applying $D{\left(D-1\right)}^2$to the both sides of the given differential equation, we get

$D{\left(D-1\right)}^2\left(D^3-2D^2+D\right)y=D{\left(D-1\right)}^2\left[x{e}^x+5\right]$

i.e,$D^2{\left(D-1\right)}^{4}y=0...\left(1\right)$

The auxiliary equation is:

$\begin{array}{rcl}{m}^2{\left(m-1\right)}^4& =& 0\\ m_1& =& 0=m_2;m_3=1=m_4=m_5=m_6\end{array}$

Thus, the general solution must be $y=\underset{}{\underbrace{C_1+C_2{e}^x+C_3{e}^x}}+C_{4}x+C_5{x}^2{e}^x+C_6{x}^3{e}^x$.

Particular solution

The last three terms denotes particular solution.

$y_p=Ax+B{x}^2{e}^x+C{x}^3{e}^x$

Then

$\begin{array}{rcl}{y}_p^{\text{'}}& =& A+2Bx{e}^x+B{x}^2{e}^x+3C{x}^2{e}^x+C{x}^3{e}^x\\ y_p^{\text{'}\text{'}}& =& 2B{e}^x+6Cx{e}^x+4Bx{e}^x+6C{x}^2{e}^x+B{x}^2{e}^x+C{x}^3{e}^x\\ y_p^{\text{'}\text{'}\text{'}}& =& 6B{e}^x+18Cx{e}^x+6C{e}^x+6Bx{e}^x+9C{x}^2{e}^x+B{x}^2{e}^x+C{x}^3{e}^x\end{array}$

Substituting value in differential equation gives

$\begin{array}{rcl}A+2B{e}^x+6Cx{e}^x+6C{e}^x& =& x{e}^x+5\\ A+\left(2B+6C\right)e^x+6Cx{e}^x& =& x{e}^x+5\end{array}$

On equating the coefficients of like terms, we get

$\begin{array}{rcl}A& =& 5\\ A& =& 5\\ 6C& =& 1\\ C& =& \frac{1}{6}\end{array}$

$B=-\frac{1}{2}$

So,$y_p=5x-\frac{1}{2}{x}^2{e}^x+\frac{1}{6}{x}^3{e}^x$

Thus, general solution is$y=C_1+C_2{e}^x+C_{3}x{e}^x+5x-\frac{1}{2}{x}^2{e}^x+\frac{1}{6}{x}^3{e}^x$ .