Question

In Problemssolve the given initial-value problem.

${\mathit{y}}^{\left(4\right)}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathit{x}\mathbf{+}{\mathit{e}}^{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

$y=-2e^x+2+x+x{e}^x-\frac{1}{6}{x}^3-\frac{1}{24}{e}^x$

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$y^{\left(4\right)}-y^{\text{'}\text{'}\text{'}}=x+e^x,y\left(0\right)=0,y^{\text{'}}\left(0\right)=0,y^{\text{'}\text{'}}\left(0\right)=0,y^{\text{'}\text{'}\text{'}}\left(0\right)=0$

The auxiliary equation corresponding to homogenous equation gives:

$m^4-m^3=0$

, we obtain

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

So, the complementary function is $y_c=C_1{e}^x+C_2+C_{3}x+C_4{x}^2$.

General solution

Apply the differential operator$D^2\left(D-1\right)$ to both sides of the given differential equation, which gives

$\begin{array}{rcl}{D}^2\left(D-1\right)\left[D^4-D^3\right]y& =& D^2\left(D-1\right)\left[x+e^x\right]\\ D^5{\left(D-1\right)}^{2}y& =& 0...\left(1\right)\end{array}$

The auxiliary equation is:

$$\begin{gathered} m^5{\left(m-1\right)}^2=0 \\ {m}_1=1=m_2,m_3=0=m_4=m_5=m_6=m_7 \end{gathered}$$

So, the general solution must be .$y=\underset{y_c}{\underbrace{C_1{e}^x+C_2+C_{3}x+C_4{x}^2}}+C_{5}x{e}^x+C_6{x}^3+C_7{x}^4$

Particular solution

The last 4 terms of particular solution are

$y_p=Ax{e}^x+B{x}^3+C{x}^4$

Then:

$\begin{array}{rcl}{y}_p^{\text{'}}& =& Ax{e}^x+A{e}^x+3B{x}^2+4C{x}^3\\ y_p^{\text{'}\text{'}}& =& 2A{e}^x+Ax{e}^x+6Bx+12C{x}^2\\ y_p^{\text{'}\text{'}\text{'}}& =& 3A{e}^x+Ax{e}^x+6B+24Cx\\ y_p^{\left(4\right)}& =& 4A{e}^x+Ax{e}^x+24C\end{array}$

The given differential equation gives .$A{e}^x+\left(24C-6B\right)-24Cx=e^x+x$

Find coefficient

On equating the coefficients of like terms, we get

$\begin{array}{rcl}A& =& 1\\ -24C& =& 1\\ 24C-6B& =& 0\end{array}$

$A=1,B=-\frac{1}{6}$and$C=-\frac{1}{24}$

So,$y_p=x{e}^x-\frac{1}{6}{x}^3-\frac{1}{24}{x}^4$

Thus, the general solution is $y=C_1{e}^x+C_2+C_{3}x+C_4{x}^2+x{e}^x-\frac{1}{6}{x}^3-\frac{1}{24}{x}^4$.

Apply initial condition

After differentiating we have:

$\begin{array}{rcl}{y}^{\text{'}}& =& C_1{e}^x+C_3+2C_{4}x+e^x+x{e}^x-\frac{1}{2}{x}^2-\frac{1}{6}{x}^3\\ y^{\text{'}\text{'}}& =& C_1{e}^x+2C_4+2e^x+x{e}^x-x-\frac{x^2}{2}\\ y^{\text{'}\text{'}\text{'}}& =& C_1{e}^x+3e^x+x{e}^x-1-x\end{array}$

Put, $y\left(0\right)=0$we get$0=C_1+C_2\Rightarrow C_2=-C_1$

Put,role="math" localid="1664171621213" $y\text{'}\left(0\right)=0$we get$0=C_1+C_3+1$

Put,$y^{\text{'}\text{'}}\left(0\right)=0$ we get$0=C_1+2C_4+2$

Put, $y^{\text{'}\text{'}\text{'}}\left(0\right)=0$we get $0=C_1+3-1\Rightarrow C_1=-2$

So, we get $C_1=-2,C_2=2,C_3=1$and$C_4=0$

Thus, the solution of the given initial-value problem is$y=-2e^x+2+x+x{e}^x-\frac{1}{6}{x}^3-\frac{1}{24}{e}^x$