Question

In Problems solve the given initial-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The general solution is$y\left(x\right)=2-e^{-x}-x+\frac{1}{2}{x}^2$

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$y\text{'}+y\text{'}=x ...\left(1\right)$

The homogenous differential equation is:$y\text{'}\text{'}+y\text{'}=0$

The auxiliary equation is:

$\begin{array}{rcl}{m}^2+m& =& 0\\ m\left(m+1\right)& =& 0\\ m& =& 0\text{or}m=-1\end{array}$

Hence, the required complementary function is $y_c=c_1+c_1{e}^{-x}$.

General solution

$y\text{'}+y\text{'}=x$The given differential equation is .

Here $D^2$annihilates x.

So,

$\begin{array}{rcl}{D}^2\left(D^2+D\right)y& =& D^2\left(x\right)\\ D^2\left(D^2+D\right)y& =& D\left(D\left(x\right)\right)\\ D^2\left(D^2+D\right)y& =& D\left(1\right)\\ D^2\left(D^2+D\right)y& =& 0\end{array}$

The auxiliary equation of the fourth order equation in $D^2\left(D^2+D\right)y=0$,

$\begin{array}{rcl}{m}^2\left(m^2+m\right)y& =& 0\\ m^3\left(m+1\right)& =& 0\end{array}$

The roots of auxiliary equation have $m_1=m_2=m_3=0,m_4=-1$.

The general solution must be$y=\underset{y_c}{\underbrace{C_1+C_4{e}^{-x}}}+C_{2}x+C_3{x}^2$ .

Particular solution

The last two terms corresponds to particular solution so it can be written as:

$\begin{array}{rcl}{y}_p& =& Ax+B{x}^2\\ y{\text{'}}_p& =& A+2Bx\\ y\text{'}{\text{'}}_p& =& 2B\end{array}$

Substituting these values in equation (1) $2B+A+2Bx=x$

Find value of constant

Compare the like terms on both sides,

$\begin{array}{rcl}2B& =& 1\\ B& =& \frac{1}{2}\end{array}$

And

$\begin{array}{rcl}2B+A& =& 0\\ 2\left(\frac{1}{2}\right)+A& =& 0\\ A& =& -1\end{array}$

Substitute in $y_p=Ax+B{x}^2,y_p=-x+\frac{1}{2}{x}^2$,.

Thus, the general solution of the differential equation (1) is.$y\left(x\right)=C_1+C_2{e}^{-x}-x+\frac{1}{2}{x}^2$

Apply initial condition

Use the initial conditions,$y\left(0\right)=1,y\text{'}\left(0\right)=0$

$\begin{array}{rcl}y\left(0\right)& =& C_1+C_2{e}^{-0}-0+\frac{1}{2}{\left(0\right)}^2\\ 1& =& C_1+C_2+0\\ 1& =& C_1+C_2\dots \dots \left(\text{3}\right)\end{array}$

Using ,$y\text{'}\left(0\right)=0$

$\begin{array}{rcl}{y}^{\text{'}}\left(x\right)& =& 0-C_2{e}^{-x}-1+\frac{2}{2}x\\ y^{\text{'}}\left(x\right)& =& -C_2{e}^{-x}-1+x\\ y^{\text{'}}\left(0\right)& =& -C_2{e}^{-0}-1+0\text{Use}{y}^{\text{'}}\left(0\right)=0\\ 0& =& -C_2-1\\ C_2& =& -1\\ & & \end{array}$

Using (3) $C_1=2$,substitute values in $y\left(x\right)=C_1+C_2{e}^{-x}-x+\frac{1}{2}{x}^2$,

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

Hence, the required general solution of the given initial value problem is, $y\left(x\right)=2-e^{-x}-x+\frac{1}{2}{x}^2$.