Find the solution to the initial value problem.$\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

The differential equation is,

$\mathrm{y}\text{'}-\mathrm{y}=1\dots \left(1\right)$

The auxiliary equation for the above equation,

$\mathrm{m}-1=0$

The root of an auxiliary equation is,

$\mathrm{m}=1$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}$

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{k}\dots \left(2\right)$

Now find the first derivative of the above equation,

${\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=0$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)$the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}-\mathrm{y}=1\\ 0-\mathrm{k}=1\\ \mathrm{k}=-1\end{array}$

Substitute the value of k in the equation (2),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=-1$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-1......\left(3\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=0$

Substitute the value of y = 0 and x = 0 in the equation (3),

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-1\\ 0={\mathrm{c}}_1{\mathrm{e}}^{\left(0\right)}-1\\ {\mathrm{c}}_1=1\end{array}$

Substitute the value of ${\mathrm{c}}_1=1$in the equation (3), and we get:

$\begin{array}{c}\mathrm{y}=\left(1\right){\mathrm{e}}^{\mathrm{x}}-1\\ \mathrm{y}={\mathrm{e}}^{\mathrm{x}}-1\end{array}$

Thus, the general solution is$\mathrm{y}={\mathrm{e}}^{\mathrm{x}}-1.$