Q23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

Step-by-Step Solution

Answer

The solution to the initial value problem is:$\mathrm{y}={\mathrm{e}}^{\mathrm{x}}-1$

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

 

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

 

The auxiliary equation for the above equation,

 

$\mathrm{m}-1=0$

Step 2: Now find the complementary solution of the given equation.

The root of an auxiliary equation is,

 

$\mathrm{m}=1$

 

The complementary solution of the given equation is,

 

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

Step 3: Now find the particular solution to find a general solution for the equation.

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),

 

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

 

Step 4: Find the general solution and use the given initial condition.

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 $\mathrm{y}={\mathrm{e}}^{\mathrm{x}}-1.$