Q14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Solve the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

Answer

Thus, the solution to the given initial value problem is;$\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$

Step 1: Firstly, write the auxiliary equation of the given differential equation.

 

The given differential equation is,

 

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

 The auxiliary equation for the above equation,

 

$$\begin{gathered} {\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{m}\mathbf{=}\mathbf{0} \\ \mathbf{m}\left(\mathbf{m}\mathbf{+}\mathbf{1}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

The roots of an auxiliary equation are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{,}\& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}.$

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

 

If the auxiliary equation has distinct real roots, then the general solution is given as;

 $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{{\mathbf{m}}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{{\mathbf{m}}_{\mathbf{2}}\mathbf{t}}$

Therefore, the general solution of the given equation is;

 $$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\left(\mathbf{0}\right)\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{1}\mathbf{t}} \\ \mathbf{y}\mathbf{=}\mathbf{A}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{t}} ...\left(\mathbf{2}\right) \end{gathered}$$

Step 3: Use the given initial condition.

Given the initial condition,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}$ 

Substitute the value of y = 2 and t = 0 in the equation (2),

 $$\begin{gathered} \mathbf{y}\mathbf{=}\mathbf{A}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{t}} \\ \mathbf{2}\mathbf{=}\mathbf{A}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\left(\mathbf{0}\right)} \end{gathered}$$

Now find the derivative of the equation (2),

 

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{Be}}^{\mathbf{-}\mathbf{t}}$

 

Substitute the value of y’ = 1 and t = 0 in the above equation,

 

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}{\mathbf{Be}}^{\mathbf{-}\mathbf{t}} \\ \mathbf{1}\mathbf{=}\mathbf{-}{\mathbf{Be}}^{\mathbf{-}\left(\mathbf{0}\right)} \\ \mathbf{B}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

 

Substitute the value of B in the equation (3),

 $$\begin{gathered} \mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}\mathbf{2} \\ \mathbf{A}\mathbf{+}\left(\mathbf{-}\mathbf{1}\right)\mathbf{=}\mathbf{2} \\ \mathbf{A}\mathbf{=}\mathbf{3} \end{gathered}$$

Step 4: Final answer.

Substitute the value of A and B in the equation (2),

 $$\begin{gathered} \mathbf{y}\mathbf{=}\mathbf{A}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{t}} \\ \mathbf{y}\mathbf{=}\mathbf{3}\mathbf{+}\left(\mathbf{-}\mathbf{1}\right){\mathbf{e}}^{\mathbf{-}\mathbf{t}} \\ \mathbf{y}\mathbf{=}\mathbf{3}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}} \end{gathered}$$

Thus, the solution to the given initial value problem is;

$\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$