Q6RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given initial value problem. 

 $$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{z}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \\ \mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{;}\mathbf{z}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Step-by-Step Solution

Answer

The solution for the given initial value problem is:

 $$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \end{gathered}$$

 

Step 1: Using the elimination method

Let's rewrite the given system in operator form:

 

 $$\begin{gathered} \mathbf{D}\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{z}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{-}\mathbf{D}\left[\mathbf{y}\right]\mathbf{+}\mathbf{z}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{D}\left[\mathbf{z}\right]\mathbf{=}\mathbf{0} \end{gathered}$$

 

One will solve this system using the method of elimination. Let eliminate $\mathbf{z}$. We can obtain the first equation without $\mathbf{z}$  by adding the first and the second equation together, so one will get:

 

 $\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{D}\left[\mathbf{y}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$

 

To obtain the second equation one will add the second equation of the initial system "multiplied" by   to the third equation:

 

 $\mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}\mathbf{x}\mathbf{-}{\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$

 

So now we are solving a new system:

 $$\begin{gathered} \mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{D}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0} \\ \mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}\mathbf{x}\mathbf{-}{\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0} \end{gathered}$$

 

Step 2: Finding the roots

By subtracting the second equation from the first, one will get  ${\mathbf{D}}^{\mathbf{2}}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{-}\mathbf{D}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

The corresponding auxiliary equation is  ${\mathbf{r}}^{\mathbf{2}}\mathbf{-}\mathbf{r}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}$ and its roots are:

 

 ${\mathbf{r}}_{\mathbf{1}\mathbf{,}\mathbf{2}}\mathbf{=}\frac{\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{1}\mathbf{+}\mathbf{8}}}{\mathbf{2}}\Rightarrow {\mathbf{r}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,} {\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{2}$

 

So, the general solution for  $y$ is  $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

 

One will now find the general solution for $x$. The equation  $\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\mathbf{x}\mathbf{-}\mathbf{D}\left[\mathbf{y}\right]\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$ gives us that $\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{D}\left[\mathbf{y}\right]\mathbf{+}\mathbf{y}$.

 

So, one will find the first derivative of y and substitute it into the previous equation to obtain $\mathbf{D}\left[\mathbf{x}\right]\mathbf{-}\mathbf{x}$.

 

 $$\begin{gathered} \mathbf{D}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathbf{D}\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\right] \\ \mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{D}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{+}\mathbf{y} \\ \mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \end{gathered}$$

Step 3: Using the method of undetermined coefficients

One can find a homogeneous solution. The auxiliary equation is  $\mathbf{r}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$ and its solution is $\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{1}$  so the homogeneous solution for $x$  is  ${\mathbf{x}}_{\mathbf{h}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}$

 

Now, one can find a particular solution using the method of undetermined coefficients. 

Assume that ${\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{A}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$. One has that ${\mathbf{x}}_{\mathbf{p}}^{}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{A}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$, so

 $$\begin{gathered} \mathbf{D}\left[{\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]\mathbf{+}{\mathbf{x}}_{\mathbf{p}}{\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{x}}_{\mathbf{p}}^{\mathbf{\backslash }\mathbf{cent}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}{\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{=}\mathbf{2}{\mathbf{Ae}}^{\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{Ae}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{=}\mathbf{3}{\mathbf{Ae}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{=}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{A}\mathbf{=}{\mathbf{c}}_{\mathbf{2}} \\ {\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \end{gathered}$$

So, the general solution for $x$ is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{h}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}{\mathbf{x}}_{\mathbf{p}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$.

Step 4: Finding the value of z

Let find $z$. One has from $\mathbf{D}\left[\mathbf{x}\right]\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{z}\mathbf{=}\mathbf{0}$, one has that

 $$\begin{gathered} \mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right]}\mathbf{+}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{=}\mathbf{D}\left[{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\right]\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{=}\mathbf{-}{\mathbf{d}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\left({\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{d}}_{\mathbf{1}}\right){\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \end{gathered}$$

 

It remains to find the constants $c_1$ ,  c₂ and d₁. One will find them from the initial conditions which are  $\mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}$ and $\mathbf{z}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$, so one has to solve the following system:

 $$\begin{gathered} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{d}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2} \\ \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2} \\ \mathbf{z}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{-}{\mathbf{d}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Step 5: Finding the values of c 1 , c 2

From the first and the second equation one has that ${\mathbf{c}}_{\mathbf{1}}\mathbf{=}{\mathbf{d}}_{\mathbf{1}}$. So, substituting it into the third equation one will get that $\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$. Also, from the second equation one has that ${\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}$, so one has that:

 

 $$\begin{gathered} \mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1} \\ \mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{-}\mathbf{1} \\ \mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{2} \\ \mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{3} \\ \Rightarrow {\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{1} \\ {\mathbf{d}}_{\mathbf{1}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{-}{\mathbf{c}}_{\mathbf{1}} \\ \mathbf{=}\mathbf{2}\mathbf{-}\mathbf{1} \\ {\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{1} \end{gathered}$$

 

So, one has that the solution for the given initial value problem is:

 $$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \end{gathered}$$