Q28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

To see the effect of changing the parameter in the initial value problem. $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{b}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ . Solve the problem for b=5, 4 and 2 and sketch the solutions.

Step-by-Step Solution

Answer

The solution of the given initial value $y\text{'}\text{'}+by\text{'}+4y=0$ is $y\text{(}t\text{)}=e^{-2t}+2t{e}^{-2t}$ when $y\left(0\right)=1,y\text{'}\left(0\right)=0$.

Step 1: Using the algebraic solution

The differential equation is $\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+4y=0$ 

Initial values are $y\left(0\right)=1,y\text{'}\left(0\right)=0$ 

The above differential equation is written as $m^2+bm+4=0$ 

The algebraic solution of the above equation is as follows,

 $m=\frac{-b\pm \sqrt{b^2-16}}{2}$

Here,  $m=\frac{dy}{dx}$.

Step 2: Finding the values of c 1 and c 2

For $b=5, m$ is calculated as

$$\begin{gathered} m=\frac{-5\pm \sqrt{25-16}}{2} \\ m=\frac{-5\pm \sqrt{9}}{2} \\ m=\frac{-5\pm 3}{2} \\ {m}_1=-4 \\ {m}_2=-1 \end{gathered}$$ 

The solution of the differential equation is given below,

 $y=C_1{e}^{m_{1}x}+C_2{e}^{m_{2}x}$

Step 3: Substituting the values

Substitute the values in the above equation,

 $y=C_1{e}^{-4x}+C_2{e}^{-x}$

Using the first initial value condition,

 $$\begin{gathered} 1=C_1{e}^{-4\times 0}+C_2{e}^{-0} \\ 1=C_1+C_2 ...\mathbf{\left(}\mathbf{i}\mathbf{\right)} \end{gathered}$$

Using the second initial value condition,

  $$\begin{gathered} \frac{dy}{dx}=-4C_1{e}^{-4x}-C_2{e}^{-x} \\ 0=-4C_1{e}^{-4\times 0}-C_2{e}^{-0} \\ 0=-4C_1-C_{2 } ...\mathbf{\left(}\mathbf{ii}\mathbf{\right)} \end{gathered}$$

Step 4: Sketching the graph

Solve equations(i) and (ii)

$$\begin{gathered} C_1=-\frac{1}{3} \\ {C}_2=\frac{4}{3} \end{gathered}$$ 

Substitute $C_1$ and $C_2$ in the solution of the differential equation, Thus the solution is  

$y=-\frac{e^{-4x}}{3}+\frac{4e^{-x}}{3}$

So, the graph will be

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

For  $b=4, m$ is calculated as

$$\begin{gathered} m=\frac{-4\pm \sqrt{16-16}}{2} \\ m=\frac{-4\pm \sqrt{0}}{2} \\ m=\frac{-4\pm 0}{2} \\ {m}_1=-2 \\ {m}_2=-2 \end{gathered}$$

 

The solution to the differential is:

 $y=\left(C_1+C_{2}x\right)e^{mx}$

Step 6: Substituting the values

Substitute the values in the above equation,

 $y=\left(C_1+C_{2}x\right)e^{-2x}$

Using the first initial value condition,

  $$\begin{gathered} 1=\left(C_1+C_2\times 0\right)e^{m\times 0} \\ 1=C_1 ...\mathbf{\left(}\mathbf{iii}\mathbf{\right)} \end{gathered}$$

Using the second initial value condition,

 $$\begin{gathered} \frac{dy}{dx}=-2C_1{e}^{-2x}+C_2\left(x\left(-2e^{-2x}\right)+e^{-2x}\right) \\ 0=-2C_1{e}^{-2\times 0}+C_2\left(0\left(-2e^{-2\times 0}\right)+e^{-2\times 0}\right) \\ 0=-2C_1+C_2 ...\mathbf{\left(}\mathbf{iv}\mathbf{\right)} \end{gathered}$$

Step 7: Constructing the graph

Solve equations(iii) and (iv)

$$\begin{gathered} C_1=1 \\ {C}_2=2 \end{gathered}$$ 

Substitute C₁ and C₂ in the solution of the differential equation, Thus the solution is

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

So, the graph will be

Step 8: Finding the values of c 1 and c 2

For  $b=2, m$ is calculated as

$$\begin{gathered} m=\frac{-2\pm \sqrt{4-16}}{2} \\ m=\frac{-2\pm \sqrt{-12}}{2} \\ m=-1\pm \sqrt{3}i \\ {m}_1=-1+\sqrt{3}i \\ {m}_2=-1-\sqrt{3}i \end{gathered}$$ 

The solution of the differential equation is,

 $y=e^{\alpha x}\left(C_1\cos \beta x+C_2\sin \beta x\right)$

Step 9: Substituting the values

Substitute the values in the above equation,

$y=e^{-x}\left(C_1\cos \sqrt{3}x+C_2\sin \sqrt{3}x\right)$

 

Using the first initial value condition,

$$\begin{gathered} 1=e^{-0}\left(C_1\cos \sqrt{3}\times 0+C_2\sin \sqrt{3}\times 0\right) \\ 1=C_1 ...\mathbf{\left(}\mathit{v}\mathbf{\right)} \end{gathered}$$ 

Using the second initial value condition,

 $$\begin{gathered} \frac{dy}{dx}=e^{-x}\left(-\sqrt{3}{C}_1\sin \sqrt{3}x+\sqrt{3}{C}_2\cos \sqrt{3}x\right)-e^{-x}\left(C_1\cos \sqrt{3}x+C_2\sin \sqrt{3}x\right) \\ 0=e^{-0}\left(-\sqrt{3}{C}_1\sin \sqrt{3}\times 0+\sqrt{3}{C}_2\cos \sqrt{3}\times 0\right)-e^{-0}\left(C_1\cos \sqrt{3}\times 0+C_2\sin \sqrt{3}\times 0\right) \\ 0=\sqrt{3}{C}_2-C_1 ...\mathbf{\left(}\mathit{v}\mathbf{i}\mathbf{\right)} \end{gathered}$$

Step 10: Constructing the graph

Solve equation (v) and (vi)

$$\begin{gathered} C_1=1 \\ {C}_2=\frac{1}{\sqrt{3}} \end{gathered}$$ 

Substitute C₁ and C₂ in the solution of the differential equation,

Thus, the solution is $y=e^{-x}\left(\cos \sqrt{3}x+\frac{1}{\sqrt{3}}\sin \sqrt{3}x\right)$

So, the graph will be