Chapter 4: Q23E (page 172)

Solve the given initial value problem. $w'' - 4 w' + 2 w = 0;$ $w (0) = 0,$ $w' (0) = 1$

Short Answer

Expert verified

The solution of the given initial value $w'' - 4 w' + 2 w = 0$ is $w (t) = \frac{}{2 \sqrt{2}} e^{(2 + \sqrt{2}) t} - \frac{}{2 \sqrt{2}} e^{(2 - \sqrt{2}) t}$ when $w (0) = 0$ and $w' (0) = 1$ .

Step by step solution

Differentiate the value of y.

Given differential equation is $w'' - 4 w' + 2 w = 0$

Let $w = e^{rt}$

Therefore,

$\begin{matrix} w' (t) = r e^{rt} \\ w'' (t) = r^{2} e^{rt} \end{matrix}$

Finding the general solution.

Then the auxiliary equation is $r^{2} - 4 r + 2 = 0$

Solve the auxiliary equation to obtain the roots.

role="math" localid="1654842815439" $\begin{matrix} r = \frac{4 \pm \sqrt{4^{2} - 4 \times 1 \times 2}}{2 \times 1} \\ r = \frac{4 \pm \sqrt{16 - 8}}{} \\ r = \frac{4 \pm \sqrt{8}}{} \\ r = \frac{4 \pm 2 \sqrt{2}}{} \\ r = 2 \pm \sqrt{2} \end{matrix}$

Therefore, the general solution is $w (t) = c_{1} e^{(2 + \sqrt{2}) t} + c_{2} e^{(2 - \sqrt{2}) t}$

Finding the values of c1 and c2

Given initial conditions are $w (0) = 0$ and $w' (0) = 1$

$\begin{array}{l} w (0) = c_{1} e^{(2 + \sqrt{2}) \times 0} + c_{2} e^{(2 - \sqrt{2}) \times 0} \\ c_{1} + c_{2} = 0 \end{array}$

And

$w' (t) = (2 + \sqrt{2}) c_{1} e^{(2 + \sqrt{2}) t} + (2 - \sqrt{2}) c_{2} e^{(2 - \sqrt{2}) t}$

Then $w' (0) = (2 + \sqrt{2}) c_{1} e^{(2 + \sqrt{2}) \times 0} + (2 - \sqrt{2}) c_{2} e^{(2 - \sqrt{2}) \times 0}$

$\begin{matrix} 2 (c_{1} + c_{2}) + \sqrt{2} (c_{1} - c_{2}) = 1 \\ \sqrt{2} (c_{1} + c_{1}) = 1 \\ 2 \sqrt{2} c_{1} = 1 \\ c_{1} = \frac{}{2 \sqrt{2}} \end{matrix}$

Substitute $c_{1}$ in $(1)$ $(1)$

$\begin{matrix} \frac{}{2 \sqrt{2}} + c_{2} = 0 \\ c_{2} = - \frac{}{2 \sqrt{2}} \end{matrix}$

Therefore, the solution is $w (t) = \frac{}{2 \sqrt{2}} e^{(2 + \sqrt{2}) t} - \frac{}{2 \sqrt{2}} e^{(2 - \sqrt{2}) t}$ .

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Solve the given initial value problem. $w'' - 4 w' + 2 w = 0;$ $w (0) = 0,$ $w' (0) = 1$

Short Answer

Step by step solution

Differentiate the value of y.

Finding the general solution.

Finding the values of c1 and c2

One App. One Place for Learning.

Most popular questions from this chapter

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Solve the given initial value problem. w''-4w'+2w=0;w(0)=0,w'(0)=1

Short Answer

Step by step solution

Differentiate the value of y.

Finding the general solution.

Finding the values of c1 and c2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Statistics

Geometry

Theoretical and Mathematical Physics

Mechanics Maths

Decision Maths

Study anywhere. Anytime. Across all devices.

Solve the given initial value problem. $w'' - 4 w' + 2 w = 0;$ $w (0) = 0,$ $w' (0) = 1$