Answered: u" (x) = -u(x), 0 < x < T, и(0) — и(т)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Consider the attached two-point BVP.

Verify that u(x) = c*sin(x) is a solution for any constant c.

u" (x) = -u(x),
u(0) = u(ㅠ) = 0.
0 <x <T,

Expert Solution

Step 1

Step:-1

Given differential equation of second order is $u'' (x) = - u (x)$ , ----(1), with $u (0) = u (π) = 0$ and $0 < x < π$

Solving $u'' (x) = - u (x)$

$\Rightarrow u'' = - u \Rightarrow u'' + u = 0$

This is homogeneous differential equation of second order.

The auxiliary equation is

$m^{2} + 1 = 0 \Rightarrow (m + i) (m - i) = 0 \Rightarrow m = + i, - i$

the complementary function (C.F.) = $c_{1} \cos (x) + c_{2} \sin (x)$

Here, $c_{1}, c_{2}$ is arbitrary constant.

Since, it is homogeneous differential equation then Particular integral is 0

Then, General solution = complementary function

So, the General solution is

$u (x) = c_{1} \cos (x) + c_{2} \sin (x)$ ---- (2)

Step:-2

Now, given boundary conditions are $u (0) = u (π) = 0$

when x=0, (put in (2))

$u (0) = c_{1} \cos (0) + c_{2} \sin (0) S i n c e, u (0) = 0 a n d \cos (0) = 0, \sin (0) = 0, t h e n c_{1} + c_{2} \times 0 = 0 \Rightarrow c_{1} = 0$

So, we have

$u (x) = c_{2} \sin (x)$

when $x = π$ , then

$u (π) = c_{2} \sin (π) as given u (π) = 0, then c_{2} \sin (π) = 0 Since, we know \sin (π) = 0 So, taking \sin (π) = 0 and c_{2} \neq 0$

Note:- If $c_{2} = 0$ then given differential equation has trivial solution only. For non trivial solution, we take $c_{2} \neq 0$

For required solution replacing $c_{2}$ by $c *$ .

Hence, the solution is

$u (x) = c * \sin (x)$ and $c *$ is arbitrary constant.

Hence, it is clear that $u (x) = c * \sin (x)$ is the solution of given differential equation.

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