# Differential equations

Consider the following problem for u = u(x,t):

,

,

a) Seeking a solution of this problem of the form

,

show that f and g satisfy the coupled system

; where and ;

, ; , .

b) Eliminating g between the differential equations in a), show that f satisfies

c) Assuming that

and noting that

conclude that

d) Applying the boundary conditions for f in a), deduce that

where

e) Now use a) to show that

.

f) Applying the boundary conditions for g in a), finally conclude that

and hence obtain the result.

since

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attached file.

Consider the following problem for u = u(x,t):

,

,

a) Seeking a solution of this problem of the form

,

show that f and g satisfy the coupled system

; where and ;

, ; , .

Now,

Dividing throughout by

=

Comparing the cos terms, and comparing the sin terms,

For u(0, t) = cos(t), f(0) = 1 and g(0) = 0

b) Eliminating g between the differential equations in a), show that f satisfies

Given that f'' = 2k2g,

Differentiating with respect to x on both sides, f''' = 2k2g'

Differentiating again with respect to x on both sides, f(4) = 2k2g''

But g'' = -2k2f, so f(4) = 2k2(-2k2f) = -4k4f

Therefore,

c) Assuming that

and noting that

conclude that

f'(x) = r erx

f''(x) = r2 erx

f'''(x) = r3 erx

f(4)(x) = r4 erx

Plugging in the equation , we get r4 erx + 4k4 erx = 0

Or (r4 + 4k4) erx = 0, which gives (r4 + 4k4) = 0

But

Therefore,

Taking

Or

Or (r + k)2 = -k2

Or r + k = -ki, where i =

Or r = -k - ki

Therefore, f(x) = e-k-ki

By Euler's formula,

Similarly, by taking , it can be shown that r = k - ki

Or f(x) = ek-ki =

Combining the two,

d) Applying the boundary conditions for f in a), deduce that

where

Given that ,

so

or

or c1 + c3 = 1

Similarly, c3 = c4= 0

Plugging in c3 = 0 in c1 + c3 = 1 gives c1 = 1

Therefore,

=

=

If , we get

e) Now use a) to show that

.

f'' = 2k2 g(x) g(x) = f"/2k2

But f = e-kx h(x)

Therefore, f'(x) = -ke-kx h(x) + h'(x) e-kx = e-kx(-kh + h') = -kf + e-kx h'(x)

Or f' + kf = e-kx h'(x)

But g(x) = -(f' + kf)/k, therefore,

f) Applying the boundary conditions for g in a), finally conclude that

and hence obtain the result.

since

Given the boundary condition g(0) = 0, 0 = , which means that

h'(0) = 0

But , therefore, h'(x) = -k sin(kx) + c2 cos(kx)

At x = 0, h'(0) = -k*sin(0) + c2*cos(0)

Or 0 = 0 + c2, which gives c2 = 0

Therefore,

Therefore, =

And =

Now, = cos(t) + sin(t) Or .

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