1. ## Linear diff equation help... please!

Hello everyone, I have an exam in Linear differential equations and transforms soon and could do with some basic-ish help on how to solve second order diff equations.

The question states:

The function u(x,t) satisfies the equation

(d2U/dU2) = (1/c^2) (d2u/dt2) + (2b/c) (du/dt)

For 0 < x < l , t > 0 , and c > 0 , 0 < b < pi/l are constants.

Suppose u(x,t) is subject to homogeneous boundary conditions u(0,t) = u(l,t) = 0 and initial condition u(x,0) = 0 for 0 < x < l.

Use separation of variables to derive general solution:

u(x,t) = SUM(n-inf) [ Bn * exp(-bct) * sin(wn)t * sin((n pi x)/l)

Where Bn is the nth value of the constant B

And wn = c * sqrt ( (n^2 pi^2)/l^2 - b^2)

I so far have separated variables to find

X''T = 1/c^2 XT'' + 2b/c XT'

So

X''/X = 1/c^2 T''/T + 2b/c T'/T = -k^2

And

X'' + k^2 X = 0 => X(x) = Acoskx + Bsinkx

And similarly

T''+2bcT' + c^2*k^2*T = 0

Could anybody confirm that this is the right approach? And where do I go from here, how do I solve T?

Thanks very much, its much appreciated, and I apologise for the crude writing of equations I tried my best to make them look clear!

2. Originally Posted by rak
Hello everyone, I have an exam in Linear differential equations and transforms soon and could do with some basic-ish help on how to solve second order diff equations.

The question states:

The function u(x,t) satisfies the equation

(d2U/dU2) = (1/c^2) (d2u/dt2) + (2b/c) (du/dt)

For 0 < x < l , t > 0 , and c > 0 , 0 < b < pi/l are constants.

Suppose u(x,t) is subject to homogeneous boundary conditions u(0,t) = u(l,t) = 0 and initial condition u(x,0) = 0 for 0 < x < l.

Use separation of variables to derive general solution:

u(x,t) = SUM(n-inf) [ Bn * exp(-bct) * sin(wn)t * sin((n pi x)/l)

Where Bn is the nth value of the constant B

And wn = c * sqrt ( (n^2 pi^2)/l^2 - b^2)

I so far have separated variables to find

X''T = 1/c^2 XT'' + 2b/c XT'

So

X''/X = 1/c^2 T''/T + 2b/c T'/T = -k^2

And

X'' + k^2 X = 0 => X(x) = Acoskx + Bsinkx

And similarly

T''+2bcT' + c^2*k^2*T = 0

Could anybody confirm that this is the right approach? And where do I go from here, how do I solve T?

Thanks very much, its much appreciated, and I apologise for the crude writing of equations I tried my best to make them look clear!
Right so far! The $k$ is found from the BC (it's $k=\frac{\pi}{l}$ )

The ODE $T''+2bcT' + c^2 k^2 T = 0$ is constant coefficient so use the usual method, seek a solution of the form $T = e^{mt}$. See the tutorial at the top of this forum for a refresher. A note, since $0 < b < \frac{\pi}{l},\;m$ will be complex.