Answer to Question #284508 in Differential Equations for Tejas

Question

If a string of length l is initially at rest in equilibrium position and each of its points is given the velocity dy/dt= b sin^3πx/l find the displacement

Answer 1

The displacement y(x,t) is given by the equation:

$y_{tt}=a^2y_{xx}$

boundary conditions:

$y(0,t)=y(l,t)=0$

$y_t(0)=b sin(3πx/l)$

$y(x,0)=0$

solution:

$y(x,t)=Bsin(n\pi x/l)(Ccos(n\pi at/l)+Dsin(n\pi at/l))$

then:

$y(x,0)=Bsin(n\pi x/l)C=0 \implies C=0$

$y(x,t)=Bsin(n\pi x/l)Dsin(n\pi at/l)$

the most general solution:

$y(x,t)=\sum B_nsin(n\pi x/l)sin(n\pi at/l)$

then:

$y_t=\sum B_n(n\pi a/l)sin(n\pi x/l)sin(n\pi at/l)$

$y_t(0)=b sin(3πx/l)=\sum B_n(n\pi a/l)sin(n\pi x/l)sin(n\pi at/l)$

$b=B_3(3\pi a/l),B_1=B_2=B_4=...=0$

$B_3=\frac{bl}{3\pi a}$

$y(x,t)=B_3sin(3\pi x/l)sin(3\pi at/l)=\frac{bl}{3\pi a}sin(3\pi x/l)sin(3\pi at/l)$