Answer :
Answer: u(x,t) = (Acos(√(λ/k)x) + Bsin(√(λ/k)x))Ce^(λt/kt)
Step-by-step explanation: A PDE problem!
The given partial differential equation is:
k ∂²u/∂x² = ∂u/∂t, k > 0
This is a wave equation with a constant wave speed. We can solve it using separation of variables.
Let's assume the solution has the form:
u(x,t) = X(x)T(t)
Substituting this into the PDE, we get:
k T(t) X''(x) = T'(t) X(x)
Now, divide both sides by kT(t)X(x):
X''(x)/X(x) = T'(t)/kT(t)
Since the left-hand side depends only on x, and the right-hand side depends only on t, we can set them equal to each other:
-X''(x) = λ X(x) / k
And:
T'(t) = λ kT(t)
where λ is a constant.
The first equation is a second-order linear ODE in x, and the second equation is a first-order linear ODE in t. We can solve them separately:
X''(x) + (λ/k)X(x) = 0
This is a harmonic oscillator equation, and its general solution is:
X(x) = A cos(√(λ/k)x) + B sin(√(λ/k)x)
And:
T(t) = Ce^(λt/kt)
where A, B, and C are constants.
Now, we can combine these solutions to get the general solution for the PDE:
u(x,t) = (Acos(√(λ/k)x) + Bsin(√(λ/k)x))Ce^(λt/kt)
This is the product solution for the given partial differential equation.
Note that we have found a general solution, and we can use the initial and boundary conditions to determine the values of the constants A, B, and C.
