Question

5. Solve the boundary value problem \frac{\partial^2 y}{\partial t^2} = 9 \frac{\partial^2 y}{\partial x^2} + x^2 \quad (0 < x < 4, t > 0) y(0, t) = y(4, t) = 0 \quad (t > 0), y(x, 0) = \frac{\partial y}{\partial t}(x, 0) = 0 \quad (0 < x < 4). Hint: Set Y(x, t) = y(x, t) + h(x) and obtain a boundary value problem for Y(x, t). Now choose h(x) so that this boundary value problem is a familiar form.

          5. Solve the boundary value problem
\frac{\partial^2 y}{\partial t^2} = 9 \frac{\partial^2 y}{\partial x^2} + x^2 \quad (0 < x < 4, t > 0)
y(0, t) = y(4, t) = 0 \quad (t > 0),
y(x, 0) = \frac{\partial y}{\partial t}(x, 0) = 0 \quad (0 < x < 4).
Hint: Set Y(x, t) = y(x, t) + h(x) and obtain a boundary value problem for Y(x, t). Now
choose h(x) so that this boundary value problem is a familiar form.

5. Solve the boundary value problem
(∂^2 y)/(∂t^2) = 9 (∂^2 y)/(∂x^2) + x^2 (0 < x < 4, t > 0)
y(0, t) = y(4, t) = 0 (t > 0),
y(x, 0) = (∂y)/(∂t)(x, 0) = 0 (0 < x < 4).
Hint: Set Y(x, t) = y(x, t) + h(x) and obtain a boundary value problem for Y(x, t). Now
choose h(x) so that this boundary value problem is a familiar form.

Added by Jose Q.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Babita Kumari

Step 1

Step 1: Set Y(x,t) = y(x,t) + h(x) By substituting Y(x,t) into the given equation, we have: 2(Y(x,t) - h(x)) = 9g(Y(x,t) - h(x)) + 2 Show more…

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Solve the boundary value problem: 2y = 9gy + 2(0 â‰¤ t â‰¤ 4, t > 0) y(0,t) = y(4,t) = 0 (t > 0) y(x,0) = 0 (0 < x < 4) Hint: Set Y(x,t) = y(x,t) + h(x) and obtain a boundary value problem for Y(x,t). Now choose h(x) so that this boundary value problem is in a familiar form.