Question
$$\frac{\partial^{2} u}{\partial t^{2}}$$$=16 \frac{\partial^{2} u}{\partial x^{2}}$$ 0<$x$<\pi, \quad t>0$$$u(0, t)=u(\pi, t)=0, \quad t>0$$$$u(x, 0)$$$=\sin ^{2} x, $$0<$x$<\pi$$$\frac{\partial u}{\partial t}(x, 0)$$$1-\cos x$$0<$x$<\pi$
Step 1
Step 1: We start by assuming that the solution to the given partial differential equation is of the form $$u(x,t) = \sum_{n=1}^{\infty} (a_n \cos(n t) + b_n \sin(n t)) \sin(n x)$$ Show more…
Show all steps
Your feedback will help us improve your experience
Sajin Shajee and 73 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$$ \frac{\partial^{2} u}{\partial t^{2}} $$ $=\frac{\partial^{2} u}{\partial x^{2}}+x \sin t$, $0$<$x$<$\pi$, $t>0$ $$ u(0, t) $$ $=u(\pi, t)=0$ $t>0$, $$ u(x, 0)=0 $$, $0$<$x$<$\pi$, $$ \frac{\partial u}{\partial t}(x, 0) $$ $=0$, $0$<$x$<$\pi$
Partial Differential Equations
The Wave Equation
$$ \frac{\partial^{2} u}{\partial t^{2}} $$4$\frac{\partial^{2} u}{\partial x^{2}}$,$0<$x$<\pi$,$t>0$ $$ u(0, t) $$$=u(\pi, t)$$=0, \quad t>0$ $$ u(x, 0) $$$=x^{2}(\pi-x)$,$0<$x$<\pi$, $$ \frac{\partial u}{\partial t}(x, 0) $$$=0$,$0<$x$<\pi$
$$ \frac{\partial^{2} u}{\partial t^{2}} $$$=\frac{\partial^{2} u}{\partial x^{2}}+t x$, $0$<$x$<$\pi$, $t>0$ $$u(0, t)=u(\pi, t)=0, \quad t>0$$ $$ u(x, 0) $$$=\sin x$, $0$<$x$<$\pi$, $$ \frac{\partial u}{\partial t}(x, 0) $$ $=5 \sin 2 x-3 \sin 5 x$, $0$<$x$<$\pi$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD