Consider the partial differential equation1 2²u4π²0x²¹together with the boundary conditions u(0, t) = 0 and u(, t) = 0 fort≥ 0 and the initial condition u(x, 0) = x(x) for 0 < x < 1/(a) If n is a positive integer, show that the functionu(x, t) = e-n²t sin(2nx),satisfies the given partial differential equation and boundaryconditions.duƏt(b) The general solution of the partial differential equation thatsatisfies the boundary conditions isu(x, t) =Bne-n²t sin(2πnx).n=1Write down (but do not evaluate) an integral that can be used todetermine the constants Bn.

In part (a), we need to check whether the given function satisfies the given partial differential…

Answered: Consider the partial differential…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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For a linear first-order differential equation dy/dt=F(y,t) , we know that the existence and uniqueness theorem tells us that a solution to an initial value problem exists if F is continuous on some interval (because of the intermediate value theorem), and that the solution is unique if the partial derivative with respect to y of F is also continuous on the same interval. However, partial derivatives measure the change of a function in one direction while the other(s) are held constant, which in this case would mean t is held constant while y varies. But if we are talking about an "interval," how can this be? Wouldn't our t interval be restricted? I'm presuming the reasoning behind the uniqueness part has to do with the fact that coplanar curves are said to be parallel and do not intersect, thus meaning only one of a family of coplanar curves can pass through a given point (our initial point). But I'm having trouble understanding how the partial derivative with respect to y relates to…
Natural numbers are arranged in a triangular shape as shown below. If the pattern continues, what is the tenth number in the 10th row?
Find the general solution of the given differential equation using variation of parameters. y" + 4y = 4sec² 2t, 0 < t < y = c1cos2t + c2 sin2t – 1+ sin(2t)ln|sec2f + tan2t| y = c1cos2t + c2sin2t – sec2t + In|sec2t + tan2t| - y = cị cos2t + c2 sin2t – sec2t + sin(2t)ln|sec2t + tan2t| y = c1cos2t + c2 sin2t – cos2t + sin(2f)ln|sec2t + tan2t|
Using the substitution y=vxanddy=vdx + xdv, solve the homogeneous differential equation y [xsin (--)-> ]dx- X y dx + xdy=0 + y y @ x (exe ² - - co₁²-) = c A CSC- X X y y ℗ x(exe²= - co²=-=) = c B y csc X X y y Ⓒy[ese ² + + co₁² ) = c C X X © x (exe ² + + co₁=² ) = c y y D X X
Find the linear, nonhomogeneous differential equation with constant coefficients whose solution is y (t) = c1e' + cze²t – 3te' + cost - A) y" + 3y' + 2y = 3et + 3 cos t – 3 sin t – 3e-t B) y" – 3y' + 2y = 3et + cos t + 3 sin t – 3e¬t C) y" + 3y' + 2y = 3te' + cos t – 3 sint D) y" – 2y' – 3y = -3e' + 3e-t + sin t E) y" – 3y' + 2y = -3e' – cos t +3 sin t – že-t

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