Applying the Existence and Uniqueness theorem to the initial value problem involving second order linear equations with constant coefficients, that is, for ay" + by' + cy= 0, a + 0, y (to) = Yo, y' (to) = d, we can conclude that The longest interval for which a unique solution exist is (to, do). The initial value problem has infinitely many solutions. A unique solution, which is valid over the entire real line, always exist. We cannot apply the theorem without knowing the numbers a, b, c, to. O O 00

Applying the Existence and Uniqueness theorem to the initial value problem involving second order linear equations with constant coefficients, that is, for ay" + by' + cy= 0, a + 0, y (to) = Yo, y' (to) = d, we can conclude that The longest interval for which a unique solution exist is (to, do). The initial value problem has infinitely many solutions. A unique solution, which is valid over the entire real line, always exist. We cannot apply the theorem without knowing the numbers a, b, c, to. O O 00

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Similar questions

(13) Prove that the equation x² + 3y² = 2 has no integer solutions.
7) Solve the system <-4 12 X subject to the initial condition x(0)=(1).
A solution of the form (g(t), h(t), t) -∞ < t < ∞ represents a line in three dimensions only if g(t) and h(t) are linear equations. Select one: True False
Let 2 6 3 A -4 2 3 9 3 (a) Compute det(A) -| (b) The following linear system 3 13 + 6x2 + 4x2 + 3r3 -3T1 2 23 3r + 9I2 + -5 has the unique solution (x1, x2, E3). Use Cramer's Rule to find the value for the solution variable 2. Il | |
II. Linear Equations (Elimination method)] 2. 2x - 11y+ 3z = 2 x - 3y +z = 1 -8y +2z = -1 X-
Q:2) a) If f(x) = x* and g(x) = x² then show that f •g(x) is odd or even function. b) solve the systems of linear equation by Inversion method. 2a + 2b + c = 3 За - 2b — 2с — 1 5а + b - Зс %3D 2
For the equation ay" + by' + cy = 0, if b² – 4ac = 0 (the repeated root case where the root is r = -b/2a) we learned that one solution is Y1 = e-bt/2a, and the second linearly independent solution is y, = te-bt/2a We were told that the second solution is found via the Reduction of Order method. Apply the Reduction of Order method to the equation using the first solution (Do the calculations, this will take about 5 or so minutes) Which of the following equations does it reduces to? au" + 2bu' - u + cu = 0 2a 62 àu" – 2bu' +u + cu = 0 au" = 0 au" + 2bu' + Eu + cu = 0 O O O
A characteristic equation has the following solutions: m1 = 5, m2 = -3, m3 = 2 %3D What is the general solution of the corresponding homogeneous linear equation with constant coefficients? 3x O y = cie + c2e¯ 5x + cze2 + c3e 2.2r O y= cie + c2xe + c3x²e2x Dy 5x-3x2 + 2x y = 5c1e"- 3c2e" + 2cze" %3D 梦 @ 99+
Find the general solution to (D5-2D³-2D²-3D-2)y=0 Oy=ex(C₁+C₂x) + C3e2x + C4cosx + С5sinx Oy=ex(C₁+C₂x) + C3e2x + C4cOSX + Cçsinx y=eX(C₁+C₂x) + C3e-2x + C4cosx + Cçsinx y=eX(C₁ + C₂x) + C3e2x + C4cos2x + С-sin2x
Which of the following is a solution of the following equation valid near the origin for x>0? 2x2(1- х)у" - x(1+7x)у +у30 1 xn+1 y= X+ > 15 n=1 (2n+3)(2n+5) 1 2 (2n- 3)(2n- 5)xn+1 15 n =1 y= x + O (-1)"x"+1 (2n- 3)(2n- 5) 1 15 n =1 1 E (2n+ 3)(2n+ 5)x+1 15 n =1 y=x+