Question

Solve the given initial-value problem in which the input
function *g*(*x*) is discontinuous. [*Hint*:
Solve the problem on two intervals, and then find a solution so
that *y* and *y'* are continuous at

x = π/2.]

y'' + 4y = g(x), y(0) = 1, y'(0) = 3, where

g(x) =

sin(x), | 0 ≤ x ≤ π/2 | |

0, | x > π/2 |

Find y(x) for each of the intervals.

, | 0 ≤ x ≤ π/2 | |

, | x > π/2 |

Answer #1

I've attached Handwritten solution for the given problem

Solve the given initial-value problem.
y'' + 4y = −3, y(π/8) =
1
4
, y'(π/8) = 2
y(x) =

transform the given initial value problem into an algebraic
equation for Y = L{y} in the s-domain. Then find the Laplace
transform of the solution of the initial value problem.
y'' + 4y = 3e^(−2t) * sin 2t,
y(0) = 2, y′(0) = −1

Solve the initial value problem: 4y′′+3y=0, y(π/2)=2 ,
y'(π/2)=−1.
Give your answer as y=... Use x as the independent variable.

Solve the given initial value problem using the method of
undetermined coefficients
yII + 4yI + 4y = (3+x)e-2x y(0)
= 2, yI(0) = 5

Solve the given initial-value problem. (Enter the first three
nonzero terms of the solution.)
(x + 2)y'' +
3y = 0, y(0) =
0, y'(0) = 1

Consider the following initial value problem, in which an input
of large amplitude and short duration has been idealized as a delta
function.
y′′−4y′=δ(t−4), y(0)=6, y′(0)=0.
a. Find the Laplace transform of the solution.
b. Obtain the solution y(t).
c. Express the solution as a piecewise-defined function and
think about what happens to the graph of the solution at t=4.
y(t)= ___ if 0 (less than or equal to) t <4, AND ___ if 4
(less than or equal to)...

Solve the given initial value problem and determine at least
approximately where the solution is valid.
(12x2+y−1)dx−(18y−x)dy=0, y(1)=0
y= ,
the solution is valid as long
as ≥0

Find the solution of the given initial value problem: y " + y =
f(t); y(0) = 6, y' (0) = 3 where f(t) = 1, 0 ≤ t < π 2 0, π 2 ≤
t < ∞

Solve the following initial value problem,
dy/dx = 8cos(x) / (y2 + 4y + 4)
y(π/4) = 4

Use the Laplace transform to solve the given initial-value
problem. y'' + y = f(t), y(0) = 0, y'(0) = 1, where f(t) = 0, 0 ≤ t
< π 5, π ≤ t < 2π 0, t ≥ 2π

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 16 minutes ago

asked 18 minutes ago

asked 26 minutes ago

asked 38 minutes ago

asked 38 minutes ago

asked 41 minutes ago

asked 50 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago